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MOTION SAMPLE BOOKLET CLASS XI • ROTATIONAL • STOICHIOMETRY - I • TRIGONOMETRIC RATIOS & IDENTITIES Copyright © reserved with Motion Edu. Pvt. Ltd. and Publications All rights reserved. No part of this work herein should be reproduced or used either graphically, electronically, mechanically or by recording, photocopying, taping, web distributing or by storing in any form and retrieving without the prior written permission of the publisher. Anybody violating this is liable to be legally prosecuted. Corporate Head Office 394 - Rajeev Gandhi Nagar Kot a, (Raj.) Ph. No. : 08003899588, 0744-2209671 IVRS No : 0744-2439051, 52, 53, www. motioniitjee.com , info@motioniitjee.com THEORY AND EXERCISE BOOKLET CONTENTS ROTATIONAL S.NO. TOPIC PAGE NO. THEORY WITH SOLVED EXAMPLES ..................................................... 5 – 24 EXERCISE - I ............................................................................................ 25 – 37 EXERCISE - II ........................................................................................... 38 – 54 EXERCISE - III .......................................................................................... 55 – 68 EXERCISE -IV ........................................................................................... 69 - 86 ANSWER KEY .......................................................................................... 87 - 89 STOICHIOMETRY-I S.NO. TOPIC PAGE NO. THEORY WITH SOLVED EXAMPLES ..................................................... 9 – 114 EXERCISE - I ............................................................................................ 120 – 126 EXERCISE - II ........................................................................................... 127 – 135 EXERCISE - III .......................................................................................... 136 – 143 EXERCISE - IV .......................................................................................... 144 – 145 ANSWER KEY .......................................................................................... 146 – 148 TRIGONOMETRIC RATIOS & IDENTITIES (PHASE–I) S.NO. TOPIC PAGE NO. THEORY WITH SOLVED EXAMPLES ..................................................... 149 – 165 EXERCISE - I ............................................................................................ 166 – 172 EXERCISE - II ........................................................................................... 173 – 179 EXERCISE - III .......................................................................................... 180 – 192 EXERCISE - IV .......................................................................................... 193 – 198 ANSWER KEY .......................................................................................... 199 - 200 SYLLABUS • ROTATIONAL Rigid body, moment of inertia, parallel and perpendicular axes theorems, moment of inertia of uniform bodies with simple geometrical shapes; Angular momentum; Torque; Conservation of angular momentum; Dynamics of rigid bodies with fixed axis of rotation; Rolling without slipping of rings, cylinders and spheres; Equilibrium of rigid bodies; Collision of point masses with rigid bodies. • STOICHIOMETRY-I Mole concept; Chemical formulae; Balanced chemical equations; Calculations (based on mole concept) involving common oxidation-reduction, neutralisation, and displacement reactions; Concentration in terms of mole fraction, molarity, molality and normality. • TRIGONOMETRIC RATIOS & IDENTITIES (PHASE–I) Trigonometric functions, their periodicity and graphs, addition and subtraction formulae, formulae involving multiple and sub-multiple angles TOTAL NO. OF QUESTIONS • PHYSICS No. of Unsolved Example : 1000 (Approx) No. of Solved Example : 5000 (Approx) • CHEMISTRY No. of Unsolved Example : 450 (Approx) No. of Solved Example : 5500 (Approx) • MATHEMATICS No. of Unsolved Example : 1000 (Approx) No. of Solved Example : 4000 (Approx) Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS Page # 5 ROTATIONAL DYNAMICS 1. RIGID BODY : Rigid body is defined as a system of particles in which distance between each pair of particles remains constant (with respect to time) that means the shape and size do not change, during the motion. Eg. Fan, Pen, Table, stone and so on. Our body is not a rigid body, two blocks with a spring attached between them is also not a rigid body. For every pair of particles in a rigid body, there is no velocity of seperation or approach between the particles. In the figure shown velocities of A and B with respect to ground are VA and VB respectively A VA sin A A VA cos VA 1 1 1 B B VB VBA B 2 VB sin 2 VB cos 2 If the above body is rigid VA cos 1 = VB cos 2 Note : With respect to any particle of rigid body the motion of any other particle of that rigid body is circular. VBA = relative velocity of B with respect to A. Types of Motion of rigid body Pure Translational Motion 1.1. Pure Rotational Motion Combined Translational and Rotational Motion Pure Translational Motion : A body is said to be in pure translational motion if the displacement of each particle is same during any time interval however small or large. In this motion all the particles have same s, v & a at an instant. example. A box is being pushed on a horizontal surface. 10 6 6 10 16 Vcm Scm V of any particle, acm a of any particle S of any particle For pure translational motion :v m1 v m4 v m2 v m7 v m5 v m3 v vm6 m8 m2 m1 m3 m4 m5 m7 m6 m8 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 6 Fext ROTATIONAL DYNAMICS m1a1 m 2 a2 m3 a 3 ............. Where m1, m2, m3, ......... are the masses of different particles of the body having accelerations a1, a2 , a3 ,............... respectively.. But acceleration of all the particles are same So, a1 Fext a2 a3 ......... a Ma Where M = Total mass of the body a = acceleration of any particle or of centre of mass of body P m1v1 m2 v 2 m3 v 3 ............. Where m1, m2, m3 ...... are the masses of different particles of the body having velocities v1, v2 , v3 ............. respectively But velocities of all the particles are same so v1 v2 v 3 .......... v P Mv Where v = velocity of any particle or of centre of mass of the body.. Total Kinetic Energy of body = 1 m1v12 2 1 m2 v 22 2 .......... . 1 Mv 2 2 1.2. Pure Rotational Motion : A body is said to be in pure rotational motion if the perpendicular distance of each particle remains constant from a fixed line or point and do not move parallel to the line, and that line is known as axis of rotation. In this motion all the particles have same , instant. Eg. : - a rotating ceiling fan, arms of a clock. For pure rotation motion :s Where = angle rotated by the particle r s = length of arc traced by the particle. r = distance of particle from the axis of rotation. m2 = angular speed of the body.. at an m1 m3 d Where dt and m4 m5 m6 d Where = angular acceleration of the body.. dt All the parameters , and are same for all the particles. Axis of rotation is perpendicular to the plane of rotation of particles. Special case : If = constant, = 0+ t Where 0 = initial angular speed 1 2 t t = time interval 2 2 = 02 + 2 0t Total Kinetic Energy 1 m1v12 2 1 [m1r12 2 1 m2 v 22 ................. 2 m2 r22 ................] 2 1 2 I Where I = Moment of Inertia = m1r12 2 = angular speed of body. m2r22 ....... Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS Page # 7 1.3 Combined translation and rotational Motion A body is said to be in translation and rotational motion if all the particles rotates about an axis of rotation and the axis of rotation moves with respect to the ground. 2. MOMENT OF INERTIA Like the centre of mass, the moment of inertia is a property of an object that is related to its mass distribut ion. The moment of inert ia (denoted by I) is an important quantity in the study of system of particles that are rotating. The role of the moment of inertia in the study of rotational motion is analogous to that of mass in the study of linear motion. Moment of inertia gives a measurement of the resistance of a body to a change in its rotaional motion. If a body is at rest, the larger the moment of inertia of a body the more difficuilt it is to put that body into rotational motion. Similarly, the larger the moment of inertia of a body, the more difficult to stop its rotational motion. The moment of inertia is calculated about some axis (usually the rotational axis). Moment of inertia depends on : (i) density of the material of body (ii) shape & size of body (iii) axis of rotation In totality we can say that it depends upon distribution of mass relative to axis of rotation. Note : 2.1 2.2 Moment of inertia does not change if the mass : (i) is shifted parallel to the axis of the rotation (ii) is rotated with constant radius about axis of rotation Moment of Inertia of a Single Particle For a very simple case the moment of inertia of a single particle about an axis is given by, I = mr2 ...(i) Here, m is the mass of the particle and r its distance from the axis under consideration. Moment of Inertia of a System of Particles The moment of inertia of a system of particles about an axis is given by, m iri2 I= ...(ii) i r1 m1 r2 m2 r3 m3 Ex.1 r where ri is the perpendicular distance from the axis to the ith particle, which has a mass mi. Two heavy particles having masses m1 & m2 are situated in a plane perpendicular to line AB at a distance of r1 and r2 respectively. C A E (i) (ii) (iii) r1 m1 r2 m2 F D B What is the moment of inertia of the system about axis AB? What is the moment of inertia of the system about an axis passing through m1 and perpendicular to the line joining m1 and m2 ? What is the moment of inertia of the system about an axis passing through m1 and m2? : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 8 Sol. (i) (ii) (iii) Ex.2 Sol. ROTATIONAL DYNAMICS Moment of inertia of particle on left is I1 = m1r12. Moment of Inertia of particle on right is I2 = m2r22. Moment of Inertia of the system about AB is I = I1+ I2 = m1r12 + m2r22 Moment of inertia of particle on left is I1 = 0 Moment of Inertia of the system about CD is I = I1 + I2 = 0 + m2 (r1 + r2)2 Moment of inertia of particle on left is I1 = 0 Moment of inertia of particle on right is I2 = 0 Moment of Inertia of the system about EF is I = I1 + I2 = 0 + 0 Three light rods, each of length 2 , are joined together to form a triangle. Three particles A, B, C of masses m, 2m, 3m are fixed to the vertices of the triangle. Find the moment of inertia of the resulting body about (a) an axis through A perpendicular to the plane ABC, (b) an axis passing through A and the midpoint of BC. X A (a) B is at a distant 2 from the axis XY so the moment of m inertia of B (IB) about XY is 2 m (2 )2 Y Similarly Ic about XY is 3m (2 )2 and IA about XY is m(0)2 2l 2l Therefore the moment of inertia of the body about XY is 2m (2 )2 + 3 m(2 )2 + m(0)2 = 20 m 2 B (b) IA about X' Y' = m(0)2 C 2 3m IB about X' Y' = 2m ( ) 2m IC about X' Y' = 3m ( )2 Therefore the moment of inertia of the body about X' Y' is m(0)2 + 2m( )2 + 3m( )2 = 5 m 2 X' A m B C 3m 2m Y' Ex.3 Sol. Four particles each of mass m are kept at the four corners of a square of edge a. Find the moment of inertia of the system about a line perpendicular to the plane of the square and passing through the centre of the square. The perpendicular distance of every particle from the given line is a / 2 . The moment of inertia of m m 1 2 2 ma one particle is, therefore, m( a / 2 ) = . The 2 moment of inertia of the system is, therefore, 4 1 2 ma = 2 ma2. 2 m m Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS 2.3 Page # 9 Moment of Inertia of Rigid Bodies For a continuous mass distribution such as found in a rigid body, we replace the summation of m iri2 by an integral. If the system is divided I i r into infinitesimal element of mass dm and if r is the distance from a mass element to the axis of rotation, the moment of inertia is, I= r 2 dm where the integral is taken over the system. (A) Uniform rod about a perpendicular bisector Consider a uniform rod of mass M and length l figure and suppose the moment of inertia is to be calculated about the bisector AB. Take the origin at the middle point O of the rod. Consider the element of the rod between a distance x and x + dx from the origin. As the rod is uniform, Mass per unit length of the rod = M / l A x dx so that the mass of the element = (M/l)dx. The perpendicular distance of the element from 0 the line AB is x. The moment of inertia of this B element about AB is dI M dx x2 . l When x = – l/2, the element is at the left end of the rod. As x is changed from – l/2 to l/2, the elements cover the whole rod. Thus, the moment of inertia of the entire rod about AB is l/2 I M 2 x dx l l /2 (B) M x3 l 3 l /2 Ml 2 12 –l / 2 Moment of inertia of a rectangular plate about a line parallel to an edge and passing through the centre The situation is shown in figure. Draw a line parallel to AB at a distance x from it and another at a distance x + dx. We can take the strip enclosed between the two lines as the small element. x A b dx l It is “small” because the perpendiculars from different points of the strip to AB differ by not more than dx. As the plate is uniform, B its mass per unit area = Mass of the strip = M bl M b dx bl M dx . l : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 10 ROTATIONAL DYNAMICS The perpendicular distance of the strip from AB = x. The moment of inertia of the strip about AB = dI = plate is, therefore, l /2 I M 2 x dx l l /2 M dx x2 . The moment of inertia of the given l Ml 2 12 The moment of inertia of the plate about the line parallel to the other edge and passing through the centre may be obtained from the above formula by replacing l by b and thus, Mb2 . 12 Moment of inertia of a circular ring about its axis (the line perpendicular to the plane of the ring through its centre) Suppose the radius of the ring is R and its mass is M. As all the elements of the ring are at the same perpendicular distance R from the axis, the moment of inertia of the ring is I (C) r 2 dm I (D) R 2 dm R2 dm MR2 . Moment of inertia of a uniform circular plate about its axis Let the mass of the plate be M and its radius R. The centre is at O and the axis OX is perpendicular to the plane of the plate. X dx 0 x R Draw two concentric circles of radii x and x + dx, both centred at O and consider the area of the plate in between the two circles. This part of the plate may be considered to be a circular ring of radius x. As the periphery of the ring is 2 x and its width is dx, the area of this elementary ring is 2 xdx. The area of the plate is R2. As the plate is uniform, M Its mass per unit area = R2 M 2Mxdx 2 xdx R2 R2 Using the result obtained above for a circular ring, the moment of inertia of the elementary ring about OX is Mass of the ring 2Mxdx dI 2 x . R2 The moment of inertia of the plate about OX is R I 0 (E) 2M R x 3 dx 2 MR 2 . 2 Moment of inertia of a hollow cylinder about its axis Suppose the radius of the cylinder is R and its mass is M. As every element of this cylinder is at the same perpendicular distance R from the axis, the moment of inertia of the hollow cylinder about its axis is I r 2 dm R2 dm MR2 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS (F) Page # 11 Moment of inertia of a uniform solid cylinder about its axis Let the mass of the cylinder be M and its radius R. Draw two cylindrical surface of radii x and x + dx coaxial with the given cylinder. Consider the part of the cylinder in between the two surface. This part of the cylinder may be considered to be a hollow cylinder of radius x. The area of cross-section of the wall of this hollow cylinder is 2 x dx. If the length of the cylinder is l, the volume of the material of this elementary hollow cylinder is 2 x dxl. The volume of the solid cylinder is R2 l and it is uniform, hence its mass per unit volume is M R2 l The mass of the hollow cylinder considered is M 2 R l 2M 2 xdx l R2 xdx . dx As its radius is x, its moment of inertia about the given axis is x 2M dI R2 2 xdx x . The moment of inertia of the solid cylinder is, therefore, R I 0 (G) 2M R 2 x3 dx MR2 2 . Note that the formula does not depend on the length of the cylinder. Moment of inertia of a uniform hollow sphere about a diameter Let M and R be the mass and the radius of the sphere, O its centre and OX the given axis (figure). The mass is spread over the surface of the sphere and the inside is hollow. Let us consider a radius OA of the sphere at an angle with the axis OX and rotate this radius about OX. The point A traces a circle on the sphere. Now change to + d and get another circle of somewhat larger radius on the sphere. The part of the sphere between these two circles, shown in the figure, forms a ring of radius R sin . The width of this ring is Rd and its periphery is 2 R sin . Hence, the area of the ring = (2 R sin ) (Rd ). x Mass per unit area of the sphere M The mass of the ring 4 R 2 M 4 R2 R sin A Rd . ( 2 R sin )(Rd ) R M sin d . 2 0 d The moment of inertia of this elemental ring about OX is M sin d . (R sin ) 2 2 dI M 2 R sin3 d 2 As increases from 0 to , the elemental rings cover the whole spherical surface. The moment of inertia of the hollow sphere is, therefore, I 0 M 2 R sin3 d 2 MR2 cos 2 MR2 2 cos 3 3 0 (1 cos2 ) sin d 0 MR2 2 (1 cos2 ) d(cos ) 0 2 MR2 3 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 12 (H) ROTATIONAL DYNAMICS Moment of inertia of a uniform solid sphere about a diameter Let M and R be the mass and radius of the given solid sphere. Let O be centre and OX the given axis. Draw two spheres of radii x and x + dx concentric with the given solid sphere. The thin spherical shell trapped between these spheres may be treated as a hollow sphere of radius x. The mass per unit volume of the solid sphere = M 4 3 R 3 x 0 x 3M 4 R3 The thin hollow sphere considered above has a surface area 4 x2 and thickness dx. Its volume i s 4 x2 dx and hence its mass is = 3M 4 R 3 ( 4 x2 dx) = 3M R3 x2 dx Its moment of inertia about the diameter OX is, therefore, 2 3M 2 2M 4 dl = 3 3 x dx x 2 = 3 x dx R R If x = 0, the shell is formed at the centre of the solid sphere. As x increases from 0 to R, the shells cover the whole solid sphere. The moment of inertia of the solid sphere about OX is, therefore, R I= 0 Ex.4 2M R3 x 4 dx = 2 MR2 . 5 Find the moment of Inertia of a cuboid along the axis as shown in the figure. I b a c M(a2 b 2 ) 12 Sol. After compressing the cuboid parallel to the axis I = 3. THEOREMS OF MOMENT OF INERTIA There are two important theorems on moment of inertia, which, in some cases enable the moment of inertia of a body to be determined about an axis, if its moment of inertia about some other axis is known. Let us now discuss both of them. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS 3.1 * Page # 13 Theorem of parallel axes A very useful theorem, called the parallel axes theorem relates the moment of inertia of a rigid body about two parallel axes, one of which passes through the centre of mass. COM Two such axes are shown in figure for a body of mass M. If r is r the distance between the axes and ICOM and I are the respective moments of inertia about them, these moments are related by, I = ICOM + Mr2 Theorem of parallel axis is applicable for any type of rigid body whether it is a two dimensional or three dimensional Ex 5. Three rods each of mass m and length l are joined together to form an equilateral triangle as shown in figure. Find the moment of inertia of the system about an axis passing through its centre of mass and perpendicular to the plane of triangle. Sol. Moment of inertia of rod BC about an axis perpendicular to plane of triangle ABC and passing through the midpoint of rod BC (i.e., D) is A COM B C ml 2 12 From theorem of parallel axes, moment of inertia of this rod about the asked axis is I1 = ml 2 I2 = I1 + mr = 12 2 m l 2 2 3 A ml 2 6 COM r Moment of inertia of all the three rod is B I 3I2 ml 2 3 6 ml 2 2 30° D C Ex.6. Find the moment of inertia of a solid sphere of mass M and radius R about an axis XX shown in figure. x x x Sol. From theorem of parallel axis, IXX = ICOM + Mr2 = 2 MR2 MR2 5 = 7 MR2 5 COM x r=R : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 14 ROTATIONAL DYNAMICS Ex.7. Consider a uniform rod of mass m and length 2l with two particles of mass m each at its ends. Let AB be a line perpendicular to the length of the rod passing through its centre. Find the moment of inertia of the system about AB. A Sol. IAB = Irod + I both particles m( 2l ) 2 12 2(ml 2 ) 7 2 ml 3 3.2 I I m m Ans. B Theorem of perpendicular axes The theorem states that the moment of inertia of a plane lamina about an axis perpendicular to the plane of the lamina is equal to the sum of the moments of inertia of the lamina about two axes perpendicular to each other, in its own plane and intersecting each other, at the point where the perpendicular axis passes through it. Let x and y axes be chosen in the plane of the body and z-axis perpendicular, to this plane, three axes being mutually perpendicular, then the theorem states that. z y xi ri P yi x O Iz = Ix + Iy (i) (ii) (iii) Ex.8 Important point in perpendicular axis theorem This theorem is applicable only for the plane bodies (two dimensional). In theorem of perpendicular axes, all the three axes (x, y and z) intersect each other and this point may be any point on the plane of the body (it may even lie outside the body). Intersection point may or may not be the centre of mass of the body. Find the moment of inertia of uniform ring of mass M and radius R about a diameter. B Z C Sol. 0 D A Let AB and CD be two mutually perpendicular diameters of the ring. Take them ax X and Yaxes and the line perpendicular to the plane of the ring through the centre as the Z-axis. The moment of inertia of the ring about the Z-axis is I = MR2. As the ring is uniform, all of its diameter equivalent and so Ix = Iy, From perpendicular axes theorem, Iz = Ix + Iy Hence Ix = Iz MR2 = 2 2 Similarly, the moment of inertia of a uniform disc about a diameter is MR2/4 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS Page # 15 Ex.9 Two uniform identical rods each of mass M and length are joined to form a cross as shown in figure. Find the moment of inertia of the cross about a bisector as shown dotted in the figure. Sol. Consider the line perpendicular to the plane of the figure through the centre of the cross. The moment of inertia of each rod about this line is M 2 and hence the moment of inertia of the 12 M 2 . The moment of inertia of the cross about the two bisector are equal by 6 symmetry and according to the theorem of perpendicular axes, the moment of inertia of the cross is cross about the bisector is M 2 . 12 Ex.10 In the figure shown find moment of inertia of a plate having mass M, length and width b about axis 1,2,3 and 4. Assume that C is centre and mass is uniformly distributed 1 4 2 C 3 Sol. b Moment of inertia of the plate about axis 1 (by taking rods perpendicular to axis 1) l1 = Mb2/3 Moment of inertia of the plate about axis 2 (by taking rods perpendicular to axis 2) I 2 = M 2/12 Moment of inertia of the plate about axis 3 (by taking rods perpendicular to axis 3) Mb2 12 Moment of inertia of the plate about axis 4(by taking rods perpendicular to axis 4) I4 = M 2/3 I3 3.3 Moment of Inertia of Compound Bodies Consider two bodies A and B, rigidly joined together. The moment of inertia of this compound body, about an axis XY, is required. If IA is the moment of inertia of body A about XY. I B is the moment of inertia of body B about XY.Then, moment of Inertia of compound body I = IA + IB Extending this argument to cover any number of bodies rigidly joined together, we see that the moment of inertia of the compound body, about a specified axis, is the sum of the moments of inertia of the separate parts of the body about the same axis. A Y X B : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 16 ROTATIONAL DYNAMICS Ex.11 Two rods each having length l and mass m joined together at point B as shown in figure.Then findout moment of inertia about axis passing thorugh A and perpendicular to the plane of page as shown in figure. A B × Sol. C We find the resultant moment of inertia I by dividing in two parts such as I = M.I of rod AB about A + M.I of rod BC about A I = I 1 + I2 ... (1) first calculate I1 : B A × m 2 ...(2) 3 Calculation of I2 : use parallel axis theorem I2 = ICM + md2 I1 = m 2 = 12 2 × /2 m 2 5 2 m 4 12 4 Put value from eq. (2) & (3) into (1) I= m 2 3 I= 4. 2 m m 2 12 d COM × = ...(3) 5 2m 4 m 2 ( 4 1 15) 12 I= 5m 3 2 CAVITY PROBLEMS : Ex.12 A uniform disc having radius 2R and mass density as shown in figure. If a small disc of radius R is cut from the disc as shown. Then find out the moment of inertia of remaining disc around the axis that passes through O and is perpendicular to the plane of the page. 2R O Sol. R We assume that in remaning part a disc of radius R and mass density ± M2 2R O R 2R when is placed. Then ( 2R) 2 M1 × I1 is taken + I2 × – R2 R when – is takes Total Moment of Inertia I = I1 + I2 I1 = M1( 2R) 2 2 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS Page # 17 4R2 .4R2 =8 R4 2 To calculate I2 we use parallel axis theorem. I2 = ICM + M2R2 I1 = I2 = M2R2 + M2R2 2 3 3 M2R2 = (– 2 2 Now I = I1 + I2 I2 = R4 – I= 8 3 2 R2 )R 2 I2 = – R4 I= 3 2 R4 13 2 R4 Ex.13 A uniform disc of radius R has a round disc of radius R/3 cut as shown in Fig. The mass of the remaining (shaded) portion of the disc equals M. Find the moment of inertia of such a disc relative to the axis passing through geometrical centre of original disc and perpendicular to the plane of the disc. O Sol. Let the mass per unit area of the material of disc be . Now the empty space can be considered as having density – and . Now I0 = I + I– ( R2)R 2/2 = M.I of about O – I– = (R / 3) 2 (R / 3) 2 2 = M.I of – I0 = 5. R 4 9 [– (R / 3) 2 ]( 2R / 3) 2 about 0 R4 Ans. TORQUE : Torque represents the capability of a force to produce change in the rotational motion of the body Line of action of force P F r r sin Q 5.1 Torque about point : Torque of force F about a point r F where F = force applied P = point of application of force Q = point about which we want to calculate the torque. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 18 ROTATIONAL DYNAMICS r = position vector of the point of application of force from the point about which we want to determine the torque. rF sin where = r F = rF = angle between the direction of force and the position vector of P wrt. Q. r = perpendicular distance of line of action of force from point Q. F = force arm SI unit to torque is Nm Torque is a vector quantity and its direction is determined using right hand thumb rule. Ex.14 A particle of mass M is released in vertical plane from a point P at x = x0 on the x-axis it falls vertically along the y-axis. Find the torque acting on the particle at a time t about origin? x0 P O x r Sol. mg Torque is produced by the force of gravity rF sin k or r F x0 mg Ex.15 Calculate the total torque acting on the body shown in figure about the point O 10N 15N 37° 90° O 30° 150° 5N 20N 15N 10N 15sin37° 37° 90° Sol. O 4cm 5N 30° 20N 150° 20sin30° = 15sin37 × 6 + 20 sin 30° × 4 0 = 54 + 40 – 40 = 54 N-cm = 0.54 N-m 0 – 10 × 4 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS Page # 19 Ex.16 A particle having mass m is projected with a velocity v0 from a point P on a horizontal ground making an angle with horizontal. Find out the torque about the point of projection acting on the particle when (a) it is at its maximum height ? V0 P Q (b) It is just about to hit the ground back ? Sol. (a) Particle is at maximum height then about point P is F = mg ; r = P p (b) = p v0 r F r R 2 mg P v 20 sin 2 R mg = mg 2g 2 mv20 sin 2 2 when particle is at point Q then about point P is ' p r F R ; F = mg r ' p mgR = mg Q P v02 sin2 g mg Ex.17 In the previous question, during the motion of particle from P to Q. Torque of gravitational force about P is : (A) increasing (B) decreasing (C) remains constant (D) first increasing then decreasing Sol. Torque of gravitational force about P is increasing because r (Range) 5.2 Torque about axis : is increasing from O to R. r F where = torque acting on the body about the axis of rotation r = position vector of the point of application of force about the axis of rotation. F = force applied on the body.. net 1 2 3 ..... To understand the concept of torque about axis we take a general example which comes out in daily life. Figure shows a door ABCD. Which can rotate about axis AB. Now if we apply force. F at point. in inward direction then AB = r F and direction of this is along y axis from right hand thumb rule. Which AB is parallel to AB so gives the resultant torque. Now we apply force at point C in the direction as shown figure. At this time r & F are perpendicular to each other which gives AB D A r × y x B C F rF : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 20 ROTATIONAL DYNAMICS But door can’t move when force is applied in this direction because the direction of perpendicular to AB according to right hand thumb rule. So there is no component of along AB which gives res AB is 0 Now conclude Torque about axis is the component of r F parallel to axis of rotation. Note : The direction of torque is calculated using right hand thumb rule and it is always perpendicular to the plane of rotation of the body. F2 r2 F3 r3 × r1 F1 If F1 or F2 is applied to body, body revolves in anti-clockwise direction and F3 makes body revolve in clockwise direction. If all three are applied. F1r1 F2r2 – F3 r3 (in anti-clockwise direction) resul tan t 6. BODY IS IN EQUILIBRIUM : We can say rigid body is in equillibrium when it is in (a) Translational equilibrium i.e. Fnet 0 Fnet x = 0 and Fnet y = 0 and (b) Rotational equillibrium net 0 i.e., torque about any point is zero Note : (i) If net force on the body is zero then net torque of the forces may or may not be zero. example. A pair of forces each of same magnitude and acting in opposite direction on the rod. F B C A 2 F (2) 2F A If net force on the body is zero then torque of the forces about each and every point is same about B about C B F +F B 2F C 2F Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS Page # 21 Ex.18 Determine the point of application of third force for which body is in equillibrium when forces of 20 N & 30 N are acting on the rod as shown in figure 20N A 10cm C 20cm B Sol. 30N Let the magnitude of third force is F, is applied in upward direction then the body is in the equilibrium when (i) Fnet (ii) 0 (Translational Equillibrium) 20 + F = 30 F = 10 N So the body is in translational equilibrium when 10 N force act on it in upward direction. Let us assume that this 10 N force act. 10N 20N Then keep the body in rotational equilibrium x So Torque about C = 0 i.e. =0 B A C 20cm c 30 × 20 = 10 x 30N x = 60 cm so 10 N force is applied at 70 cm from point A to keep the body in equilibrium. Ex.19 Determine the point of application of force, when forces are acting on the rod as shown in figure. 10N 5N 5cm 5cm 3N Sol. Since the body is in equillibrium so we conclude F net i.e., net 0 and torque about any point is zero 0 10N 5N 6 F2 A x 37° 8N F F1 3N Let us assume that we apply F force downward at A angle from B From F net from the horizontal, at x distance 0 Fnet x = 0 which gives F2 = 8 N From Fnet y = 0 5 + 6 = F1 + 3 F1 = 8 N If body is in equillibrium then torque about point B is zero. 3 × 5 + F1. x – 5 × 10 = 0 15 + 8x – 50 = 0 x= 35 9 x = 4.375 cm : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 22 ROTATIONAL DYNAMICS Ex.20 A uniform rod length , mass m is hung from two strings of equal length from a ceiling as shown in figure. Determine the tensions in the strings ? /4 Sol. B A Let us assume that tension in left and right string is TA and TB respectively. Then Rod is in equilibrium then Fnet From Fnet From 0& net 0 0 mg = TA + TB ...(1) = 0 about A net mg 3 TB 4 2 TB TA 0 /2 B A mg 2mg TB = 3 2mg = mg 3 from eq. (1) TA /4 TA = mg 3 Ladder Problems : Ex.21 A stationary uniform rod of mass ‘m’, length ‘ ’ leans against a smooth vertical wall mak ing an angle with rough horizontal floor. Find the normal force & frictional force that is exerted by the floor on the rod? smooth Sol. rough As the rod is stationary so the linear acceleration and angular acceleration of rod is zero. i.e., acm = 0 ; = 0. A N2 N2 = f acm =0 N = mg 1 N1 Torque about any point of the rod should also be zero =0 mg A =0 N1 cos f= mg cos = sin mgcos 2 sin = f+ 2 +f mgcos 2 sin = N1 cos . B f Free Body Diagram mgcot 2 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS Page # 23 Ex.22 The ladder shown in figure has negligible mass and rests on a frictionless floor. The crossbar connects the two legs of the ladder at the middle. The angle between the two legs is 60°. The fat person sitting on the ladder has a mass of 80 kg. Find the contanct force exerted by the floor on each leg and the tension in the crossbar. W 1m 60° T N 1m Sol. N The forces acting on different parts are shown in figure. Consider the vertical equilibrium of “the ladder plus the person” system. The forces acting on this system are its weight (80 kg) g and the contact force N + N = 2 N due to the floor. Thus 2 N = (80 kg) g or N = (40 kg) (9.8 m/s2) = 392 N Next consider the equilibrium of the left leg of the ladder. Taking torques of the forces acting on it about the upper end, N (2m) tan 30° = T (1m) or T=N 2 3 = (392 N) × 2 3 = 450 N Ex.23 A thin plank of mass m and length is pivoted at one end and it is held stationary in horizontal position by means of a light thread as shown in the figure then find out the force on the pivot. Sol. (i) Free body diagram of the plank is shown in figure. Plank is in equilibrium condition So Fnet & net on the plank is zero from Fnet = 0 Fnet x = 0 N1 = 0 Now Fnet y N2 N1 T O A mg 0 N2 + T = mg from net = 0 about point A is zero net so N2 . = mg . /2 N2 ...(i) mg 2 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 24 ROTATIONAL DYNAMICS Ex.24 A square plate is hinged as shown in figure and it is held stationary by means of a light thread as shown in figure. Then find out force exerted by the hinge. square plate Sol. T F.B.D. Body is in equilibrium and T and mg force passing through one line so from net = 0, N N= 0 mg Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS Exercise - I Page # 25 OBJECTIVE PROBLEMS (JEE MAIN) (A) MOMENT OF INERTIA Sol. 1. The moment of inertia of a body depends upon (A) mass only (B) angular velocity only (C) distribution of particles only (D) mass and distribution of mass about the axis Sol. 4. The M.I. of a disc about its diameter is 2 units. Its M.I. about axis through a point on its rim and in the plane of the disc is (A) 4 units. (B) 6 units (C) 8 units (D) 10 units Sol. 2. Two spheres of same mass and radius are in contact with each other. If the moment of inertia of a sphere about its diameter is I, then the moment of inertia of both the spheres about the tangent at their common point would be (A) 3I (B) 7I (C) 4I (D) 5I Sol. 5. A solid sphere and a hollow sphere of the same mass have the same moments of inertia about their respective diameters, the ratio of their radii is (A) (5)1/2 : (3)1/2 (B) (3)1/2 : (5)1/2 (C) 3 : 2 (D) 2 : 3 Sol. 3. A disc of metal is melted to recast in the form of a solid sphere. The moment of inertias about a vertical axis passing through the centre would (A) decrease (B) increase (C) remains same (D) nothing can be said : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 26 ROTATIONAL DYNAMICS 6. A stone of mass 4kg is whirled in a horizontal circle of radius 1m and makes 2 rev/sec. The moment of inertia of the stone about the axis of rotation is (A) 64 kg × m 2 (B) 4 kg × m 2 2 (C) 16 kg × m (D) 1 kg × m 2 Sol. 8. A circular disc A of radius r is made from an iron plate of thickness t and another circular disc B of radius 4r is made from an iron plate of thickness t/4. The relation between the moments of inertia IA and IB is (A) IA > I B (B) IA = IB (C) IA < IB (D) depends on the actual values of t and r. Sol. 7. Three rings, each of mass P and radius Q are arranged as shown in the figure. The moment of inertia of the arrangement about YY’ axis will be 9. The moment of inertia of a uniform semicircular wire of mass M and radius r about a line perpendicular to the plane of the wire through the centre is (A) Mr2 (B) 1 2 Mr 2 (C) 1 2 Mr 4 (D) 2 2 Mr 5 Sol. (A) 7 PQ2 2 (B) 2 PQ2 7 (C) 2 PQ2 5 (D) 5 PQ2 2 Sol. 10. Let IA and IB be moments of inertia of a body about two axes A and B respectively. The axis A passes through the centre of mass of the body but B does not. (A) IA < IB (B) If IA < IB, the axes are parallel. (C) If the axes are parallel, IA < IB (D) If the axes are not parallel, IA IB Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS Sol. Page # 27 Sol. 11. For the same total mass which of the following will have the largest moment of inertia about an axis passing through its centre of mass and perpendicular to the plane of the body (A) a disc of radius a (B) a ring of radius a (C) a square lamina of side 2a (D) four rods forming a square of side 2a Sol. 12. Moment of inertia of a thin semicircular disc (mass = M & radius = R) about an axis through point O and perpendicular to plane of disc, is given by : O 13. A rigid body can be hinged about any point on the x-axis. When it is hinged such that the hinge is at x, the moment of inertia is given by I = 2x 2 – 12x + 27 The x-coordinate of centre of mass is (A) x = 2 (B) x = 0 (C) x = 1 (D) x = 3 Sol. 14. Consider the following statements Assertion (A) : The moment of inertia of a rigid body reduces to its minimum value as compared to any other parallel axis when the axis of rotation passes through its centre of mass. Reason (R) : The weight of a rigid body always acts through its centre of mass in uniform gravitational field. Of these statements : (A) both A and R are true and R is the correct explanation of A (B) both A and R are true but R is not a correct explanation of A (C) A is true but R is false (D) A is false but R is true Sol. R 1 MR2 4 1 2 (C) MR 8 (A) (B) 1 MR2 2 (D) MR2 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 28 ROTATIONAL DYNAMICS 15. A body is rotating uniformly about a vertical axis fixed in an inertial frame. The resultant force on a particle of the body not on the axis is (A) vertical (B) horizontal and skew with the axis (C) horizontal and intersecting the axis (D) none of these Sol. 16. One end of a uniform rod of mass m and length I is clamped. The rod lies on a smooth horizontal surface and rotates on it about the clamped end at a uniform angular velocity . The force exerted by the clamp on the rod has a horizontal component (A) m 2 l (B) zero 1 2 (C) mg (D) m 2 Sol. (B) TORQUE AND PURE ROTATIONAL MOTION 18. A disc of radius 2m and mass 200kg is acted upon by a torque 100N-m. Its angular acceleration would be (A) 1 rad/sec2 (B) 0.25 rad/sec 2 (C) 0.5 rad/sec2 . (D) 2 rad/sec2 . Sol. 19. On applying a constant torque on a body(A) linear velocity increases (B) angular velocity increases (C) it will rotate with constant angular velocity (D) it will move with constant velocity Sol. 17. A rod of length 'L' is hinged from one end. It is brought to a horizontal position and released. The angular velocity of the rod when it is in vertical position is (A) 2g L (B) 3g L (C) g 2L (D) g L Sol. 20. A wheel starting with angular velocity of 10 radian/sec acquires angular velocity of 100 radian/sec in 15 seconds. If moment of inertia is 10kg-m2 , then applied torque (in newtonmetre) is (A) 900 (B) 100 (C) 90 (D) 60 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS Page # 29 23. The an gular velocity of a body is Sol. = 2i + 3 j + 4k and a torqu e = i + 2 j + 3 k acts on it. The rotational power will be (A) 20 watt (B) 15 watt (C) 17 watt (D) 14 watt Sol. 21. An automobile engine develops 100H.P. when rotating at a speed of 1800 rad/min. The torque it delivers is (A) 3.33 W-s (B) 200W-s (C) 248.7 W-s (D) 2487 W-s Sol. 24. A torque of 2 newton-m produces an angular acceleration of 2 rad/sec 2 a body. If its radius of gyration is 2m, its mass will be: (A) 2kg (B) 4 kg (C) 1/2 kg (D) 1/4 kg Sol. 22. The moment of inertia and rotational kinetic energy of a fly wheel are 20kg-m 2 and 1000 joule respectively. Its angular frequency per minute would be (A) 600 (B) 25 2 (C) 5 (D) 300 Sol. 25. A particle is at a distance r from the axis of rotation. A given torque produces some angular acceleration in it. If the mass of the particle is doubled and its distance from the axis is halved, the value of torque to produce the same angular acceleration is (A) /2 (B) (C) 2 (D) 4 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 30 ROTATIONAL DYNAMICS 28. In an experiment with a beam balance on unknown mass m is balanced by two known mass m is balanced by two known masses of 16 kg and 4 kg as shown in figure. Sol. l1 16kg l2 l1 m m l2 4kg The value of the unknown mass m is (A) 10 kg (B) 6 kg (C) 8 kg (D) 12 kg Sol. 26. A weightless rod is acted on by upward parallel forces of 2N and 4N ends A and B respectively. The total length of the rod AB = 3m. To keep the rod in equilibrium a force of 6N should act in the following manner : (A) Downwards at any point between A and B. (B) Downwards at mid point of AB. (C) Downwards at a point C such that AC = 1m. (D) Downwards at a point D such that BD = 1m. Sol. 29. A homogeneous cubical brick lies motionless on a rough inclined surface. The half of the brick which applies greater pressure on the plane is : 27. A right triangular plate ABC of mass m is free to rotate in the vertical plane about a fixed horizontal axis through A. It is supported by a string such that the side AB is horizontal. The reaction at the support A is : A l (A) left half (B) right half (C) both applies equal pressure (D) the answer depend upon coefficient of friction Sol. B l C (A) Sol. mg 3 (B) 2 mg 3 (C) mg 2 (D) mg Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS Page # 31 30. Consider the following statements Assertion (A) : A cyclist always bends inwards while negotiating a curve Reason (R) : By bending he lowers his centre of gravity Of these statements, (A) both A and R are true and R is the correct explanation of A (B) both A and R are true but R is not the correct explanation of A (C) A is true but R is false (D) A is false but R is true Sol. 32. A pulley is hinged at the centre and a massless thread is wrapped around it. The thread is pulled with a constant force F starting from rest. As the time increases, F (A) its angular velocity increases, but force on hinge remains constant (B) its angular velocity remains same, but force on hinge increases (C) its angular velocity increases and force on hinge increases (D) its angular velocity remains same and force on hinge is constant. 31. A rod is hinged at its centre and rotated by applying a constant torque starting from rest. The power developed by the external torque as a function of time is : P ex t P ex t (A) (B) time time P ex t Pex t (C) (D) time time 33. The angular momentum of a flywheel having a moment of inertia of 0.4 kg m2 decreases from 30 to 20 kg m2/s in a period of 2 second. The average torque acting on the flywheel during this period is : (A) 10 N.m (B) 2.5 N.m (C) 5 N.m (D) 1.5 N.m Sol. Sol. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 32 ROTATIONAL DYNAMICS 34. A particle starts from the point (0m, 8m) and Sol. moves with uniform velocity of 3 i m/s. After 5 seconds, the angular velocity of the particle about the origin will be : y 3m/s 8m x O (A) 8 rad / s 289 (B) 3 rad / s 8 (C) 24 rad / s 289 (D) 8 rad / s 17 Sol. (C) ANGULAR MOMENTUM 35. The rate of change of angular momentum is called (A) angular velocity (B)angular acceleration (C) force (D) torque Sol. 37. The angular velocity of a body changes from one revolution per 9second to 1 revolution per second without applying any torque. The ratio of its radius of gyration in the two cases is (A) 1 : 9 (B) 3 : 1 (C) 9 : 1 (D) 1 : 3 Sol. 38. A dog of mass m is walking on a pivoted disc of radius R and mass M in a circle of radius R/2 with an angular frequency n: the disc will revolve in opposite direction with frequency R/2 R (A) mn M (B) mn 2M (C) 2mn M (D) 2Mn M Sol. 36. The rotational kinetic energy of a rigid body of moment of inertia 5 kg-m2 is 10 joules. The angular momentum about the axis of rotation would be (A) 100 joule-sec (B) 50 joule-sec (C) 10 joule-sec (D) 2 joule -sec Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS 39. A particle moves with a constant velocity parallel to the X-axis. Its angular momentum with respect to the origin. (A) is zero (B) remains constant (C) goes on increasing (D) goes on decreasing. Sol. Page # 33 Sol. 42. A particle of mass 2 kg located at the position ( i j ) m has a velocity 2( i – j k) m/s. Its angular momentum about z-axis in kg-m2 /s is : (A) zero (B) +8 (C) 12 (D) – 8 Sol. 40. A person sitting firmly over a rotating stool has his arms streatched. If he folds his arms, his angular momentum about the axis of rotation (A) increases (B) decreases (C) remains unchanged (D) doubles. Sol. 43. A ball of mass m moving with velocity v, collide with the wall elastically as shown in the figure. After impact the change in angular momentum about P is : P d (A) 2 mvd (C) 2 mvd sin Sol. (B) 2 mvd cos (D) zero 41. A man, sitting firmly over a rotating stool has his arms streched. If he folds his arms, the work done by the man is (A) zero (B) positive (C) negative (D) may be positive or negative. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 34 ROTATIONAL DYNAMICS 44. A uniform rod of mass M has an impulse applied at right angles to one end. If the other end begins to move with speed V, the magnitude of the impulse is (A) MV (B) MV 2 (C) 2MV (D) Sol. 2MV 3 Sol. (D) COMBINED TRANSLATIONAL + ROTATIONAL MOTION 47. A circular disc has a mass of 1kg and radius 40 cm. It is rotating about an axis passing through its centre and perpendicular to its plane with a speed of 10rev/s. The work done in joules in stopping it would be(A) 4 (B) 47.5 (C) 79 (D) 158 Sol. 45. A circular ring of wire of mass M and radius R is making n revolutions/sec about an axis passing through a point on its rim and perpendicular to its plane. The kinetic energy of rotation of the ring is given by(A) 4 2MR2 n2 (B) 2 2MR2 n2 (C) 1 2 MR2n2 2 (D) 8 2 MR2n2 Sol. 48. A disc rolls on a table. The ratio of its K.E. of rotation to the total K.E. is (A) 2/5 (B) 1/3 (C) 5/6 (D) 2/3 Sol. 46. Rotational kinetic energy of a disc of constant moment of inertia is (A) directly proportional to angular velocity (B) inversely proportional to angular velocity (C) inversely proportional to square of angular velocity (D) directly proportional to square of angular velocity Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS Page # 35 49. A disk and a ring of the same mass are rolling to have the same kinetic energy. What is ratio of their velocities of centre of mass (A) (4:3) 1/2 (B) (3 : 4)1/2 1/2 1/2 (C) (2) : (3) (D) (3)1/2 : (2)1/2 Sol. Sol. 52. A solid sphere, a hollow sphere and a disc, all having smooth incline and released. Least time will be taken in reaching the bottom by (A) the solid sphere (B) the hollow sphere (C) the disc (D) all will take same time. Sol. 50. If the applied torque is directly proportional to the angular displacement , then the work done in rotating the body through an angle would be - (C is constant of proportionality) (A) C (B) 1 C 2 (C) 1 C 2 2 (D) C 2 Sol. 51. The centre of a wheel rolling without slipping in a plane surface moves with speed v0 . A particle on the rim of the wheel at the same level as the centre will be moving at speed. (A) zero (B) v 0 (C) 2v0 (D) 2v0 53. A wheel of radius r rolling on a straight line, the velocity of its centre being v. At a certain instant the point of contact of the wheel with the grounds is M and N is the highest point on the wheel (diametrically opposite to M). The incorrect statement is : (A) The velocity of any point P of the wheel is proportional to MP. (B) Points of the wheel moving with velocity greater than v form a larger area of the wheel than points moving with velocity less than v. (C) The point of contact M is instantaneously at rest. (D) The velocities of any two parts of the wheel which are equidistant from centre are equal. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 36 ROTATIONAL DYNAMICS Sol. Sol. 54. There is rod of length l. The velocities of its two ends are v1 and v2 in opposite directions n ormal to t he rod . Th e di stan ce of th e instantaneous axis of rotation from v1 is : v2 v1l (A) zero (B) v v l (C) (D) l/2 v1 v2 1 2 Sol. 56. The linear speed of a uniform spherical shell after rolling down an inclined plane of vertical height h from rest, is : (A) 10gh 7 (B) 4gh 5 (C) 6gh 5 (D) 2gh Sol. 55. A ring of radius R rolls without sliding with a constant velocity. The radius of curvature of the path followed by any particle of the ring at the highest point of its path will be (A) R (B) 2R (C) 4R (D) none 57. A body kept on a smooth horizontal surface is pulled by a constant horizontal force applied at the top point of the body. If the body rolls purely on the surface, its shape can be : (A) thin pipe (B) uniform cylinder (C) uniform sphere (D) thin spherical shell Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS Page # 37 Sol. 60. A uniform circular disc placed on a rough horizontal surface has initially a velocity v0 and an angular velocity 0 as shown in the figure. The disc comes to rest after moving some distance v0 in the direction of motion. Then r is 0 0 v0 58. A solid sphere with a velocity (of centre of mass) v and angular velocity is gently placed on a rough horizontal surface. The frictional force on the sphere : (A) must be forward (in direction of v) (B) must be backward (opposite to v) (C) cannot be zero (D) none of the above Sol. 59. A cylinder is pure rolling up an incline plane. It stops momentarily and then rolls back. The force of friction. (A) on the cylinder is zero throughout the journey (B) is directed opposite to the velocity of the centre of mass throughout the journey (C) is directed up the plane throughout the journey (D) is directed down the plane throughout the journey Sol. (A) Sol. 1 2 (B) 1 (C) 3 2 (D) 2 61. A Cubical bloc of mass M and edge a slides down a rough inclined plane of inclination with a uniform velocity. The torque of the normal force on the block about its centre has a magnitude. (A) zero (B) Mga (C) Mga sin (D) 1 Mgasin 2 Sol. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 38 ROTATIONAL DYNAMICS Exercise - II (A) MOMENT OF INERTIA Sol. 1. Three bodies have equal masses m. Body A is solid cylinder of radius R, body B is a square lamina of side R, and body C is a solid sphere of radius R. Which body has the smallest moment of inertia about an axis passing through their centre of mass and perpendicular to the plane (in case of lamina) (A) A (B) B (C) C (D) A and C both Sol. 4. A thin uniform rod of mass M and length L has its moment of inertia I1 about its perpendicular bisector. The rod is bend in the form of a semicircular arc. Now its moment of inertia through the centre of the semi circular arc and perpendicular to its plane is I2. The ratio of I1 : I2 will be _________________ (A) < 1 (B) > 1 (C) = 1 (D) can’t be said Sol. 2. Two rods of equal mass m and length l lie along the x axis and y axis with their centres origin. What is the moment of inertia of both about the line x = y : ml 2 3 ml 2 (C) 12 Sol. (A) ml 2 4 ml 2 (D) 6 (B) 5. A square plate of mass M and edge L is shown in figure. The moment of inertia of the plate about the axis in the plane of plate passing through one of its vertex making an angle 15° from horizontal is. axis 15° L L 3. Moment of inertia of a rectangular plate about an axis passing through P and perpendicular to the plate is I. Then moment of PQR about an axis perpendicular to the plane of the plate : Q P S (A) about P = I/2 (C) about P > I/2 ML2 (A) 12 Sol. 11ML2 (B) 24 (C) 7 ML2 12 (D) none R (B) about R = I/2 (D) about R > I/2 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS Page # 39 Question No. 6 to 9 (4 questions) The figure shows an isosceles triangular plate of mass M and base L. The angle at the apex is 90°. The apex lies at the origin and the base is parallel to X - axis. Y Sol. M X 6. The moment of inertia of the plate about the z-axis is (A) ML2 12 (B) ML2 24 (C) ML2 (D) none of these 6 Sol. 9. The moment of inertia of the plate about the y-axis is ML2 6 ML2 (C) 24 Sol. (A) 7. The moment of inertia of the plate about the x-axis is (A) ML2 8 (B) ML2 32 (C) ML2 24 (D) ML2 6 (B) ML2 8 (D) none of these 10. ABCD is a square plate with centre O. The moments of inertia of the plate about the perpendicular axis through O is I and about the axes 1, 2, 3 & 4 are I1, I2, I3 & I4 respectively. It follows that : Sol. 1 2 A B 3 O D (A) I2 = I3 (C) I = I2 + I4 Sol. C 4 (B) I = I1 + I4 (D) I1 = I3 8. The moment of inertia of the plate about its base parallel to the x-axis is (A) ML2 18 (B) ML2 36 (C) ML2 (D) none of these 24 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 40 ROTATIONAL DYNAMICS (B) TORQUE & PURE ROTATIONAL MOTION 11. A horizontal force F = mg/3 is applied on the upper surface of a uniform cube of mass ‘m’ and side ‘a’ which is resting on a rough horizontal surface having s = 1/2. The distance between lines of action of ‘mg’ and normal reaction ‘N’ is : (A) a/2 (B) a/3 (C) a/4 (D) None Sol. 13. A uniform cube of side ‘b’ and mass M rest on a rough horizontal table. A horizontal force F is applied normal to one of the face at a point, at a height 3b/4 above the base. What should be the coefficient of friction ( ) between cube and table so that is will tip about an edge before it starts slipping? F b (A) Sol. 12. A man can move on a horizontal plank supported symmetrically as shown. The variation of normal reaction on support A with distance x of the man from the end of the plank is best represented by : x=0 A B 1m 4m N (A) 1 3 (B) 3 2 (D) none at the origin O, as shown. If i , j and k are unit vectors, and a, b, and c are positive constants, which of the following forces F applied to the rim of the cone at a point P results in a torque on the cone with a negative component Z ? z ko x N (C) 14. A solid cone hangs from a frictionless pivot (B) x y i j x c N (C) (D) x Sol. 1m N 2 3 3b/4 b x (A) F = a k , P is (0, b, –c) (B) F = –a k , P is (0, –b, –c) (C) F = a j , P is (–b, 0, –c) (D) None Sol. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS 15. A block of mass m is attached to a pulley disc of equal mass m, radius r by means of a slack string as shown. The pulley is hinged about its centre on a horizontal table and the block is projected with an initial velocity of 5 m/s. Its velocity when the string becomes taut will be (A) 3 m/s (C) 5/3 m/s Sol. (B) 2.5 m/s (D) 10/3 m/s 16. A rod of weight w is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at a distance d from each other. The centre of mass of the rod is at a distance x from A. wx (A) the normal reaction at A is d w(d x) (B) the normal reaction at A is d wx (C) the normal reaction at B is d w(d x) (D) the normal reaction at B is d Sol. Page # 41 17. A block with a square base measuring axa and height h, is placed on an inclined plane. The coefficient of friction is . The angle of inclination ( ) of the plane is gradually increased. The block will a (A) topple before sliding if h a (B) topple before sliding if h a (C) slide before toppling if h a (D) slide before toppling if h Sol. 18. A body is in equilibrium under the influence of a number of forces. Each force has a different line of action. The minimum number of forces required is (A) 2, if their lines of action pass through the centre of mass of the body (B) 3, if their lines of action are not parallel (C) 3, if their lines of action are parallel (D) 4, if their lines of action are parallel and all the forces have the same magnitude Sol. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 42 ROTATIONAL DYNAMICS 19. A block of mass m moves on a horizontal rough surface with initial velocity v. The height of the centre of mass of the block is h from the surface. Consider a point A on the surface (A) angular momentum about A is mvh initially (B) the velocity of the block decreases at time passes (C) torque of the forces acting on block is zero about A (D) angular momentum is not conserved about A Sol. Sol. 21. A particle falls freely near the surface of the earth. Consider a fixed point O (not vertically below the particle) on the ground. (A) Angular momentum of the particle about O is increasing (B) Torque of the gravitational force on the particle about O is decreasing (C) The moment of inertia of the particle about O is decreasing (D) The angular velocity of the particle about O is increasing Sol. 20. Four point masses are fastened to the corners of a frame of negligible mass lying in the xy plane. Let w be the angular speed of rotation. Then y-axis m b M M x-axis m a z-axis (A) rotational kinetic energy associated with a given angular speed depends on the axis of rotation. (B) rotational kinetic energy about y-axis is independent of m and its value is Ma2 2 (C) rotational kinetic energy about z-axis depends on m and its value is (Ma2 + mb2) 2 (D) rotational kinetic energy about z-axis is independent of m and its value is Mb2 2 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS Page # 43 22. A rod hinged at one end is released from the horizontal position as shown in the figure. When it becomes vertical its lower half separates without exerting any reaction at the breaking point. Then the maximum angle ‘ ’ made by the hinged upper half with the vertical is : C B A B (A) 30° Sol. (B) 45° Sol. B C (C) 60° (D) 90° 25. A thin circular ring of mass 'M' and radius 'R' is rotating about its axis with a constant angular velocity . Two objects each of mass m, are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velcoity. (C) ANGULAR MOMENTUM 23. If a person sitting on a rotating stool with his hands outstretched, suddenly lowers his hands, then his (A) Kinetic energy will decrease (B) Moment of inertia will decrease (C) Angular momentum will increase (D) Angular velocity will remain constant Sol. 24. A man spinning in free space changes the shape of his body, eg. by spreading his arms or curling up. By doing this, he can change his (A) moment of inertia (B) angular momentum (C) angular velocity (D) rotational kinetic energy (A) M (M m) (B) (C) M (M – 2m) (D) M (M 2m) (M 3m) M Sol. 26. A small bead of mass m moving with velocity v gets threaded on a stationary semicircular ring of mass m and radius R kept on a horizontal table. The ring can freely rotate about its centre. The bead comes to rest relative to the ring. What will be the final angular velocity of the system? R O (A) v/R (B) 2v/R : 0744-2209671, 08003899588 | url : www.motioniitjee.com, v m (C) v/2R (D) 3v/R :info@motioniitjee.com Page # 44 ROTATIONAL DYNAMICS Sol. Sol. 27. A thin uniform straight rod of mass 2 kg and length 1 m is free to rotate about its upper end when at rest. It receives an impulsive blow of 10 Ns at its lowest point, normal to its length as shown in figure. The kinetic energy of rod just after impact is Question No. 29& 30 (2 questions) A uniform rod is fixed to a rotating turntable so that its lower end is on the axis of the turntable and it makes an angle of 20° to the vertical. (The rod is thus rotating with uniform angular velocity about a vertical axis passing through one end.) If the turntable is rotating clockwise as seen from above. 20° 10 NS (A) 75 J (C) 200 J Sol. (B) 100 J (D) none 29. What is the direction of the rod’s angular momentum vector (calculated about its lower end) (A) vertically downwards (B) down at 20° to the horizontal (C) up at 20° to the horizontal (D) vertically upwards Sol. 28. A child with mass m is standing at the edge of a disc with moment of inertia I, radius R, and initial angular velocity . See figure given below. The child jumps off the edge of the disc with tangential velocity v with respect to the ground. The new angular velocity of the disc is v (A) (C) I 2 – mv2 I I – mvR I (B) (D) (I + mR 2 ) I ( I + mR2 ) I 2 30. Is there a torque acting on it, and if so in what direction? (A) yes, vertically (B) yes, horizontally (C) yes at 20° to the horizontal (D) no Sol. – mv2 mvR Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS Page # 45 31. One ice skater of mass m moves with speed 2v to the right, while another of the same mass m moves with speed v toward the left, as shown in figure I. Their paths are separated by a distance b. At t = 0, when they are both at x = 0, they grasp a pole of length b and negligible mass. For t > 0, consider the system as a rigid body of two masses m separated by distance b, as shown in figure II. Which of the following is the correct formula for the motion after t = 0 of the skater initially at y = b/2 ? Sol. y y m 2v b/2 b x x t=0 v (t<0) m Figure 1 Figure II (A) x = 2vt, y = b/2 (B) x = vt + 0.5b sin (3vt/b), y = 0.5b cos(3vt/b) (C) x = 0.5vt + 0.5b sin (3vt/b), y = 0.5b cos(3vt/b) (D) x = 0.5vt + 0.5b sin (6vt/b), y = 0.5b cos(6vt/b) Sol. 32. A uniform rod AB of length L and mass M is lying on a smooth table. A small particle of mass m strike the rod with a velocity v0 at point C at a distance x from the centre O. The particle comes to rest after collision. The value of x, so that point A of the rod remains ststionary just after collision is : B m 33. A uniform rod AB of mass m and length l is at rest on a smooth horizontal surface. An impulse J is applied to the end B, perpendicular to the rod in the horizontal direction. Speed of particle P at l a distance from the centre towards A of the 6 ml rod after time t is 12J J J J J (A) 2 (B) (C) (D) 2 2m m m m Sol. 34. A uniform rod of mass M is hinged at its upper end. A particle of mass m moving horizontally strikes the rod at its mid point elastically. If the particle comes to rest after collision find the value of M/m = ? v0 C v x m O M (A) L/3 (B) L/6 A (C) L/4 (D) L/12 (A) 3/4 (C) 2/3 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, (B) 4/3 (D) none :info@motioniitjee.com Page # 46 ROTATIONAL DYNAMICS Sol. (D) COMBINED TRANSLATIONAL + ROTATIONAL MOTION 37. A ring rolls without slipping on the ground. Its centre C moves with a constant speed u. P is any point on the ring. The speed of P with respect to the ground is v. (A) 0 v 2u (B) v = u, if CP is horizontal (C) v = u, if CP makes an angle of 30º with the horizontal and P is below the horizontal level of C 35. Two equal masses each of mass M are joined by a massless rod of length L. Now an impulse MV is given to the mass M making an angle of 30º with the length of the rod. The angular velocity of the rod just after imparting the impulse is M M (D) v 2 u , if CP is horizontal Sol. 30° MV 2v (B) L v (A) L v (C) 2L Sol. (D) none of these 36. Two particles of equal mass m at A and B are connected by a rigid light rod AB lying on a smooth horizontal table. An impulse J is applied at A in the plane of the table and perpendicular at AB. Then the velocity of particle at A is : (A) J 2m (B) J m (C) 2J m 38. A yo-yo is resting on a perfectly rough horizontal table. Forces F1, F2 and F3 are applied separately as shown.The F statement is F correct 3 2 (D) zero F1 Sol. (A) when F3 is applied move to the right (B) when F2 is applied move to the left (C) when F1 is applied move to the right (D) when F2 is applied move to the right the centre of mass will the centre of mass will the centre of mass will the centre of mass will Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS Page # 47 40. A plank with a uniform sphere placed on it, rests on a smooth horizontal plane. Plank is pulled to right by a constant force F. If the sphere does not slip over the plank. Sol. F 39. A disc of circumference s is at rest at a point A on a horizontal surface when a constant horizontal force begins to act on its centre. Between A and B there is sufficient friction to prevent slipping, and the surface is smooth to the right of B. AB = s. The disc moves from A to B in time T. To the right of B, (A) acceleration of centre of sphere is less than that of the plank (B) acceleration of centre of sphere is greater than the plank because friction acts rightward on the sphere (C) acceleration of the centre of sphere may be towards left (D) acceleration of the centre of sphere relative to plank may be greater than that of the plank relative to floor Sol. Force A B (A) the angular acceleration of the disc will disappear, linear acceleration will remain unchanged (B) linear acceleration of the disc will increase (C) the disc will make one rotation in time T/2 (D) the disc will cover a distance greater than s in further time T. Sol. 41. A hollow sphere of radius R and mass m is fully filled with water of mass m. It is rolled down a horizontal plane such that its centre of mass moves with a velocity v. If it purely rolls 5 2 (A) Kinetic energy of the sphere is mv 6 4 2 (B) Kinetic energy of the sphere is mv 5 (C) Angular momentum of the sphere about a 8 fixed point on ground is mvR 3 (D) Angular momentum of the sphere about a 14 mvR fixed point on ground is 5 Sol. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 48 ROTATIONAL DYNAMICS 42. In the figure shown, the plank is being pulled to the right with a constant speed v. If the cylinder does not slip then : R v (A) the speed of the centre of mass of the cylinder is 2v (B) the speed of the centre of mass of the cylinder is zero (C) the angular velocity of the cylinder is v/R (D) the angular velocity of the cylinder is zero Sol. 43. If a cylinder is rolling down the incline with sliding (A) after some time it may start pure rolling (B) after sometime it will start pure rolling (C) it may be possible that it will never start pure rolling (D) none of these Sol. , 44. Which of the following statements are correct (A) friction acting on a cylinder without sliding on an inclined surface is always upward along the incline irrespective of any external force acting on it. (B) friction acting on a cylinder without sliding on an inclined surface is may be upward may be downwards depending on the external force acting on it. (C) friction acting on a cylinder rolling without sliding may be zero depending on the external force acting on it. (D) nothing can be said exactly about it as it depends on the friction coefficient on inclined plane Sol. Question No. 45 to 47 (3 Questions) A cylinder and a ring of same mass M and radius R are placed on the top of a rough inclined plane of inclination . Both are released simultaneously from the same height h. 45. Choose the correct statement(s) related to the motion of each body (A) The friction force acting on each body opposes the motion of its centre of mass (B) The friction force provides the necessary torque to rotate the body about its centre of mass (C) without friction none of the two bodies can roll (D) The friction force ensures that the point of contact must remain stationary Sol. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS 46. Identify the correct statement(s) (A) The friction force acting on the cylinder may be more than that acting on the ring (B) The friction force acting on the ring may be more than that acting on the cylinder (C) If the friction is sufficient to roll the cylinder then the ring will also roll (D) If the friction is sufficient to roll the ring then the cylinder will also roll Sol. Page # 49 Question No. 48 to 51 (4 Questions) A ring of mass M and radius R sliding with a velocity v0 suddenly enters into rough surface where the coefficient of friction is , as shown in figure. v0 Rough ( ) 48. Choose the correct statement(s) (A) As the ring enters on the rough surface, the limiting friction force acts on it (B) The direction of friction is opposite to the direction of motion (C) The friction force accelerates the ring in the clockwise sense about its centre of mass (D) As the ring enters on the rough surface it starts rolling Sol. 47. When these bodies roll down to the foot of the inclined plane, then (A) the mechanical energy of each body is conserved (B) the velocity of centre of mass of the cylinder gh 3 (C) the velocity of centre of mass of the ring is is 2 gh (D) the velocity of centre of mass of each body is 2 gh Sol. 49. Choose the correct statement(s) (A) The momentum of the ring is conserved (B) The angular momentum of the ring is conserved about its centre of mass (C) The angular momentum of the ring conserved about any point on the horizontal surface (D) The mechanical energy of the ring is conserved Sol. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 50 ROTATIONAL DYNAMICS 50. Choose the correct statement(s) (A) The ring starts its rolling motion when the centre of mass stationary (B) The ring starts rolling motion when the point of contact becomes stationary (C) The time after which the ring starts rolling is v0 2 g v (D) The rolling velocity is 0 2 Sol. 52. Consider a sphere of mass ‘m’ radius ‘R’ doing pure rolling motion on a rough surface having velocity v 0 as shown in the Figure. It makes an elastic impact with the smooth wall and moves back and starts pure rolling after some time again. v0 O (A) Change in angular momentum about ‘O’ in the entire motion equals 2mv0 R in magnitude. (B) Moment of impulse provided by wall during impact about O equals 2mv0R in magnitude 3 (C) Final velocity of ball will be v0 7 3 v0 (D) Final velocity of ball will be 7 Sol. 51. Choose the correct alternative(s) (A) The linear distance moved by the centre of mass before the ring starts rolling is 3v 20 8 g 3 mv20 8 mv20 (C) The loss is kinetic energy of the ring is 4 mv20 (D) The gain in rotational kinetic energy is 8 Sol. (B) The net work done by friction force is 53. A solid sphere, a hollow sphere and a disc, all having same mass and radius, are placed at the top of an incline and released. The friction coefficients between the objects and the incline are same and not sufficient to allow pure rolling. The smallest kinetic energy at the bottom of the incline will be achieved by (A) the solid sphere (B) the hollow sphere (C) the disc (D) all will achieve same kinetic energy. Sol. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS Page # 51 54. Fig. shows a smooth inclined plane fixed in a car accelerating on a horizontal road. The angle of incline is related to the acceleration a of the car as a = g tan . If the sphere is set in pure rotation on the incline. a 56. A ladder of length L is slipping with its ends against a vertical wall and a horizontal floor. At a certain moment, the speed of the end in contact with the horizontal floor is v and the ladder makes an angle = 30º with the horizontal. Then the speed of the ladder’s center must be (A) 2v / 3 (C) v Sol. (B) v/2 (D) none (A) it will continue pure rolling (B) it will slip down the plane (C) its linear velocity will increase (D) its linear velocity will decrease. Sol. 57. In the previous question, if dv/dt = 0, then the angular acceleration of the ladder when = 45º is (A) 2v2/L2 (B) v2/2L2 (C) (D) None 2[ v2 / L2 ] Sol. 55. A straight rod of length L is released on a frictionless horizontal floor in a vertical position. As it falls + slips, the distance of a point on the rod from the lower end, which follows a quarter circular locus is (A) L/2 (B) L/4 (C) L/8 (D) None Sol. 58. A time varying force F = 2t is applied on a spool rolling as shown in figure. The angular momentum of the spool at time t about bottom most point is : F=2t r R r 2t 2 R (C) (R + r)t2 (A) : 0744-2209671, 08003899588 | url : www.motioniitjee.com, (R r ) 2 2 t r (D) data is insufficient (B) :info@motioniitjee.com Page # 52 ROTATIONAL DYNAMICS Sol. Sol. 59. A ball rolls down an inclined plane, figure. The ball is first released from rest from P and then later from Q. Which of the following statement is/ are correct ? Q P 2h h O (i) The ball takes twice as much time to roll from Q to O as it does to roll from P to O. (ii) The acceleration of the ball at Q is twice as large as the acceleration at P. (iii) The ball has twice as much K.E. at O when rolling from Q as it does when rolling from P. (A) i, ii only (B) ii, iii only (C) i only (D) iii only Sol. 61. In the figure shown a ring A is initially rolling without sliding with a velocity v on the horizontal surface of the body B (of same mass as A). All surfaces are smooth. B has no initial velocity. What will be the maximum height reached by A on B. A v B (A) 3v 2 4g (B) v2 4g (C) v2 2g (D) v2 3g Smooth Sol. 60. Starting from the rest, at the same time, a ring, a coin and a solid ball of same mass roll down an incline without slipping. The ratio of their translational kinetic energies at the bottom will be (A) 1 : 1 : 1 (B) 10 : 5 : 4 (C) 21 : 28 : 30 (D) none Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS Page # 53 62. Inner and outer radii of a spool are r and R respectively. A thread is wound over its inner surface and placed over a rough horizontal surface. Thread is pulled by a force F as shown in fig. then in case of pure rolling Sol. F (A) Thread unwinds, spool rotates anticlockwise and friction act leftwards (B) Thread winds, spool rotates clockwise and friction acts leftwards (C) Thread winds, spool moves to the right anf friction act rightwards (D) Thread winds, spool moves to the right and friction does not come into existence. Sol. 64. A plank of mass M is placed over smooth inclined plane and a sphere is also placed over the plank. Friction is sufficient between sphere and plank. If plank and sphere are released from rest, the frictional force on sphere is : (A) up the plane (C) horizontal Sol. (B) down the plane (D) zero 63. Portion AB of the wedge shown in figure is rough and BC is smooth. A solid cylinder rolls without slipping from A to B. The ratio of translational kinetic energy to rotational kinetic energy, when the cylinder reaches point C is : A B D (A) 3/4 (B) 5 AB=BC (C) 7/5 C (D) 8/3 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 54 ROTATIONAL DYNAMICS 65. A plank with a uniform sphere placed on it rests on a smooth horizontal plane. Plank is pulled to right by a constant force F. If sphere does not slip over the plank. Which of the following is incorrect. Sol. F (A) Acceleration of the centre of sphere is less than that of the plank (B) Work done by friction acting on the sphere is equal to its total kinetic energy. (C) Total kinetic energy of the system is equal to work done by the force F (D) None of the above Sol. 67. A uniform sphere of radius R is placed on a rough horizontal surface and given a linear velocity v0 angular velocity 0 as shown. The sphere comes to rest after moving some distance to the right. It follows that : v0 0 (A) v0 = 0R (C) 5v0 = 2 0R Sol. (B) 2v0 = 5 0R (D) 2v0 = 0R 66. A ring of mass m and radius R has three particles attached to the ring as shown in the figure. The centre of the ring has speed v0. The kinetic energy of the system is (Slipping is absent) m 2m (A) 6mv02 (C) 4 mv 02 m (B) 12 mv02 (D) 8 mv02 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS Exercise - III (A) MOMENT OF INERTIA 1. Find the moment of inertia of a uniform halfdisc about an axis perpendicular to the plane and passing through its centre of mass. Mass of this disc is M and radius is R. Sol. Page # 55 (JEE ADVANCED) 3. Find the radius of gyration of a circular ring of radius r about a line perpendicular to the plane of this ring and tangent to the ring. Sol. 2. Find the moment of inertia of a pair of solid spheres, each having a mass m and radius r, kept in contact about the tangent passing through the point of contact. Sol. 4. Moment of inertial of a triangle plane of mass M shown in figure about vertical axis AB is : A l 45° l m B Sol. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 56 ROTATIONAL DYNAMICS 5. A uniform rod of mass m is bent into the form of a semicircle of radius R. The moment of inertia of the rod about an axis passing through A and perpendicular to the plane of the paper is A R 7. Two forces F1 2i – 5 j – 6k and F2 – i 2 j – k are acting on a body at the points (1, 1, 0) and (0, 1, 2). Find torque acting on the body about point (–1, 0, 1). Sol. Sol. (B) TORQUE & PURE ROTATIONAL MOTION 6. A simple pendulum of length is pulled aside to made an angle with the vertical. Find the magnitude of the torque of the weight w of the bob about the point of suspension. When is the torque zero ? Sol. 8. Assuming frictionless contacts, determine the magnitude of external horizontal force P applied at the lower end for equilibrium of the rod. The rod is uniform and its mass is 'm'. Wall P Sol. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS Page # 57 9. A rod of mass m and length L, lying horizontally, is free to rotate about a vertical axis through its centre. A horizontal force of constant magnitude F acts on the rod at a distance of L/4 from the centre. The force is always perpendicular to the rod. Find the angle rotated by the rod during the time t after the motion starts. Sol. 11. Figure shows two blocks of mass m and m connected by a string passing over a pulley. The horizontal table over which the mass m slides is smooth. The pulley (uniform disc) has mass m and it can freely rotate about this axis. Find the acceleration of the mass m assuming that the string does not slip on the pulley. m m m Sol. 10. The uniform rod AB of mass m is released from rest when = 60°. Assuming that the friction force between end A and the surface is large enough to prevent sliding, determine (for the instant just after release) B L A 12. A solid cylinder of mass M = 1kg & radius R = 0.5m is pivoted at its centre & has three particles of mass m = 0.1 kg mounted at its perimeter as shown in the figure. The system is originally at rest. Find the angular speed of the cylinder, when it has swung through 90° in anticlockwise direction. (a) The angular acceleration of the rod (b) The normal reaction and the friction force at A. (c) The minimum value of , compatible with the described motion. Sol. Sol. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 58 13. A cube is in limiting equilibrium on an inclined plane forming an angle of 30° with the horizontal. The line of action of the normal reaction of the plane on the cube is Sol. ROTATIONAL DYNAMICS 15. An inverted “V” is made up of two uniform boards each weighing 200 N. Each side has the same length and makes an angle 30° with the vertical as shown in figure. The magnitude of the static frictional force that acts on each of the lower end of the V is P l 30°30° Sol. 14. A body weighs 6 gms when placed in one pan and 24 gms when placed on the other pan of a false balance. If the beam is horizontal when both the pans are ampty, the true weight of the body is : Sol. 16. A uniform sphere of weight W and radius 5 cm is being held by a string as shown in the figure. The wall is smooth. The tension in the string will be 8cm Sol. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS 17. A light string is wrapped around a cylinder of mass ‘m and radius ‘R’. The string is pulled vertically upward to prevent the centre of mass from falling as the cylinder unwinds the string. Then length of the string unwound when the cylinder has reached a speed will be : Sol. Page # 59 20. A rectangular plate of mass 20 kg is suspended from points A and B as shown. If pin B is removed determine the initial angular acceleration (in rad/ s2) of plate. (g = 10m/s2) A B 0.15m Sol. 0.2m 18. The moment of inertia of the pulley system as shown in the figure is 4 kgm2. The radii of bigger and smaller pulleys 2m and 1m respectively. The angular acceleration of the pulley system is 1m 2m 4kg Sol. 5kg 21. A solid homogeneous cylinder of height h and base radius r is kept vertically on a conveyer belt moving horizontally with an increasing velocity v = a + bt2. If the cylinder is not allowed to slip find the time when the cylinder is about to topple. Sol. 19. The two small spheres each have a mass of 3 kg and are attached to the rod of negligible mass. A torque M = 8t Nm, where t is in seconds is applied to the rod. Find the value of time when each sphere attains a speed of 3 m/s starting from rest. 3kg 3kg 1m 1m Sol. M : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 60 ROTATIONAL DYNAMICS 22. A square frame made up of a wire of mass m & length l is held in horizontal plane. It is free to rotate about AD. If the frame is released, the work done by gravity during the time frame rotates through an angle of 90° is equal to P D A C Sol. 24. In the figure A & B are two blocks of mass 4 kg & 2 kg respectively attached to the two ends of a light string passing over a disc C of mass 40 kg and radius 0.1m. The disc is free to rotate about a fixed horizontal axes, coinciding with its own axis. The system is released from rest and the string does not slip over the disc. Find : B A B (i) the linear acceleration of mass B. (ii) the number of revolutions made by the disc at the end of 10 sec. from the start. (iii) the tension in the string segment supporting the block A. Sol. 23. Three equal masses m are rigidly connected to each other by massless rods of length l forming an equilateral triangle, as shown above. The assembly is to be given an angular velocity about an axis perpendicular to the triangle. For fixed , the ratio of the kinetic energy of the assembly for an axis through B compared with that for an axis through A is equal to m l l A Sol. Bm l m Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS Page # 61 25. A mass m is attached to a pulley through a cord as shown in the fig. The pulley is a solid disk with radius R. The cord does not slip on the disk. The mass is released from rest at a height h from the ground and at the instant the mass reaches the ground, the disk is rotating with angular velocity . Find the mass of the disk. 27. A particle having mass 2 kg is moving with velcoity ( 2 i 3 j )m / s . Find angular momentum of the particle about origin when it is at (1, 1, 0). Sol. R m h Sol. (C) ANGULAR MOMENTUM 28. A uniform square plate of mass 2.0 kg and edge 10 cm rotates about one of its diagonals under the action of a constant torque of 0.10 N.m. Calculate the angular momentum and the kinetic energy of the plate at the end of the fifth second after the start. Sol. 26. A particle having mass 2 kg is moving along straight line 3x+ 4 y = 5 with speed 8m/s. Find angular momentum of the particle about origin, x and y are in meters. Sol. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 62 ROTATIONAL DYNAMICS 31. Two identical disks are positioned on a vertical axis. The bottom disk is rotating at angular velocity 0 and has rotational kinetic energy KE0. The top disk is initially at rest. It is allowed to fall, and sticks to the bottom disk. What is the rotational kinetic energy of the system after the collision? 29. A wheel of moment of inertia 0.500 kg-m2 and radius 20.0 cm is rotating about its axis at an angular speed of 20.0 rad/s. It picks up a stationary particle of mass 200 g at its edge. Find new angular speed of the wheel. Sol. 30. A uniform circular disc can rotate freely about a rigid vertical axis through its centre O. A man stands at rest at A on the edge due east of O. The mass of the disc is 22 times the mass of the man. The man starts walking anticlockwise. When he reaches the point A after completing one rotation relative to the disc he will be : Sol. 0 Sol. 32. A uniform ring is rotating about vertical axis with angular velocity initially. A point insect (S) having the same mass as that of the ring starts walking from the lowest point P1 and finally reaches the point P2 (as shown in figure). The final angular velocity of the ring will be equal to axis of rotation O 90° P1 P2 S Sol. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS 33. A particle of mass 10 kg is moving with a uniform speed of 6m/sec. in x-y plane along the line 3y = 4x+ 10 the magnitude of its angular momentum about the origin in kg –m2/s is Sol. (D) COMBINED TRANSLATIONAL + ROTATIONAL MOTION 34. A sphere of mass m rolls on a plane surface. Find its kinetic energy at an instant when its centre moves with speed v. Sol. Page # 63 35. A cylinder rolls on a horizontal plane surface. If the speed of the centre is 25 m/s, what is the speed of the highest point ? Sol. 36. A small spherical ball is released from a point at a height h on a rough track shown in figure. Assuming that it does not slip anywhere, find its linear speed when it rolls on the horizontal part of the track. h Sol. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 64 ROTATIONAL DYNAMICS 39. Two small spheres A & B respectively of mass m & 2m are connected by a rigid rod of length & negligible mass. The two spheres are resting on a horizontal, frictionless surface. When A is suddenly given the velocity v0 as shown. Find velocities of A & B after the rod has rotated through 180°. A 37. A sphere starts rolling down an incline of inclination . Find the speed of its centre when it has covered a distance . Sol. v0 B Sol. 38. A solid uniform sphere of mass m is released from rest from the rim of a hemispherical cup so that it rolls without sliding along the surface. If the rim of the hemisphere is kept horizotnal, find the normal force exerted by the cup on the ball when the ball reaches the bottom of the cup. Sol. 40. A uniform rod of mass m and length is struck at an end by a force F perpendicular to the rod for a short time interval t. Calculate (a) the speed of the centre of mass, (b) the angular speed of the rod about the centre of mass, (c) the kinetic energy of the rod and (d) the angular momentum of the rod about the centre of mass after the force has stopped to act. Assume that t is so small that the rod does not appreciably change its direction while the force acts. Sol. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS Page # 65 42. The cylinder shown, with mass M and radius R, has a radially dependent density. The cylinder starts from rest and rolls without slipping down an inclined plane of height H. At the bottom of the plane of height H. At the bottom of the plane its translational speed is (8gH/7)1/2. Which of the following is the rotational inertia of the cylinder? R M H 41. A hollow cylinder with inner radius R, outer radius 2R mass M is rolling with speed of its axis v. Its kinetic energy is Sol. R Sol. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 66 ROTATIONAL DYNAMICS 1. A thin uniform rod of mass M and length L is hinged at its upper end, and released from rest in a horizontal position. The tension at a point located at a distance L/3 from the hinge point, when the rod becomes vertical, will be 2. A rigid horizontal smooth rod AB of mass 0.75 kg and length 40 cm can rotate freely about a fixed vertical axis through its mid point O. Two rings each of mass 1 kg are initially at rest a distance of 10 cm from O on either side of the rod. The rod is set in rotation with an angular velocity of 30 radians per second. The velocity of each ring along the length of the rod in m/s then they reach the ends of the rod is A D C R/2 7. A slightly loosely fit window is balanced by two strings which are connected to weights w/2 each. The strings pass over the frictionless pulleys as shown in the figure. The strings are tied almost at the corner of the window. The string on the right is cut and then the window accelerates downwards. If the coefficients of friction between the window and the side supports is then calculate the acceleration of the window in terms of , a, b and g, where a is width and b is the length of the window. B O 3. A straight rod AB of mass M and length L is placed on a frictionless horizontal surface. A horizontal force having constant magnitude F and a fixed direction starts acting at the end A. The rod is initially perpendicular to the force. The initial acceleration of end B is 4. A wheel is made to roll without slipping, towards right, by pulling a string wrapped around a coaxial spool as shown in figure. With what velocity the string should be pulled so that the centre of the wheel moves with a velocity of 3 m/s? 0.3m w/2 w b w/2 a fixed window support 8. A uniform wood door has mass m, height h, and width w. It is hanging from two hinges attached to one side; the hinges are located h/3 and 2h/3 from the bottom of the door. Suppose that m = 20.0 kg, h = 2.20 m, and w = 1.00 m and the bottom smooth hinge is not screwed into the door frame. Find the forces acting on the door. W C 0.1m B A String 5. A solid uniform disk of mass m rolls without slipping down a fixed inclined plane with an acceleration a. The frictional force on the disk due to surface of the plane is : 6. A carpet of mass ‘M’ made of inextensible material is rolled along its length in the form of a cylinder of radius ‘R’ and is kept on a rough floor. The carpet starts unrolling without sliding on the floor when a negligibly small push is given to it. The horizontal velocity of the axis of the cylindrical part of the carpet when its radius reduces to R/2 will be : com Hinges h 9. A hole of radius R/2 is cut from a solid sphere of radius R. If the mass of the remaining plate is M, then moment of inertia of the body about an axis through O perpendicular to plan e is _________. R O R/2 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS Page # 67 10. A uniform beam of length L and mass m is supported as shown. If the cable suddenly breaks, determine ; (1/4)L A B 14. A spool of inner radius R and outer radius 3R has a moment of inertia = MR2 about an axis passing through its geometric centre, where M is the mass of the spool. A thread woudn on the inner surface of the spool is pulled horizontally with a constant force = Mg. Find the acceleration of the point on the thread which is being pulled assuming that the spool rolls purely on the floor. L (a) the acceleration of end B. (b) the reaction at the pin support. 11. A thin rod AB of length a has variable mass x per unit length 0 1 where x is the distance a measured from A and 0 is a constant. (a) Find the mass M of the rod. (b) Find the position of centre of mass of the rod. (c) Find moment of inertia of the rod about an axis passing through A and perpendicular to AB. Rod is freely pivoted at A and is hanging in equilibrium when it is struck by a horizontal impulse of magnitude P at the point B. (d) Find the angular velocity with which the rod begins to rotate. (e) Find minimum value of impulse P if B passes through a point vertically above A. 12. Two separate cylinders of masses m (= 1kg) and 4m and radii R(=10cm) and 2R rotating in clockwise direction with 1 = 100 rad/sec and 2 = 200 rad/sec. Now they are held in contact with each other as in fig. Determine their angular velocities after the slipping between the cylinders stops. 15. A sphere of mass m and radius r is pushed onto the fixed horizontal surface such that it rolls without slipping from the beginning. Determine the minimum speed v of its mass centre at the bottom so that it rolls completely around the loop of radius (R + r) without leaving the track in between. (R+r) Sphere r V 16. Two uniform cylinders, each of mass m = 10 kg and radius r = 150 mm, are connected by a rough belt as shown. If the system is released from rest, determine r r 13. A circular disc of mass 300 gm and radius 20 cm can rotate freely about a vertical axis passing through its centre of O. A small insect of mass 100 gm is initially at a point A on the disc (which is initially stationary) the insect starts walking from rest along the rim of the disc with such a time varying relative velocity that the disc rotates in the opposite direction with a constant angular acceleration = 2 rad/s2. After some time T, the insect is back at the point A. By what angle has the disc rotated till now ; as seen by a stationary earth observer ? Also find the time T. (a) the velocity of the centre of cylinder A after it has moved through 1.2 m & (b) the tension in th e portion of the b elt connecting the two cylinders. 17. A uniform rod of mass m and length l is resting on a smooth horizontal surface. A particle of mass m/2 travelling with a speed v0 hits the rod normally and elastically. Find final velocity of particle and the angular velocity of the rod. l/4 C Rod (m, l) v0 m/2 Top view : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 68 ROTATIONAL DYNAMICS 18. One side of a spring of initial, unstretched length l0 = 1m, lying on a frictionless table, is fixed, the other one is fastened to a small puck of mass m = 0.1kg. The puck is given velocity in a direction perpendicular to the spring, at an initial speed v0 = 11 m/s. In the course of the motion, the maximum elongation of the spring is l = l0/10. What is the force constant of the spring (in SI units) ? v0 m l0 19. A block X of mass 0.5 kg is held by a long massless string on a frictionless inclined plane of inclination 30º to the horizontal. The string is wound on a uniform solid cylindrical drum Y of mass 2kg and of radius 0.2 m as shown in the figure. The drum is given an initial angular velocity such that the block X starts moving up the plane. Y X (i) Find the tension in the string during the motion (ii) At a certain instant of time the magnitude of the angular velocity of Y is 10 rad/sec. Calculate the distance travelled by X from that instant of time until it comes to rest. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS Exercise - IV Page # 69 PREVIOUS YEAR QUESTIONS JEE MAIN LEVEL - I 1. A solid sphere, a hollow sphere and a ring are released from top of an inclined plane (frictionless) so that they slide down the plane. Then maximum acceleration down the plane is for (no rolling) [AIEEE 2002] (A) solid sphere (B) hollow sphere (C) ring (D) All same Sol. 3. A particle of mass m moves along line PC with veloc ity v as shown . What is th e angular momentum of the particle about O ? [AIEEE 2002] C L P r l (A) mvL Sol. 2. Moment of inertia of a circular wire of mass M and radius R about its diameter is [AIEEE 2002] MR 2 (A) 2 Sol. (B) MR2 (C) 2MR2 MR2 (D) 4 O (B) mvl (C) mvr (D) zero 4. Initial angular velocity of a circular disc of mass M is 1. Then two small spheres of mass m are attached gently to two diametrically opposite points on the edge of the disc. What is the final angular velocity of the disc ? [AIEEE 2002] (A) (C) M m M M M 4m m m M (D) M 2m (B) 1 1 M Sol. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com 1 1 Page # 70 5. Let ROTATIONAL DYNAMICS F be the force acting on a particle having position vector r and be the torque of this force about the origin. Then (A) r . 0 and F. 0 (B) r . 0 and F. 0 (C) r . 0 and F. 0 (D) r . Sol. 0 and F. 0 [AIEEE 2003] 8. A solid sphere is rotating in free space. If the radius of the sphere is increased keeping mass same, which one of the following will not be affected ? [AIEEE 2004] (A) Moment of inertia (B) Angular momentum (C) Angular velocity (D) Rotational kinetic energy Sol. 6. A particle performing uniform circular motion has angular momentum L. If its angular frequency is doubled and its kinetic energy halved, then the new angular momentum is [AIEEE 2003] (A) L 4 (B) 2L Sol. (C) 4L (D) L/2 Sol. 9. One solid sphere A and another hollow sphere B are of same mass and same outer radii. Their moment of inertia about their diameters are respectively IA and IB such that [AIEEE 2004] (A) IA IB (B) IA IB (C) IA IB (D) IA IB dA dB Where d A and dB are their densities. Sol. 7. A circular disc X of radius R is made from an iron plate of thickness t, and another disc y of radius 4R is made from an iron plate of thickness t/4. Then the relation between the moment of inertia Ix and IY is [AIEEE 2003] (A) IY 32I x (B) IY 16 I X (C) IY IX (D) IY 64 I X Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS Page # 71 10. A T shaped object with dimensions shown in the figure, is lying on a smooth floor. A force F is applied at the point P parallel to AB, such that the object has only the translational motion without rotation. Find the location of P with respect to C. [AIEEE 2005] l A B 12. An angular ring with inner and outer radii R1 and R 2 is rolling without slipping with a uniform angular speed. The ratio of the forces experienced by the two particles situated on the inner and outer parts of the ring, R2 (A) R1 R1 R2 (B) F1 F2 is [AIEEE 2005] 2 (C) 1 (D) R1 R2 Sol. P 2l F C 2 (A) l 3 3 (B) l 2 (C) 4 l 3 (D) l Sol. 13. A thin circular ring of mass m and radius R is rotating about its axis with a constant angular velocity . Two objects each of mass M are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity ' = [AIEEE 2006] ( m 2M ) m (A) 11. The moment of inertia of uniform semicircular dis c of mass M and rad ius r about a line perpendicular to the plane of the disc through the centre is [AIEEE 2005] (A) 1 Mr 2 4 (B) 2 Mr 2 5 (C) Mr 2 (D) (C) (B) m (m M ) (D) (m 2 M ) (m 2M ) m ( m 2M ) Sol. 1 Mr 2 2 Sol. 14. Four point masses, each of value m, are placed at the corners of a square ABCD of side l. The moment of inertia of this system about an axis passing through A and parallel to BD is [AIEEE 2006] (A) 2ml 2 (B) : 0744-2209671, 08003899588 | url : www.motioniitjee.com, 3ml 2 (C) 3ml 2 (D) ml 2 :info@motioniitjee.com Page # 72 ROTATIONAL DYNAMICS 17.For the given uniform square lamina ABCD, whose centre is O [AIEEE 2007] Sol. D F C O A 15. Angular momentum of the particle rotating with a central force is constant due to [AIEEE 2007] (A) constant force (B) constant linear momentum (C) zero torque (D) constant torque Sol. 16. A round uniform body of radius R, mass M and moment of inertia I, rolls down (without slipping) an inclined plane making an angle with the horizontal. Then its acceleration is [AIEEE 2007] (A) g sin 1 I / MR 2 (B) g sin 1 MR2 / I (C) g sin 1 I / MR 2 (D) g sin 1 MR2 / I (A) (C) 2I AC I AD I EF 4I EF B E (B) I AD (D) I AD 3I EF 2I EF Sol. 18. Consider a uniform square plate of side a and mass m. The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is [AIEEE 2008] (A) 5 1 7 2 ma 2 (B) ma 2 (C) ma 2 (D) ma 2 6 12 12 3 Sol. Sol. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS Page # 73 19. A thin uniform rod of length l and mass m is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is . Its centre of mass rises to maximum height of [AIEEE 2009] 1 l2 2 (A) 3 g (B) 1l 6 g 1 l2 2 2 g (D) 1 l2 2 6 g (C) Sol. 20. A pulley of radius 2 m is rotated about its axis by a force F = (20t – 5t2) N (where t is measured in seconds) applied tangentially. It the moment of inertia of the pulley about its axis of rotation is 10 kg-m2 the number of rotations made by the pulley before its direction of motion if reserved, is [AIEEE 2011] (A) more than 3 but less than 6 (B) more than 6 but less than 9 (C) more than 9 (D) less than 3 Sol. 21. A thin horizontal circular disc is rotating about a vertical axis passing through its centre. An insect is at rest at a point near the rim of the disc. The insect now moves along a diameter of the disc to reach its other end. During the journey of the insect, the angular speed of the disc. [AIEEE 2011] (A) continuously decreases (B) continuously increases (C) first increases and then decreases (D) remains unchanged Sol. 22. A hoop of radius r and mass m rotating with an angular velocit y 0 is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. What will be the velocity of the centre of the hoop when it cases to slip ? [JEE Mains 2013] (1) r 0 2 (2) r 0 (3) r 0 4 (4) Sol. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com r 0 3 Page # 74 ROTATIONAL DYNAMICS JEE ADVANCED LEVEL - II 1. Three particles A, B and C, each of mass m, are connected to each other by three massless rigid rods to form a rigid, equilateral triangular body of side l. This body is placed on a horizontal frictionless table (x-y plane) and is hinged to it at the point A so that it can move without friction about the vertical axis through A (see figure). The body is set into rotational motion on the table about A with a constant angular velocity . y 2. A particle is moving in a horizontal uniform circular motion. The angular momentum of the particle is conserved about the point : [JEE’(Scr) 2003] (A) Centre of the circle (B) Outside the circle (C) Inside the circle (D) Point on circumference Sol. A x F B C [JEE’(Scr) 2002] (a) Find the magnitude of the horizontal force exerted by the hinge on the body (b) At time T, when the side BC is parallel to the x-axis, a force F is applied on B along BC (as shown). Obtain the x-component and the ycomponent of the force exterted by the hinge on the body, immediately after time T. Sol. l 3. Two particles each of mass M are connected by a massless rod of length l. The rod is lying on the smooth surface. If one of the particle is given an impulse MV as shown in the figure then angular velocity of the rod would be Mv M (A) v/l (B) 2v/l (C) v/2 l (D) none [JEE’(Scr) 2003] Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS Page # 75 Sol. 5. A child is standing with folded hands at the centre of a platform rotating about its central axis. The kinetic energy of the system is K. The child now stretches his arms so that the moment of inertia of the system doubles. The kinetic energy of the system now is [JEE’(Scr) 2004] (A) 2K (B) K/2 (C) K/4 (D) 4K Sol. 4. A disc is rolling (without slipping) on a horizontal surface. C is its center and Q and P are two points equidistant from C. Let Vp, VQ and VC be the magnitude of velocities of points P, Q and C respectively, then Q C P (A) VQ > VC > VP (C) VQ = Vp, VC (B) VQ < VC < VP 1 VP 2 6. A block of mass m is held fixed against a wall by a applying a hor izontal force F. Which of the following option is incorrect : (D) VQ < VC > VP F [JEE’(Scr) 2004] a 2a Sol. 2a (A) friction force = mg (B) F will not produce torque (C) normal will not produce torque (D) normal reaction = F [JEE’(Scr) 2005] Sol. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 76 ROTATIONAL DYNAMICS 7. A disc has mass 9m. A hole of radius R/3 is cut from it as shown in the figure. The moment of inertia of remaining part about an axis passing through the centre ‘O’ of the disc and perpendicular to the plane of the disc is : R/3 Sol. 2R/3 O R (A) 8 mR2 (C) 40 mR2 9 (B) 4 mR2 (D) 37 mR2 9 [JEE’(Scr) 2005] Sol. 9. A wooden log of mass M and length L is hinged by a frictionless nail at O. A bullet of mass m strikes with velocity v and sticks to it. Find angular velocity of the system immediately after the collision about O. O M L m v [JEE’ 2005] Sol. 8. A particle moves in circular path with decreasing speed. Which of the following is correct (A) L is constant (B) only direction of L is constant (C) acceleration a is towards the centre (D) it will move in a spiral and finally reach the centre [JEE’(Scr) 2005] Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS Page # 77 10. A cylinder of mass m and radius R rolls down an inclined plane of inclination . Calculate the linear acceleration of the axis of cylinder. [JEE’ 2005] Sol. Sol. 12. A solid sphere of mass M, radius R and having moment of inertia about an axis passing through the centre of mass as I, is recast into a disc of thickness t, whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains I. Then, radius of the disc will be [JEE’ 2006] (A) 2R / 15 (B) R 2 / 15 (C) 4R / 15 Sol. (D) R/4 11. Two identical ladders, each of mass M and length L are resting on the rough horizontal surface as shown in the figure. A block of mass m hangs from P. If the system is in equilibrium, find the magnitude and the direction of frictional force at A and B. [JEE’ 2005] P L A m B : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 78 ROTATIONAL DYNAMICS 13. A solid cylinder of mass m and radius r is rolling on a rough inclined plane of inclination . The coefficient of friction between the cylinder and incline is . Then [JEE’ 2006] (A) frictional force is always mg cos (B) friction is a dissipative force (C) by decreasing , frictional force decreases (D) friction opposes translation and supports rotation Sol. Sol. 15. There is a rectangular plate of mass M kg of dimensions (a × b). The plate is held in horizontal position by striking n small balls each of mass m per unit area per unit time. These are striking in the shaded half region of the plate. The balls are colliding elastically with velocity v. What is v ? [JEE’ 2006] b a It is given n = 100, M = 3 kg, m = 0.01 kg; b = 2 m, a = 1m; g = 10 m/s2. Sol. 14. A ball moves over a fixed track as shown in the figure. From A to B the ball rolls without slipping. Surface BC is frictionless. KA, KB and KC are kinetic energies of the ball at A, B and C, respe0ctively. Then [JEE’ 2006] A C hA B (A) hA > hC ; KB > K C (C) hA = h C ; KB = K C hC (B) hA > hC ; K C > KA (D) hA < hC ; KB > KC Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS Paragraph Q.16 to Q.18 (3 questions) Two discs A and B are mounted coaxially on a vertical axle. The discs have moments of inertia I and 2I respectively about the common axis. Disc A is imparted an initial angular velocity 2 using the entire potential energy of a spring compressed by a distance x 1. Disc B is imparted an angular velocity by a spring having the same spring constant and compressed by a distance x 2. Both the discs rotate in the clockwise direction. 16. The ratio x 1/x2 is [JEE’ 2007] (A) 2 (B) 1/2 (C) 2 (D) 1/ 2 Sol. Page # 79 Sol. 18. The loss of kinetic energy during the above process is [JEE’ 2007] (A) I 2 /2 (B) I 2 /3 (C) I 2 /4 (D) I 2 /6 Sol. 17. When disc B is brought in contact with disc A, they acquire a common angular velocity in time t. The average frictional torque on one disc by the other during this period is [JEE’ 2007] (A) 2I /(3t) (B) 9I /(2t) (C) 9I /(4t) (D) 3I /(2t) : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 80 ROTATIONAL DYNAMICS 19. A small object of uniform density rolls up a curved surface with an initial velocity v. It reaches up to a maximum height of 3v2 / (4g) with respect to the initial position. The object is [JEE’ 2007] Sol. v (A) ring (C) hollow sphere Sol. (B) solid sphere (D) disc 20. STATEMENT-1 If there is no external torque on a body about its center of mass, then the velocity of the center of mass remains constant because STATEMENT-2 The linear momentum of an isolated system remains constant. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1 (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True [JEE 2007] 21. STATEMENT-1 Two cylinders, one hollow (metal) and the other solid (wood) with the same mass and identical dimensions are simultaneously allowed to roll without slipping down an inclined plane from the same height. The hollow cylinder will reach the bottom of the inclined plane first. STATEMENT-2 By the principle of conservation of energy, the total kinetic energies of both the cylinders are identical when they reach the bottom of the incline. (A) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1 (B) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1 (C) STATEMENT-1 is True, STATEMENT-2 is False (D) STATEMENT-1 is False, STATEMENT-2 is True [JEE-2008] Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS Sol. Page # 81 23. A sphere is rolling without slipping on a fixed horizontal plane surface. In the figure A is the point of contact, B is the centre of the sphere and C is its topmost point Then, [JEE 2009] C B A (A) VC – VA 2( VB – VC ) (B) VC – VB VB – VA (C) | VC – VA | 2| VB – VC | (D) | VC – VA | 4| VB | Sol. 22. If the resultant of all the external forces acting on a system of particles is zero, then from an inertial frame, one can surely say that [JEE 2009] (A) linear momentum of the system does not change in time (B) kinetic energy of the system does not change in time (C) angular momentum of the system does not change in time (D) potential energy of the system does not change in time Sol. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 82 ROTATIONAL DYNAMICS 24. A boy is pushing a ring of mass 2 kg and radius 0.5 m with a stick as shown in the figure. The stick applies a force of 2 N on the ring and rolls it without slipping with an acceleration of 0.3 m/s2. The coefficient of friction between the ground and ring is large enough that rolling always occurs and the coefficient of friction between the stick and the ring is (P/10). The value of P is? stick [JEE 2011] 25. A thin uniform rod, pivoted at O is rotating in the horizontal plane with constant angular speed , as shown in the figure. At time t = 0, small insect starts from O and moves with constant speed with respect to the rod towards the other end. it reaches the end of the rod at t = T and stops. The angular speed of the system remains throughout. The magnitude of the torque on the system about O, as a function of time is best represented by which plot? Z O Ground Sol. (A) (B) 0 T t (C) 0 T t (D) 0 T t 0 t T [JEE 2012] Sol. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS Page # 83 26. A small mass m is attached to a massless string whose other end is fixed at P as shown in the figure. The mass is undergoing circular motion in the x-y plane with centre at O and constant angular speed .If the angular momentum of the system, calculated about O and P are denoted by L0 and LP respectively, then. z 27. A lamina is made by removing a small disc of diameter 2R from a bigger disc of uniform mass density and radius 2R, as shown in the figure. The moment of inertia of this lamina about axes passing through O and P is Io and IP, respectively. Both these axes are perpendicular to the IP plane of the lamina. The ratio to the nearest Io integer is P O m (A) L0 and LP do not vary with time (B) L 0 varies with time while LP remains constant (C) L 0 remains constant while LP varies with time (D) L0 and LP both vary with time. [JEE 2012] Sol. [JEE 2012] Sol. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 84 ROTATIONAL DYNAMICS 28. Consider a disc rotating in the horizontal plane with a constant angular speed about its centre O. The disc has a shaded region on one side of the diameter and an unshaded region on the other side as shown in the figure. When the disc is in the orientation as shown, two pebbles P and Q are simultaneously projected at an angle towards R. The velocity of projection is in the y-z plane and is same for both pebbles with respect to the disc. Assume that (i) they land back on the disc before the disc has completed 1/8 rotation, (ii) their range is less than half the disc radius, and (iii) remains constant throughout. Then R y x Q O P (A) P lands in the shaded region and Q in the unshaded region (B) P lands in the unshaded region and Q in the shaded region (C) Both P and Q land in the unshaded region (D) Both P and Q land in the shaded region [JEE 2012] Sol. Paragraph for Question Nos. 29 to 30 The general motion of a rigid body can be considered to be a combination of (i) a motion of its centre of mass about an axis, and (ii) its motion about an instantaneous axis passing through the centre of mass . These axes need not be stationary. Consider, for example, a thin uniform disc welded (rigidly fixed) horizontally at its rim to a massless stick, as shown in the figure. When the disc-stick system is rotated about the origin on a horizontal frictionless plane with angular speed , the motion at any instant can be taken as a combination of (i) a rotation of the centre of mass of the disc about the z-axis, and (ii) a rotation of the disc through an instantaneous vertical axis passing through its centre of mass (as is seen from the changed orientation of points P and Q). Both these motions have the same angular speed in this case. Now consider two similar systems as shown in the figure: Case (a) the disc with its face vertical and parallel to x-z plane; case (b) the disc with its face making an angle of 45o with x-y plane and its horizontal diameter parallel to x-axis. In both the cases, the disc is welded at point P, and the systems are rotated with constant angular speed about the z-axis. Case (a) Case (b) 29. Which of the following statements about the instantaneous axis (passing through the centre of mass) is correct ? (A) It is vertical for both the cases (a) and (b). (B) It is vertical for case (a); and is at 45o to the x-z plane and lies in the plane of the disc for case (b). (C) It is horizontal for case (a); and is at 45o to the x-z plane and is normal to the plane of the disc for case (b). (D) It is vertical for case (a); and is at 45o to the x-z plane and is normal to the plane of the disc for case (b). [JEE 2012] Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS Page # 85 Sol. Sol. 31. The figure shows a system consisting of (i) a ring of outer radius 3R rolling clockwise without slipping on a horizontal surface with angular speed and (ii) an inner disc of radius 2R rotating anticlockwise with angular speed 2. The ring and disc are separated by frictionless ball bearings. The system is in the x-z plane. The point P on the inner disc is at a distance R from the origin, where OP makes an angle of 30 o with the horizontal. Then with respect to the horizontal surface. (A) the point O has a linear velocity 3R î (B) th e poin t P has a li near v el oc it y 30. Which of the following statements regarding the angular speed about the instantaneous axis (passing through the centre of mass) is correct (A) It is 2 for both the cases. (B) It is for case (a); and (C) It is for case (a); and (D) It is for both the cases. 2 2 for case (b). for case (b). 11 ˆ R i 4 3 ˆ R k 4 (C) the poin t P has a li near v el oc it y 13 3 R î R k̂ 4 4 (D) th e poin t P has a li near v el oc it y 3 3 R î 4 1 R k̂ 4 [JEE 2012] : 0744-2209671, 08003899588 | url : www.motioniitjee.com, [JEE 2012] :info@motioniitjee.com Page # 86 Sol. ROTATIONAL DYNAMICS 33. A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angular velocity of 10 rad s-1 about its own axis, which is vertical. Two uniform circular rings, each of mass 6.25 kg and radius 0.2 m, are gently placed symmetrically on the disc in such a manner that they are touching each other along the axis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at rest relative to the disc and the system rotates about the original axis. The new angular velocity (in rad s-1) of the system is : [JEE 2013] Sol. 32. Two solid cylinders P and Q of same mass and same radius start rolling down a fixed inclined plane from the same height at the same time. Cylinder P has most of its mass concentrated near its surface, while Q has most of its mass concentrated near the axis. Which statement(s) is(are) correct? (A) Both cylinders P and Q reach the ground at the same time. (B) Cylinder P has larger linear acceleration than cylinder Q. (C) Both cylinders reach the ground with same translational kinetic energy. (D) Cylinder Q reaches the ground with larger angular speed. [JEE 2012] Sol. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS Page # 87 ANSWER KEY OBJECTIVE PROBLEMS (JEE MAIN) Exercise - I 1. D 2. B 3. A 4. D 5. A 6. B 7. A 8. C 9. A 10. C 11. D 12. B 13. D 14. B 15. C 16. D 17. B 18. B 19. A 20. D 21. D 22. D 23. A 24. D 25. A 26. D 27. B 28. C 29. A 30. B 31. B 32. A 33. C 34. C 35. D 36. C 37. B 38. B 39. B 40. C 41. B 42. D 43. B 44. B 45. A 46. D 47. D 48. B 49. A 50. C 51. C 52. D 53. D 54. C 55. C 56. C 57. A 58. D 59. C 60. A 61. D 5. B 6. C Exercise - II 1. B 2. C 3. C 4. A 7. A 8. C 9. C 10. ABCD 11. B 12. B 13. A 14. C 15. D 16. BC 17. AD 18. BCD 19. ABD 20. ABC 21. ACD 22. C 23. B 24. ACD 25. B 26. C 27. A 28. D 29. B 30. B 31. C 32. B 33. D 34. A 35. C 36. B 37. ACD 38. C 39. BCD 40. A 41. C 42. BC 43. AC 44. BC 45. ABCD 46. BD 47. ABC 48. ABC 49. C 50. BCD 51. ACD 52. ABD 53. B 54. A 55. B 56. C 57. A 58. C 59. D 60. C 61. B 62. B 63. B 64. D 65. D 66. A 67. C : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 88 ROTATIONAL DYNAMICS (JEE ADVANCED) Exercise - III 1. MR2 4R –M 2 3 2 2. 14mr 2 5 3. 2r 6. w sin , when the bob is at the lowest point mg cot 2 8. P= 10. (a) 3g ( cw ) 4L 11. 2g 5 9. (b) N 12. w = 100 15. 20. 48 21. gr/bh 25. M = 2m 2gh R2 2 –1 3 N Ml 2 2 7. –14 i 10 j – 9k 5. 2mR2 3Ft 2 2m 13mg ,F 16 5 rad/s 14. 12 gm 4. 3 3 mg 16 3 3 16 13. at a distance a / 2 3 from the centre down the plane. 16. 13 W / 12 22. (c) mgl 8 26.16 kg m2/s 17. R2 2 4g 23. 2 18. 2.1 rad/s2 24. (i) 10/13 m/s2, (ii) 5000/26 , (iii) 480/13 N 27. 2kkg m2 / s 28. 0.5 kg – m2 /s, 75 J 29. 19.7 rad/s 30. 60° east of south, 30° south of east. 31. (1/2)KE0 33. 120 34. 39. v0 2v ( ), 0 ( 3 3 ) 7 mv2 10 40. (a) Ft m 35. 50m/s (b) 19. 15 . 2 sec 36. 10gh 37. 7 6Ft 2F 2 t 2 F t (c) (d) m 2 m 32. 10 g sin 7 41. 13 Mv2 16 3 38. 17 mg 7 42. 3 MR2 4 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) ROTATIONAL DYNAMICS 1. 2mg Page # 89 2. 3 3. 2F/M 4. 2m/s 5. 1/2 ma 7. a b– a g 3b a 11. (a) 3 0a 5a 7a 3 0 18P M 70ag (b) , (c) , (d) , (e) 2 9 7Ma 9 12 8. Fdlx = 3mgw and Fduy = mg 2h 14. 16 m/s2 13. t = 2 / 5 sec, q = 4p/5 rad 17. – 1 v0 15 9. 57/140 MR2 10. (a) 18. 210 6. v = 9g 7 (b) 12. 300 rad/sec, 150 rad/sec 27 3 200 gR 16. (a) 4 m / s, (b) N 7 7 7 15. v = PREVIOUS YEAR QUESTIONS JEE MAIN LEVEL - I 1. D 2. A 3. B 4. C 5. D 6. A 7. D 8. B 9. C 10. C 11. D 12. D 13. D 14. C 15. C 16. A 17. C 18. D 19. D 20. A 21. C 22. A JEE ADVANCED LEVEL - II 3m 2 l, (b) Fx = F/4, Fy = 2. A 3. A 8. B 9. 3 m 4. A 3 mv ( 3m M) L 2 l 5. B 10. aaxis 6. C 7. B 2g sin 3 cot 2 12. A 13. C,D 14. A,B 16. C 17. A 18. B 19. D 20. D 21. D 22. A 23. B,C 24. 0004 25. B 26. C 27. 0003 28. C 29. A 30. D 31. A,B 32. D 33. 0008 11. f 4mg 7 19. 1.63 N, 1.224 m Exercise - IV 1. (a) 14 gR 3 (M m) g : 0744-2209671, 08003899588 | url : www.motioniitjee.com, 15. 10m/s :info@motioniitjee.com Page # 90 STOICHIOMETRY - 1 STOICHIOMETRY - 1 CLASSIFICATION OF MATTER LAWS OF CHEMICAL COMBINATION (a) Law of conservation of mass [Lavoisier] In a chemical change total mass remains conserved i.e. mass before the reaction is always equal to mass after the reaction. H2 (g) + 1/2 O2 H2O ( ) (g) 1 mole 1/2 mole 1 mole mass before the reaction = 1 × 2 + 1/2 × 32 = 18 gm mass after the reaction = 1 × 18 = 18 gm Ex. A 15.9g sample of sodium carbonate is added to a solution of acetic acid weighing 20.0g. The two substances react, releasing carbon dioxide gas to the atmosphere. After reaction, the contents of the reaction vessel weigh 29.3g. What is the mass of carbon dioxide given off during the reaction? Sol. The total mass of reactants taken = 15.9 + 20.0 = 35.9 gm. From the conservation of mass, the final mass of the contents of the vessel should also be 35.9 gm. But it is only 29.3 gm. The difference is due to the mass of released carbon dioxide gas. Hence, the mass of carbon dioxide gas released = 35.9 – 29.3 = 6.6 gm (b) Law of constant composition [Proust] All chemical compounds are found to have constant composition irrespective of their method of prepration or sources. In H2O, Hydrogen & oxygen combine in 2 : 1 molar ratio, this ratio remains constant whether it is Tap water, river water or seawater or produced by any chemical reaction. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Page # 91 STOICHIOMETRY - 1 Ex. The following are results of analysis of two samples of the same or two different compounds of phosphorus and chlorine. From these results, decide whether the two samples are from the same or different compounds. Also state the law, which will be obeyed by the given samples. Amount P Amount Cl Compound A 1.156 gm 3.971 gm Compound B 1.542 gm 5.297 gm Sol. The mass ratio of phosphorus and chlorine in compound A, m P : mCl = 1.156:3.971 = 0.2911:1.000 The mass ratio of phosphorus and chlorine in compound B, mP : mCl = 1.542:5.297 = 0.2911:1.000 As the mass ratio is same, both the compounds are same and the samples obey the law of definite proportion. (c) Law of multiple proportions [Dalton] When one element combines with the other element to form two or more different compounds, the mass of one element, which combines with a constant mass of the other bear a simple ratio to one another. Carbon is found to form two oxides which contain 42.9% & 27.3% of carbon respectively show that these figures shows the law of multiple proportion. First oxide Second oxide Carbon 42.9 % 27.3 % Oxygen 57.1 % 72.7% Given In th first oxide, 57.1 parts by mass of oxygen combine with 42.9 parts of carbon. 1 part of oxygen will combine with 42.9 part of carbon = 0.751 57.1 Similarly in 2nd oxide 27.3 part of carbon = 0.376 72.7 The ratio of carbon that combine with the same mass of oxygen = 0.751 : 0.376 = 2 : 1 This is a simple whole no ratio this means above data shows the law of multiple proportion. 1 part of oxygen will combine with Ex. Two oxide samples of lead were heated in the current of hydrogen and were reduced to the metallic lead. The following data were obtained (i) Weight of yellow oxide taken = 3.45 gm; Loss in weight in reduction = 0.24 gm (ii) Weight of brown oxide taken = 1.227 gm; Loss in weight in reduction = 0.16 gm. Show that the data illustrates the law of multiple proportion. Sol. When the oxide of lead is reduced in the current of hydrogen, metallic lead is formed. Definitely, the loss in weight in reduction is due to removal of the oxygen present in the oxide, to combine with the hydrogen. Therefore, the composition of the yellow oxide is: oxygen = 0.24 gm and lead = 3.45 – 0.24 = 3.21 gm. The mass ratio of lead and oxygen, r1 = m Pb mO 3.21 0.244 13.375 1.000 and the compositon of the brown oxide is : oxygen = 0.16 gm and lead = 1.227 – 0.16 = 1.067 gm. The mas ratio of lead and oxygen, r2 = m Pb mO 1.067 0.16 6.669 1.000 Now, r1 : r2 = 13.375 : 6.669 = 2.1 (simple ratio) and hence the data illustrates the law of multiple proportion. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 92 (d) STOICHIOMETRY - 1 Law of reciprocal proportions [Richter] When two elements combine seperately with definite mass of a third element, then the ratio of their masses in which they do so is either the same or some whole number multiple of the ratio in which they combine with each other. This law can be understood easily with the help of the following examples. At. Mass 1 H H 2S At. Mass 32 S O At. Mass 16 Let us consider three elements – hydrogen, sulphur and oxygen. Hydrogen combines with oxygen to form H 2 O whereas sulphur combines with it to form SO2 . Hydrogen and sulphur can also combine together to form H2S. The formation of these compounds is shown in fig. In H2 O, the ratio of masses of H and O is 2 : 16. In SO 2, the ratio of masses of S and O is 32 : 32. Therefore, the ratio of masses of H and S which combines with a fixed mass of oxygen (say 32 parts) will be 4 : 32 i.e. 1 : 8 ...(1) When H and S combine together, they form H 2S in which the ratio of masses of H and S is 2 : 32 i.e., 1 : 16 The two ratios (i) and (ii) are related to each other as 1 1 : 8 16 or ...(ii) 2 :1 i.e., they are whole number multiples of each other. Thus, the ratio masses of H and S which combines with a fixed mass of oxygen is a whole number multiple of the ratio in which H and S combine together. Ex. Methane contains 75 % carbon and 25% hydrogen, by mass. Carbon dioxide contains 27.27 % carbon and 72.73% oxygen, by mass. Water contains 11.11 % hydrogen and 88.89% oxygen, by mass. Show that the data illustrates the law of reciprocal proportion. Sol. Methane and carbon dioxide, both contains carbon and hence, carbon may be considered as the third element. Now, let the fixed mass of carbon = 1 gm. Then, 1 gm 3 72.73 and the mass of oxygen combined with 1 gm carbon in carbon dioxide = 27.27 the mass of hydrogen combined with 1 gm carbon in methane = 25 75 8 gm 3 Hence, the mass ratio of hydrogen and oxygen combined with the fixed mass of carbon, r1 = Now, the mass ratio of hydrogen and oxygen in water, r2 = 11.11 88.89 1 8 : 3 3 1 8 As r1 and r2 are same , the data is according to the law of reciprocal proportion. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 1 8 Page # 93 STOICHIOMETRY - 1 (e) Gay Lussac law of combining volumes : When two or more gases react with one another, their volumes bear simple whole number ratio with one another and to the volume of products (if they are also gases) provided all volumes are measured under identical conditions of temperature and pressure. When gaseous hydrogen and gaseous chlorine react together to form gaseous hydrogen chloride according to the following equation. H (g ) 2 one volume Cl ( g ) 2 one volume 2HCl ( g ) two volumes It has been observed experimentally that in this reaction, one volume of hydrogen always reacts with one volume of chlorine to form two volumes of gaseous hydrogen chloride. all reactants and products are in gaseous state and their volumes bear a ratio of 1 : 1 : 2. This ratio is a simple whole number ratio. “These are no longer useful in chemical calculations now but gives an idea of earlier methods of analysing and relating compounds by mass.” Ex. 2.5 ml of a gaseous hydrocarbon exactly requires 12.5 ml oxygen for complete combustion and produces 7.5 ml carbon dioxide and 10.0 ml water vapour. All the volumes are measured at the same pressure and temperature. Show that the data illustrates Gay Lussac’s law of volume combination. = Sol. V hydrocarbon : V oxygen : V carbon dioxide : V water v apou r 2.5 : 12.5 : 7.5 : 10.0 = 1 : 5 : 3 : 4 (simple ratio) Hence, the data is according to the law of volume combination. MOLE CONCEPT Definition of mole : One mole is a collection of that many entities as there are number of atoms exactly in 12 gm of C – 12 isotope. or 1 mole = collection of 6.02 × 1023 species 6.02 × 1023 = N A = Avogadro’s No. 1 mole of atoms is also termed as 1 gm-atom, 1 mole of ions is termed as 1 gm-ion and 1 mole of molecule termed as 1 gm-molecule. METHODS OF CALCULATIONS OF MOLE (a) If no. of some species is given, then no. of moles = Given no. NA (b) If weight of a given species is given, then no of moles = or = Given wt (for atoms), Atomic wt. Given wt. (for molecules) Molecular wt. (c) If volume of a gas is given along with its temperature (T) and pressure (P) use n = PV RT where R = 0.0821 lit-atm/mol–K (when P is in atmosphere and V is in litre.) 1 mole of any gas at STP (0°C & 1 bar) occupies 22.7 litre. 1 mole of any gas at STP (0°C & 1 atm) occupies 22.4 litre. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 94 STOICHIOMETRY - 1 Atom : Atom is smallest particle which can not be divided into its constituents. Atomic weight : It is the weight of an atom relative to one twelvth of weight of 1 atom of C-12 RELATIONSHIP BETWEEN GRAM AND AMU 1 amu = for C 1 wt of one C - 12 atom. 12 1 mole C = 12 gm = 6.023 × 10 23 atoms wt of 6.023 × 10 23 atoms = 12 gm 12 wt of 1 atom of C = N gm (N A A 1 amu = Avogadro’s number = 6.23 × 10 23 ) 1 wt of one C - 12 atom 12 1 = 12 12 N A gm 1 1 amu = N gm A ELEMENTAL ANALYSIS Ex. For n mole of a compound (C 3H7 O2) Moles of C = 3n Moles of H = 7n Moles of O = 2n Find the wt of water present in 1.61 g of Na2 SO4. 10H 2O Sol. Moles of Na2SO 4. 10H2O = wt. in gram 1.61 = = 0.005 moles molecular wt 322 Moles of water = 10 × moles of Na2 SO4 . 10H2 O = 10 × 0.05 = 0.05 wt of water = 0.5 × 18 = 0.9 gm Ans. AVERAGE ATOMIC WEIGHT = % of isotope X molar mass of isotope. The % obtained by above expression (used in above expression) is by number (i.e. its a mole%) MOLECULAR WEIGHT It is the sum of the atomic weight of all the constituent atom. niMi (a) Average molecular weight = ni where n i = no. of moles of any compound and mi = molecular mass of any compound. Make yourselves clear in the difference between mole% and mass % in question related to above. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Page # 95 STOICHIOMETRY - 1 Shortcut for % determination if average atomic weight is given for X having isotopes X A & XB. % of X A Average atomic wei ght – wt of X B difference in weight of X A & X B 100 Try working out of such a shortcut for X A, X B, XC EMPIRICAL FORMULA, MOLECULAR FORMULA Empirical formula : Formula depicting constituent atom in their simplest ratio. Molecular formula : Formula depicting actual number of atoms in one molecule of the compound. Relation between the two : Molecular formula = Empirical formula × n n Molecular mass Empirical Formula mass Check out the importance of each step involved in calculations of empirical formula. Ex. Sol. A molecule of a compound have 13 carbon atoms, 12 hydrogen atom, 3 oxygen atoms and 3.02 × 10 –23 gm of other element. Find the molecular wt. of compound. 12 1 16 wt. of the 1 molecule of a compound = 13 × N + 12 × N + 3 × N + 3.02 × 10 –23 A A A 156 12 48 3. 02 10 – 23 NA = 234.18 / N A = 234 amu. Ans. NA • Density : (a) Absolute density (b) Relative density Absolute density = Relative density = Mass volume density of subs tan ce density of s tan dard subs tan ce density of subs tan ce Specific gravity = density of H O at 4 C 2 Vapour density : It is defined only for gas. It is a density of gas with respect to H 2 gas at same temp & press PMgas / RT Mgas d gas M V.D = dH = PM / RT = M = H H 2 2 2 2 V.D = M 2 Molecular wt of gas V.D = Molecular wt of H gas 2 • density of Cl2 gas with respect to O2 gas Molecularwt of Cl2 gas = Molecular wt of O gas 2 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 96 • STOICHIOMETRY - 1 STOICHIOMETRY : Stoichiometry is the calculations of the quantities of reactants and products involved in a chemical reaction. Following methods can be used for solving problems. • (a) Mole Method (For Balance reaction) (b) POAC method } Balancing not required but common sense ------- use it with slight care. (c) Equivalent concept CONCEPT OF LIMITING REAGENT. Limiting Reagent : It is very important concept in chemical calculation. It refers to reactant which is present in minimum stoichiometry quantity for a chemical reaction. It is reactant consumed fully in a chemical reaction. So all calculations related to various products or in sequence of reactions are made on the basis of limiting reagent. • It comes into picture when reaction involves two or more reactants. For solving any such reactions, first step is to calculate L.R. Calculation of Limiting Reagent. (a) By calculating the required amount by the equation and comparing it with given amount. [Useful when only two reactant are there] (b) By calculating amount of any one product obtained taking each reactant one by one irrespective of other reactants. The one giving least product is limiting reagent. (c) Divide given moles of each reactant by their stoichiometric coefficient, the one with least ratio is limiting reagent. [Useful when number of reactants are more than two.] • PERCENTAGE YIELD : The percentage yield of product = actual yield the theoretical maximum yield 100 • The actual amount of any limiting reagent consumed in such incomplete reactions is given by [% yield × given moles of limiting reagent] [For reversible reactions] • For irreversible reaction with % yield less than 100, the reactants is converted to product (desired and waste.) Ex. A compound which contains one atom of X and two atoms of y for each three atoms of z is made of mixing 5 gm of x, 1.15 × 10 23 atoms of Y and 0.03 mole of Z atoms. Given that only 4.40 gm of compound results. Calculate the atomic weight of Y if atomic weight of X and Z are 60 and 80 respectively. Sol. Moles of x = moles of y = 5 1 = = 0.083 60 12 115 . 10 23 0.19 6.023 10 23 moles of z = 0.03 x + 2y + 3z xy2 z3 for limiting reagent, 0.083/1 = 0.083 0.03 0.19 0.095 , 3 2 Hence z is limiting reagent 0.01 wt of xy2z 3 = 4.4 gm = moles × molecular wt. moles of xy2z3 = 300 + 2 m = 440 1 3 0.03 = 0.01 2m = 440 – 300 m = 70 Ans. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Page # 97 STOICHIOMETRY - 1 P O A C Rule : P O A C is the simple mass conservation. KClO3 KCl + O2 Apply the POAC on K. moles of K in KClO3 = moles of K in KCl 1 × moles of KClO3 = 1 × moles of KCl moles of KClO 3 = moles of KCl Apply POAC on O moles O in KClO 3 = moles of O in O2 3 × moles of kClO3 = 2 × moles of O2 Ex. In the gravimetric determination of phosphorous, an aqueous solution of dihydrogen phosphate ion (H 2PO4– ) is treated with a mix of ammonium & magnesium ions to precipitate magnesium ammonium phosphate MgNH 4 PO4 .6H2 O. This is heated and decomposed to magnesium Pyrophosphate, Mg2 P2 O7 which is weighted. A solution of H 2PO4 – yielded 1.054 gm of Mg2P 2O 7 what weight of NaH2 PO 4 was present originally. NaH2PO 4 Mg 2P 2O 7 apply POAC on P Let wt of NaH2 PO 4 = w gm moles of P in NaH 2PO 4 = moles of P in Mg2 P2 O7 w 1 .054 1 120 232 w = 1.054 × 2 120 232 2 = 1.09 gm Ans. SOME EXPERIMENTAL METHODS FOR DETERMINATION OF ATOMIC MASS Dulong’s and Petit’s Law : Atomic weight × specific heat (cal/gm°C) 6.4 Gives approximate atomic weight and is applicable for metals only. Take care of units of specific heat. FOR MOLECULAR MASS DETERMINATION (a) Victor Maeyer’s process : (for volatile substance) Procedure : Some known weight of a volatile substance (w) is taken, converted to vapour and collected over water. The volume of air displaced over water is given (V) and the following expressions are used. M= w RT PV If aq. tension is not given or M w RT (P – P') V If aq. tension is P Aqueous tension : Pressure exerted due to water vapours at any given temperature. This comes in picture when any gas is collected over water. Can you guess why ? : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 98 (b) STOICHIOMETRY - 1 Silver salt method : (for organic acids) Basicity of an acid : No. of replacible H + atoms in an acid (H bonded to more electronegative atom is acidic) Procedure : Some known amount of silver salt (w 1 gm) is heated to obtain w2 gm of white shining residue of silver. Then if the basicity of acid is n, molecular weight of acid would be AgnA nAg + + A –n Agn A is the salt w2 108 1 n M salt w 1 and molecular weight of acid = M – n(108) salt This is one good practicle application of POAC. (c) Chloroplatinate salt method : (for organic bases) Lewis acid : electron pair acceptor Lewis base : electron pair donor Procedure : Some amount of organic base is reacted with H2PtCl6 and forms salt known as chloroplatinate. If base is denoted by B then salt formed. (i) with monoacidic base = B 2 H2 PtCl 6 (ii) with diacidic base = B 2 (H2 PtCl6)2 (iii) with triacidic base = B 2(H2 PtCl6)3 The known amount (w1 gm) of salt is heated and pt residue is measured. (w2 gm). If acidity of base is ‘n’ then • (a) w2 195 1 n × M salt = w1 and M base = Msalt – n(410 ) 2 For % determination of elements in organic compounds : • All these methods are applications of POAC • Do not remember the formulas, derive them using the concept, its easy. Liebig’s method : (Carbon and hydrogen) (w) Organic Compound CuO (w 1) CO 2 + H2 O (w2) w1 12 100 44 w w2 2 100 % of H = 18 w where w1 = wt. of CO 2 produced, w2 = wt. of H 2O produced, % of C = w = wt, of organic compound taken (b) Duma’s method : (for nitrogen) (w) Organic Compound CuO N2 (P, V, T given) use PV = nRT to calculate moles of N 2 , n. n 28 100 w w = wt of organic compound taken % of N = Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Page # 99 STOICHIOMETRY - 1 (c) Kjeldahl’s method : (for nitrogen) (w) O.C. + H2SO4 NaOH (NH4 ) 2SO 4 NH3 + H2SO 4 (molarity M and volume (V1) consumed given) MV1 2 14 100 w where M = molarity of H 2SO4 . % of N = • (d) Some N containing compounds do not give the above set of reaction as in Kjeldahl’s method. Sulphur : (w) O.C. + HNO 3 % of S = H2SO 4 + BaCl2 w1 233 ×1× (w1 ) BaSO4 32 × 100 w where w1 = wt. of Ba SO 4 , w = wt. of organic compound (e) Phosphorus : O.C+ HNO 3 H 3PO 4 + [NH3 + magnesia mixture ammonium molybdate] MgNH4 PO4 Mg2 P2O7 w 1 2 31 100 222 w Carius method : (Halogens) % of P = (f) O.C. + HNO 3 + AgNO3 AgX If X is Cl then colour = white If X is Br then colour = dull yellow If X is I then colour = bright yellow • Flourine can’t be estimated by this w1 (M. weight of AgX ) % of X 1 ( At. wt of X ) 100 w Ex. 0.607 g of a silver salt of a tribasic organic acid was quantitatively reduced to 0.370 g of pure silver. Calculate the molecular weight of the acid (Ag = 108) Sol. Suppose the tribasic acid is H 3A. H3 A acid Ag3 A Ag salt 0.607 g 0.37 g Since Ag atoms are conserved, applying POAC for Ag atoms, moles of Ag atoms in Ag 3A = moles of Ag atoms in the prduct 3 × moles of Ag3 A = moles of Ag in the product 3 0.607 mol. wt. of Ag3 A 0.37 108 (Ag = 108) mol. wt. of Ag3 A = 531. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 100 STOICHIOMETRY - 1 mol. weight of tribasic acid (H 3A) = mol wt. of the salt (Ag3 A) – 3 × at. wt. of Ag + 3 × at. wt. of H = 531 – 324 + 3 = 210 Ans. EUDIOMETRY [For reactions involving gaseous reactants and products] • The stoichiometric coefficient of a balanced chemical reactions also gives that ratio of volumes in which gasesous reactants are reacting and products are formed at same temperature and pressure. The volume of gases produced is often given by mentioning certain solvent which absorb contain gases. Solvent KOH gas(es) absorb CO 2, SO 2 , Cl2 Ammon Cu 2Cl2 Turpentine oil CO O3 Alkaline pyrogallol water CuSO 4/CaCl2 O2 NH3, HCl H2O Assumption : On cooling the volume of water is negligible Ex. 7.5 mL of a hydrocarbon gas was exploded with excess of oxygen. On cooling, it was found to have undergone a contraction of 15 mL. If the vapour density of the hydrocarbon is 14, determine its molecular formula. (C = 12, H = 1) Sol. CxHy + (x + y ) O2 4 y HO 2 2 X CO2 + 7.5 ml 15 on cooling the volume contraction = 15 ml i.e. The volume of H 2O (g) = 15 ml V.D. of hydrocarbon = 14 Molecular wt. of CxHy = 28 12x + y = 28 ...(1) From reaction 7.5 y = 15 2 y=4 12 x + 4 = 28 12x = 24 x= 2 Hence Hydrocalbon is C2H4. CONCENTRATION OF SOLUTION Concentration of solution can be expressed in any of the following ways. (a) % by wt amount of solute dissolved in 100 gm of solution 4.9% H2SO 4 by wt. 100 gm of solution contains 4.9 gm of H2 SO4 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Page # 101 STOICHIOMETRY - 1 (b) % by volume volume of solute dissolved in 100 ml of solution x% H 2SO4 by volume 100 ml of solution contains x ml H 2SO 4 (c) % wt by volume wt. of solute present in 100 ml of solution (d) % volume by wt. volume of solute present in 100 gm of solution. CONCENTRATION TERMS • Molarity (M) : No. of moles of solute present in 1000 ml of solution. molarity (M) = M= moles of solute volume of solution (lit) m.moles of solute volume of solution(ml) MOLALITY (m) No. of moles of solute present in 1000 gm of solvent m= moles of solute wt. of solvent in kg m= m.moles of solute wt.of solvent in gm NORMALITY (N) No of gm equivalents of solute present in 1000 ml of solution N= gm equivalents of solute m. equivalent of solute = volume of solution(lit) volume of solution in (ml) FORMALITY (f) The formality is the no. of gm -formula weights of the ionic solute present in 1000 ml of solution. wt in gm f = formula wt volume of solution (lit) MOLE FRACTION The mole fraction of a perticular component in a solution is defined as the number of moles of that component per mole of solution. If a solution has nA mole A & n B mole of B. nA mole fraction of A (X A) = n A nB mole fraction of B (X B) = n A nB nB X A + XB = 1 • Parts per million (ppm) : = Mass of solute Mass of solvent × 10 6 Mass of solute Mass of solution : 0744-2209671, 08003899588 | url : www.motioniitjee.com, 10 6 :info@motioniitjee.com Page # 102 STOICHIOMETRY - 1 VOLUME STRENGTH OF H2 O2 Strength of H2O 2 is represented as 10V, 20V, 30V etc. 20V H2 O2 means one litre of this sample of H 2 O2 on decomposition gives 20 It of O2 gas at S.T.P. Decomposition of H 2O 2 is given as H2O 2 H2 O + 1 O2 2 1 × 22.4 It O2 at S.T.P.. 2 = 34 g = 11.2 It O 2 at S.T.P. 1 mole • To obtain 11.2 litre O2 at S.T.P. at lest 34 gm H2O 2 must be decomposed • for 20 It O 2, we should decompose atleast • 1 It solution of H2 O2 contains 34 112 . 1 It solution of H2 O 2 contains 34 112 . Normality of H2 O2 = 34 112 . NH2 O 2 NH2O 2 v. f 2 20 gm H2 O2 20 equivalents of H2 O2 17 (EH2 O2 M 2 34 2 17 ) 20 5.6 Volume, strength of H2 O 2 5.6 Normality of H2 O2(N) = MH2 O 2 20 17 34 × 20 gm H 2 O2 112 . IInd Method : 1 O2 2 From law of equivalence H2 O 2 H2 O + gm eq. of O 2 = gm eq. of H2O 2 gm eq. of O 2 = moles × n factor of O 2, = gm. eq. of H2O 2 = 20 22.4 4 = 20 5.6 20 5.6 and the volume of H2 O2 is 1 lit. this means 1 lit of H 2O 2 have i.e. Normality N = 20 gm eq. 5.6 20 5.6 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Page # 103 STOICHIOMETRY - 1 NORMALITY OF H2O2 = volume strength of H2 O 2 • 5.6 Molarity of H2O2 (M) = Volume, strength of H2 O 2 11 .2 Strength (in g/ ) : Denoted by S Strength = molarity × mol. wt. = molarity × 34 strength = Normality × Eq. weight. = Normality × 17 Ex. A bottle labeled with “12V H 2O 2” contain 700 ml solution. If a sdudent mix 300 ml water in it what is the g/litre strenth & normality and volume strength o final solution. Sol. N= 12 5.6 meq. of H2 O2 = 12 5.6 700 let the normality of H2 O2 on dilution is N meq. before dilution = meq. after dilution N × 1000 = 12 5 .6 strength gm/lit = 700 N= 15 . 2 34 = 25.5 volume strength = N × 5.6 = 12 7 × = 1.5 5.6 10 M= 15 . 2 84 = 8.4 V Ans. 10 Strength of Oleum Oleum is SO 3 dissolved in 100% H2SO 4. Sometimes, oleum is reported as more then 100% by weight, say y% (where y > 100). This means that (y – 100) grams of water, when added to 100 g of given oleum sample, will combine with all the free SO 3 in the oleum to give 100% sulphuric acid. Hence weight % of free SO 3 in oleum = 80( y – 100) 18 Ex. Calculate the percentage of free SO3 in an oleum (considered as a solution of SO3 in H2SO4 ) that is labelled '109% H 2SO4 '. Sol. '109% H2 SO 4' refers to the total mass of pure H2SO 4 , i.e., 109 g that will be formed when 100 g of oleum is diluted by 9 g of H2O which (H2O) combines with all the free SO 3 present in oleum to form H2SO4 H2 O + SO 3 H2 SO4 1 mole of H2O combines with 1 mole of SO 3 or 18 g of H2 O combines with 80 g of SO 3 or 9 g of H 2O combines with 40 g of SO 3. Thus, 100 g of oleum contains 40 g of SO 3 or oleum contains 40% of free SO 3 . : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 104 STOICHIOMETRY - 1 Ex. A 62% by mass of an aqueous solution of acid has specific gravity 1.8. This solution is diluted such that the specific gravity of solutin became 1.2. Find the % by wt of acid in new solutiuon. Sol. density = mass volume 1.8 = 100 volume of solution = volume of soln 100 18 . Let x gm water is added in soluion then d= 12 . 12 . 200 3 mass volume 100 100 18 . 100 18 . x x 12 . x 12 . x 100 0.2 x = 100 – x= 3 100 x x 200 100 = 3 3 100 1000 500 = = = 166.67 0.2 6 3 mass of new solution = 100 + 166.67 = 266.67 266.67 gm solution contains 62 gm of acid % by mass = 62 266.67 100 = 23.24 % RELATION SHIP BETWEEN MOLARITY, MOLALITY & DENSITY OF SOLUTION Let the molarity of solution be 'M', molality be 'm' and the density of solution be d gm/m. Molarity implies that there are M moles of solute in 1000 ml of solution wt of solution = density × volume = 1000 d gm wt of solute = MM 1 where M1 is the molecular wt of solute wt of solvent = (1000d – MM 1) gm (1000d – MM 1 ) gm of solvent contains M moles of solute M 1000 gm of solvent have = 1000d – MM 1000 mole = Molality 1 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Page # 105 STOICHIOMETRY - 1 1000 M no. of moles of solute present in 1000 gm of solvent = 1000 d – MM = Molality 1 on simplyfying d M 1 m M1 1000 RELATION SHIP BETWEEN MOLALITY & MOLE FRACTION consider a binary solution consisting of two components A (Solute) and B (Solvent). Let xA & xB are the mole fraction of A & B respectively. nA nB xA = n , xb = n n A B A nB If molality of solution be m then m nA mass of solvent 1000 = nB nA MB 1000 where M B is the molecular wt of the solvent B m xA 1000 xB MB mole fraction of A molality = mole fraction of B m= mole fraction of solute mole fraction of solvent 1000 MB 1000 molecular wt. of solvent Ex. An aqueous solution is 1.33 molal in methanol. Determine the mole fraction of methanol & H2 O Sol. molality = mole fraction of solute 1000 mole fraction of solvent mol.wt of solvent xA 1.33 = x B MB 1000 , 1.33 18 1000 x A 23.94 xB , 1000 xA xB xA = 0.02394 xB, x A + xB = 1 1.02394 xB = 1 xB 1 = 0.98, xA = 0.02 Ans. 102394 . Second Method : Let wt of solvent = 1000 gm molality = 1.33 = moles of solute moles of solute mole fraction of solute = , moles of solute moles of solvent m 1000 m+ 18 1.33 1. 33 1000 / 18 mole fraction of solute = 0.02 mole fraction of solvent = 1 – 0.02 = 0.98 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 106 Ex. STOICHIOMETRY - 1 The density of 3 M solution of sodium thiosulphate (Na2 S2O3 ) is 1.25 g/mL. Calculate (i) amount of sodium thiosulphate (ii) mole fraction of sodium thiosulphate (iii) molality of Na + and S2 O3 2– ions Sol. (i) Let us consider one litre of sodium thiosulphate solution. wt. of the solution = density × volume (mL) = 1.25 × 1000 = 1250 g. wt. of Na2 S2 O3 present in 1 L of the solution = molarity × mol. wt. = 3 × 158 = 474 g. Ans. wt. % of Na2 S2 O 3 = 474 1250 100 = 37.92% (ii) Wt. of solute (Na2 S2 O 3) = 474 g. Moles of solute = 474 158 3 Ans. Wt. of solvent (H2 O) = 1250 – 474 = 776 g Moles of solvent = 776 18 43.11 mole fraction of Na2 S2 O 3 = (iii) Molality of Na2S 2 O3 = 3 3 43.11 0.063 moles of N a 2 S2 O 3 wt. of solvent in grams 1000 = 3 776 1000 3.865 1 mole of Na2S 2 O3 contains 2 moles of Na+ ions and 1 mole of S 2 O3 2– ions. molality of Na+ = 2 × 3.865 = 7.73 m Molality of S 2 O3 2– = 3.865 m. Ans. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Page # 107 STOICHIOMETRY - 1 Solved Objective Ex.1 Sol. 8 litre of H 2 and 6 litre of Cl 2 are allowed to react to maximum possible extent. Fi nd out the final volume of reaction mi xture. Suppose P and T remains constant throughout the course of reaction (A) 7 litre (B) 14 litre (C) 2 litre (D) None of these. (B) Volume before reaction Volume after reaction H2 + 8 lit 2 Cl 2 6 lit 0 2 HCl 0 12 Volume after reaction = Volume of H 2 left + Vol ume of HCl formed = 2 + 12 = 14 lit Ex.2 Sol. Naturally occurring chlorine is 75.53% Cl 35 which has an atomic mass of 34.969 amu and 24.47% Cl 37 which has a mass of 36.966 amu. Cal culate the average atomic mass of chlorine(A) 35.5 amu (B) 36.5 amu (C) 71 amu (D) 72 amu (A) Average atomic mass % of I isotope its atomic mass % of II isotope its atomic mass = 75.53 x 34.969 24.47 x 36.96 = 100 100 Ex.3 Sol. Ex.4 Calculate the mass in gm of 2g atom of Mg(A) 12 gm (B) 24 gm (C) 6 gm (D) 1 gm atom of Mg has mass = 24 gm 2 gm atom of Mg has mass = 24 x 2 = 48 gm. Ex.5 Sol. Ex.6 (D) None of these. In 5 g atom of Ag (At. wt. of Ag = 108), calculate the weight of one atom of Ag (A) 17.93 × 10 –23 gm (C) 17.93 × 10 23 gm Sol. = 35.5 amu. (B) 16.93 × 10 –23 gm (D) 36 × 10 –23 gm (A) N atoms of Ag weigh 108 gm 108 108 1 atom of Ag weigh = = = 17.93 × 10 –23 gm. 6.023 x 10 23 N In 5g atom of Ag (at. wt. = 108), calcul ate the no. of atoms of Ag (A) 1 N (B) 3N (C) 5 N (D) 7 N. (C) 1 gm atom of Ag has atoms = N 5 gm atom of Ag has atoms = 5N. Sol. Calculate the mass in gm of 2N molecules of CO 2 (A) 22 gm (B) 44 gm (C) 88 gm (D) None of these. (C) N molecules of CO 2 has molecular mass = 44. 2N molecules of CO 2 has molecular mass = 44 x 2 = 88 gm. Ex.7 How many carbon atoms are present in 0.35 mol of C 6H12O 6 (A) 6.023 × 10 23 carbon atoms (C) 1.26 × 10 24 carbon atoms (B) 1.26 × 10 23 carbon atoms (D) 6.023 × 10 24 carbon atoms : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 108 Sol. STOICHIOMETRY - 1 (C) 1 mol of C6 H12O 6 has = 6 N atoms of C 0.35 mol of C 6H12O 6 has = = 2.1 N atoms 6 × 0.35 N atoms of C = 2.1 × 6.023 × 10 23 = 1.26 × 10 24 carbon atoms Ex.8 How many molecules are in 5.23 gm of glucose (C 6H12O 6) - Sol. (A) 1.65 × 10 22 (B) (B) 1.75 × 10 22 (C) 1.75 × 10 21( D) None of these 180 gm glucose has = N molecules 5.23 gm glucose has = 5.23 6 .023 10 180 23 = 1.75 × 10 22 molecules Ex.9 What is the weight of 3.01 × 10 23 molecules of ammonia (A) 17 gm (B) 8.5 gm (C) 34 gm Sol. (B) 6.023 × 10 23 (D) None of these molecul es of NH 3 has weight = 17 gm 3.01 × 10 23 molecules of NH 3 has weight = 17 3 .01 10 23 6 .023 10 23 = 8.50 gm Ex.10 How many significant figures are in each of the following numbers (a) 4.003 (b) 6.023 × 10 23 (c) 5000 (A) 3, 4, 1 (B) 4, 3, 2 (C) 4, 4, 4 (D) 3, 4, 3 Sol. (C) Ex.11 How many molecules are present in one m l of water vapours at STP Sol. (A) 1.69 × 10 19 (B) 2.69 × 10 –19 (D) 22.4 litre water vapour at STP has (C) 1.69 × 10 –19 (D) 2.69 × 10 19 = 6.023 × 10 23 molecules 1 × 10 –3 litre water vapours at STP has = 6 .023 10 23 22 .4 × 10 –3 = 2.69 × 10 +19 Ex.12 How many years it would take to spend Avogadro's number of rupees at the rate of 1 million rupees in one second (A) 19.098 × 10 19 years (B) 19.098 years Sol. (C) 19.098 × 10 9 years (C) (D) None of these 10 6 rupees are spent in 1sec. 6.023 × 10 23 rupees are spent in = 1 6 .023 10 23 10 6 60 60 24 365 = 1 6 .023 10 23 10 6 sec years , = 19.098 × 10 9 year Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Page # 109 STOICHIOMETRY - 1 Ex.13 An atom of an element weighs 6.644 × 10 –23 g. Calculate g atoms of element i n 40 kg(A) 10 gm atom Sol. (B) 100 gm atom (C) 1000 gm atom (D) 10 4 gm atom (C) wei ght of 1 atom of element = 6.644 × 10 –23 gm weight of 'N' atoms of element = 6.644 × 10–23 × 6.023 × 10 23 = 40 gm 40 gm of element has 1 gm atom. 40 x 10 3 gm of element has 40 103 , = 10 3 gm atom. 40 Ex.14 Calculate the number of Cl – and Ca+2 ions in 222 g anhydrous CaCl 2 - Sol. (A) 2N ions of Ca +2 4 N ions of Cl – (C) 1N ions of Ca +2 & 1N ions of Cl – (A) mol. wt. of CaCl 2 = 111 g 111 g CaCl 2 has = N ions of Ca +2 222g of CaCl 2 has Also (B) 2N ions of Cl – & 4N ions of Ca +2 (D) None of these. N 222 111 = 2N ions of Ca +2 111 g CaCl 2 has = 2N ions of Cl – 222 g CaCl 2 has = 2N 222 ions of Cl – 111 = 4N ions of Cl – . Ex.15 The density of O 2 at NTP is 1.429g / litre. Calcul ate the standard molar volume of gas(A) 22.4 lit. (B) 11.2 lit (C) 33.6 lit (D) 5.6 l it. Sol. (A) 1.429 gm of O 2 gas occupies volume = 1 l itre. 32 gm of O 2 gas occupies = 32 ,= 22.4 litre/mol. 1429 . Ex.16 Which of the following will weigh maximum amount(A) 40 g iron (B) 1.2 g atom of N (C) 1 × 10 23 atoms of carbon (D) 1.12 litre of O2 at STP Sol. (A) (A) Mass of iron = 40 g (B) Mass of 1.2 g atom of N = 14 × 1.2 = 16.8 gm (D) Mass of 1 × 10 23 atoms of C (D) Mass of 1.12 litre of O 2 at STP = 12 1 10 23 = 1.99 gm. 6.023 10 23 32 1 .2 = = 1.6 g 22 .4 Ex.17 How many mol es of potassi um chlorate to be heated to produce 11.2 li tre oxygen (A) 1 mol 2 (B) 1 mol 3 (C) 1 mol 4 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, (D) 2 mol. 3 :info@motioniitjee.com Page # 110 Sol. STOICHIOMETRY - 1 (B) 2 KClO 3 2KCl + 3O 2 Mole for reaction 2 2 3 3 × 22.4 litre O 2 is formed by 2 mol KClO3 11.2 litre O 2 is formed by 2 11 .2 3 22 .4 = 1 mol KClO 3 3 Ex.18 Calculate the weight of lime (CaO) obtained by heating 200 kg of 95% pure lime stone (CaCO 3). (A) 104.4 kg (B) 105.4 kg (C) 212.8 kg (D) 106.4 kg Sol. (D) 100 kg impure sample has pure CaCO 3 = 95 kg 200 kg impure sample has pure CaCO 3 = 95 200 = 190 kg. CaCO 3 100 CaO + CO 2 100 kg CaCO 3 gives CaO = 56 kg. 190 kg CaCO 3 gives CaO = 56 190 = 106.4 kg. 100 Ex.19 The chloride of a metal has the formula MCl 3 . The formul a of i ts phosphate will be(A) M 2PO 4 (B) MPO 4 (C) M 3PO4 (D) M(PO 4) 2 Sol. (B) AlCl 3 as it is AlPO 4 Ex.20 A silver coin weighing 11.34 g was dissol ved i n nitric acid. When sodium chloride was added to the solution all the silver (present as AgNO 3) was precipitated as si lver chloride. The weight of the precipitated silver chloride was 14.35 g. Calculate the percentage of silver in the coin (A) 4.8 % (B) 95.2% (C) 90 % (D) 80% Sol. (B) Ag + 2HNO3 AgNO 3 + NO 2 + H 2O 108 AgNO3 + NaCl AgCl + NaNO3 143.5 143.5 gm of silver chloride would be precipi tated by 108 g of silver. or 14.35 g of silver chl oride would be precipi tated 10.8 g of si lver. 11.34 g of silver coi n contain 10.8 g of pure silver. 100 g of sil ver coin contain 10.8 × 100 1134 . = 95.2 %. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Page # 111 STOICHIOMETRY - 1 SOLVED SUBJECTIVE Ex.1 Calculate the following for 49 gm of H 2 SO 4 (a) moles (b) Molecules (c) Total H atoms (d) Total O atoms (e) Total electrons Sol. Molecular wt of H 2SO 4 = 98 (a) moles = wt in gm 49 molecular wt 98 = 1 mole 2 (b) Since 1 mole = 6.023 × 10 23 molecules. 1 1 mole = 6.023 × 10 23 × molecules = 3.011 × 10 23 molecules 2 2 (c) 1 molecule of H 2SO 4 Contains 2 H atom 3.011 × 10 23 of H2 SO4 contain 2 × 3.011 × 1023 atoms = 6.023 × 10 23 atoms (d) 1 molecules of H 2SO4 contains 4 O atoms 3.011 × 10 23 molecular of H2 SO4 contains = 4 × 3.011 × 1023 = 12.044 × 10 23 (e) 1 molecule of H 2 SO4 contains 2H atoms + 1 S atom + 4 O atom this means 1 molecule of H 2SO 4 Contains (2 + 16 + 4 × 8) e– So 3.011 × 10 23 molecules have 3.011 × 1023 × 50 electrons = 1.5055 × 10 25 e– Ex.2 Calculate the total ions & charge present in 4.2 gm of N –3 Sol. mole = wt in gm 4.2 = = 0.3 Ionic wt 14 total no of ions = 0.3 × N A ions total charge = 0.3 N A × 3 × 1.6 × 10–19 = 0.3 × 6.023 × 10 23 × 3 × 1.6 × 10–19 , = 8.67 × 10 4 C Ans. Ex.3 Find the total number of iron atom present in 224 amu iron. Sol. Since 56 amu = 1 atom therefore 224 amu = 1 × 224 = 4 atom Ans. 56 Ex.4 A compound containing Ca, C, N and S was subjected to quantitative analysis and formula mass determination. A 0.25 g of this compound was mixed with Na2 CO3 to convert all Ca into 0.16 g CaCO3. A 0.115 gm sample of compound was carried through a series of reactions until all its S was changed into SO4 2– and precipitated as 0.344 g of BaSO 4. A 0.712 g sample was processed to liberated all of its N as NH3 and 0.155 g NH 3 was obtained. The formula mass was found to be 156. Determine the empirical and molecular formula of the compound. Sol. Moles of CaCO3 = Wt of Ca = 0.16 = Moles of Ca 100 0.16 × 40 100 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 112 STOICHIOMETRY - 1 Mass % of Ca = 0.16 100 100 0.25 40 Similarly Mass % of S = 0.344 233 Similarly Mass % of N= 25.6 32 100 0.115 0.155 17 14 0.712 41 100 = 17.9 Mass % of C = 15.48 Now : Elements Ca S N C Mass % 25.6 41 17.9 15.48 Mol ratio 0.64 1.28 1.28 1.29 Simple ratio 1 2 2 2 Empirical formula = CaC2 N2S2, Molecular formula wt = 156 , n × 156 = 156 n= 1 Hence, molecular formula = CaC2 N2S 2 Ex.5 A polystyrne having formula Br 3C6H3 (C3H8 )n found to contain 10.46% of bromine by weight. Find the value of n. (At. wt. Br = 80) Sol. Let the wt of compound is 100 gm & molecular wt is M Then moles of compound = Moles of Br = 100 M 100 ×3 M 100 × 3 × 80 = 10.46 M M = 2294.45 = 240 + 75 + 44 n , Hence n = 45 Ans. wt of Br = Ex.6 A sample of clay was partially dried and then analysed to 50% silica and 7% water. The original clay contained 12% water. Find the percentage of silica in the original sample. Sol. In the partially dried clay the total percentage of silica + water = 57%. The rest of 43% must be some 50 impurity. Therefore the ratio of wts. of silica to impurity = . This would be true in the original sample 43 of silica. The total percentage of silica + impurity in the original sample is 88. If x is the percentage of silica, x 88 – x 50 ; x = 47.3% Ans. 43 Ex.7 A mixture of CuSO 4.5H2 O and MgSO 4. 7H2O was heated until all the water was driven-off. if 5.0 g of mixture gave 3 g of anhydrous salts, what was the percentage by mass of CuSO4 .5H2O in the original mixture ? Sol. Let the mixture contain x g CuSO 4.5H2O x 159.5 249.5 5–x 120 = 3 246 x = 3.56 Mass percentage of CuSO4 . 5H2O = 3 .56 100 = 71.25 % Ans. 5 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Page # 113 STOICHIOMETRY - 1 Ex .8 Sol. 367.5 gm KClO3 (M = 122.5) when heated, How many litre of oxygen gas is proudced at S.T.P. KClO 3 KCl + O2 Applying POAC on O, moles of O in KClO3 = moles of O in O2 3 × moles of KClO 3 = 2 × moles of O 2 367.5 3 367.5 = 2 × n, n = × 122.5 2 122.5 Volume of O 2 gas at S.T.P = moles × 22.4 3× 3 367.5 22.4 = 9 × 11.2 = 100.8 lit Ans. 2 122.5 0.532 g of the chloroplatinate of a diacid base on ignition left 0.195 g of residue of Pt. Calculate molecular weight of the base (Pt = 195) = Ex.9 Sol. Suppose the diacid base is B. B + diacid H2PtCl 6\ BH2 PtCl6 acid chloroplatinate base 0.532 g Pt 0.195 g Since Pt atoms are conserved, applying POAC for Pt atoms, moles of Pt atoms in BH 2PtCl 6 = moles of Pt atoms in the product 1 × moles of BH2 PtCl 6 = moles of Pt in the product 0.532 mol. wt. of BH2PtCl6 0.195 195 mol. wt. of BH 2PtCl 6 = 532 From the formula BH2 PtCl6 , we get mol. wt. of B = mol. wt. of BH 2PtCl 6 – mol. wt. of H 2 PtCl6 = 532 – 410 = 122. Ans. Ex.10 10 mL of a gaseous organic compound containing. C, H and O only was mixed with 100 mL of oxygen and exploded under conditions which allowed the water formed to condense. The volume of the gas after explosion was 90 mL. On treatment with potash solution, a further contraction of 20 mL in volume was observed. Given that the vapour density of the compound is 23, deduce the molecular formula. All volume measurements were carried out under the same conditions. Sol. CxHyOz + x y z – 4 2 O2 xCO2 + y HO 2 2 10 ml after explosion volume of gas = 90 ml 90 = volume of CO 2 gas + volume of unreacted O2 on treatment with KOH solution volume reduces by 20 ml. This means the volume of CO 2 = 20 ml the volume of unreacted O 2 = 70 ml volume of reacted O 2 = 30 ml V.D of compoud = 23 molecular wt 12x + y + 16z = 46 ...(1) from equation we can write : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 114 10 x STOICHIOMETRY - 1 y z – 4 2 30 , x + 4x + y – 2z = 12 & 10x = 20 y z – =3 4 2 ...(2) x=2 from eq. (1) & (2) ; z = 1 & y = 6; Hence C2 H6 O Ans. Ex.11 A sample of coal gas contained H 2, CH4 and CO. 20 mL of this mixture was exploded with 80 mL of oxygen. On cooling, the volume of gases was 68 mL. There was a contraction of 10 mL. When treated with KOH. Find the composition of the original mixture. Sol. H2 + CH4 + CO; at H2 = x ml CH4 = y ml; CO = (20 – x – y) ml H2 + CH4 + CO x y + O2 CO2 + H2O 20 – x – y on cooling the volume of gases = 68 ml = volume of CO 2 + unreacted O 2 volume contraction due to KOH = 10 ml this means volume of CO 2 = 10 ml volume of unreacted O 2 = 58 ml volume of reacted O 2 = 80 – 58 = 22 ml Applying POAC on C; y + 20 – x – y = volume of CO 2, 20 – x = 10 x = 10 Applying POAC on H; 2x + 4y = 2x moles of H2 O; moles of H2 O = x + 2y Applying POAC on O 1 × moles of CO + 2 × moles of O 2 = 2 × moles of CO 2 + 1 × moles of H 2 O 1 × 20 – x – y + 2 × 22 = 2 × 10 + x + 2y 20 – x – y + 44 = 20 + x + 2y; 2x + 3y = 44 3y = 44 – 20 = 24; y = 8 ml; x = 10 ml; volume of CO = 20 – x – y = 2 ml Ans. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) STOICHIOMETRY - 1 Page # 115 Class Room Problems Problem.1 From the following reaction sequence Cl 2 + 2KOH 3KClO KCl + KClO + H2O 2KCl + KClO3 4KClO3 3KClO4 + KCl Calculate the mass of chlorine needed to produce 100 g of KClO4. Problem.4 5 mL of a gaseous hydrocarbon was exposed to 30 mL of O2. The resultant gas, on cooling is found to measure 25 mL of which 10 mL are absorbed by NaOH and the remainder by pyrogallol. Determine molecular formula of hydrocarbon. All measurements are made at constant pressure and temperature. Sol. C2H 4 Sol. 205.04 gm Problem.2 Calculate the weight of FeO produced from 2 g VO and 5.75 g of Fe2O3. Also report the limiting reagent. VO + Fe2O3 FeO + V 2O 5 Sol. 5.175 gm Problem.5 A gaseous alkane is exploded with oxygen. The volume of O 2 for complete combustion to CO 2 formed is in the ratio of 7 : 4. Deduce molecular formula of alkane. Sol. C2H 6 Problem.3 A polystyrene, having formula Br 3C6H3 (C8H8 ) n was prepared by heating styrene with tribromobenzoyl peroxide in the absence of air. If it was found to contain 10.46% bromine be weight, find the value of n. P roblem.6 A sampl e of ga seous hydro carbon occupying 1.12 litre at NTP, when completely burnt in air produced 2.2 g CO2 and 1.8 g H2O. Calculate the weight of hydrocarbon taken and the volume of O 2 at NTP required for its combustion. Sol. 19 Sol. 0.8 gm, 2.24 L : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 116 Problem.7 16 mL of a gaseous aliphatic compound CnH3nO m was mixed with 60 mL O2 and sparked. The gas mixture on cooling occupied 44 mL. After treatment with KOH solution, the volume of gas remaining was 12 mL . Deduce the formul a of co mpound. All measurements are made at constant pressure and room temperature. STOICHIOMETRY - 1 Problem.10 What is the purity of conc. H 2SO 4 solution (specific gravity 1.8 g/mL), if 5.0 mL of this solution is neutralized by 84.6 mL of 2.0 N NaOH ? Sol. 92.12% Sol. C2H6O Problem.11 A sample of H2SO4 (density 1.787 g mL–1) is labelled as 86% by weight. What is molarity of acid ? What volume of acid has to be used to make 1 litre of 0.2 M H2SO4 ? Sol. 15.68 M, 0.013 L Problem.8 In what ratio should you mix 0.2M NaNO3 and 0.1M Ca(MO 3)2 solution so that in resulting solution, the concentration of –ve ion is 50% greater than the concentration of +ve ion ? Sol. 1/2 Problem.12 Mole fracti on of I 2 in C 6H6 is 0.2. Calculate molality of I2 in C6H6. Problem.9 How much BaCl2 would be needed to make 250 mL of a solution having same concentration of Cl – as the one containing 3.78 g of NaCl per 100 mL ? Sol. 3.2 Sol. 16.8 gm Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) STOICHIOMETRY - 1 Problem.13 A drop (0.05 mL) of 12 M HCl is spread over a thin sheet of aluminium foil (thickness 0.10 mm and density of Al = 2.70 g/mL). Assuming whole of the HCl is used to dissolve Al, what will be the maximum area of hole produced in foi l ? Page # 117 Problem.16 A mixture of FeO and Fe3O 4 when heated in air to constant weight, gains 5% in its weight. Find out composition of mixture. Sol. 20.25 Sol. 5.4×10–3 Problem.14 What would be the molarity of solution obtained by mixing equal volumes of 30% by weight H2SO 4 (d = 1.218 g mL –1) and 70% by weight H2SO 4 (d = 1.610g mL–1) ? If the resulting solution has density 1.425 g/mL, calculate its molality. Problem.17 25.4 g of I2 and 14.2 g of Cl 2 are made to react completely to yield a mixture of ICl and ICl 3. Calculate mole of ICl and ICl3 formed. Sol. 0.1,0.1 Sol. 7.67 M Problem.15 A mixture of Al and Zn weighing 1.67 g was completley dissolved in acid and evolved 1.69 litre of H2 at NTP. What was the weight of Al in original mixture ? Sol. 1.21 gm Problem.18 A mixture of HCOOH and H 2C2O 4 is heated wi th conc. H 2SO 4. The gas produced is collected and on treating with KOH solution the volume 1 th. Calculate molar ratio 6 of two acids in original mixture. of the gas decreases by Sol. 1/4 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 118 STOICHIOMETRY - 1 Problem.19 For the reaction, N2O5(g) 2NO 2(g) + 0.5O 2(g), cal culate the mole fraction of N2O5(g) decomposed at a constant volume and temperature, if the initial pressure is 600 mm Hg and the pressure at any time is 960 mm Hg. Assume ideal gas behaviour. Sol. 0.375 Problem.23 A sample of CaCO 3 and MgCO 3 weighed 2.21 g is ignited to constant weight of 1.152 g. What is the composition of mixture ? Also calculale the volume of CO 2 evolved at 0ºC and 76 cm of pressure. Sol. 46% 1.19 Problem.20 0.22 g sample of volatile compound, containing C, H and Cl only on combustion in O 2 gave 0.195 g CO 2 and 0.0804 g H2O. If 0.120 g of the compound occupied a volume of 37.24 mL at 105º and 768 mm of pressure, calculate molecular formula of compound. Sol. C 2H 4Cl2 P roblem.24 2.0 g of a mi xture of carbonate, bicarbonate and chl oride of sodi um on heati ng produced 56 mL of CO 2 at NTP. 1.6 g of the same mixture requi red 25 mL of N HCl sol uti on for neu tra l i za t i on . Cal cul at e per ce nt age o f each component present in mixture. Sol. 69.56% Problem.21 2.0 g sample containing Na2CO3 and NaHCO 3 loses 0.248 g when heated to 300ºC, the temperature at which NaHCO3 decomposes to Na2CO3, and H2O. What is % of Na2CO 3 in mixture ? Sol. 66.4% Problem .25 Igni ti ng M nO 2 i n ai r co nverts i t quantitatively to Mn 3O 4. A sample of pyrolusite has MnO 2 80%, Si O 2 15% and rest having water. The sample is heated in air to constant mass. What is the % of Mn in igni ted sample ? Sol. 59.37 Problem.22 10 mL of a solution of KCl containing NaCl gave on evaporation 0.93 g of the mi xed salt which gave 1.865 g of AgCl by the reaction with AgNO 3. Calculate the quantity of NaCl in 10 mL of solution. Sol. 0.67 gm Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) STOICHIOMETRY - 1 Page # 119 Problem.26 A solid mixture 5 g consits of lead nitrate and sodium nitrate was heated below 600ºC until weight of residue was constant. If the loss in weight is 28%, find the amount of lead nitrate and sodium nitrate in mixture. Sol. 3.32 gm Problem. 29 0.50 g of a mixture of K2CO 3 and Li2CO 3 required 30 mL of 0.25 N HCl solution for neturalization. What is % composition of mixture ? Sol. Problem.27 Determine the formula of ammonia form the folloiwng data : (i) Volume of ammonia = 25 mL. (ii) Volume on addition of O2 after explosion = 71.2 mL. (iii) Vol ume after explosion and reaction with O 2 on cooling = 14.95 mL. (i v) Vol ume after bein g absorbed by al kal i ne pyrogall ol = 12.5 mL. Sol. Problem.30 A mixture in which the mole ratio of H 2 and O2 is 2 : 1 is used to prepare water by the reaction, 2H2(g) + O 2(g) 2H2O(g) The total pressure in the container is 0.8 atm at 20ºC before the reaction. Determine the final pressure at 120ºC after reaction assuming 80% yield of water. Sol. 0.7865 Problem.28 A mixture of ethane (C2H6) and ethene (C 2H4) occupies 40 litre at 1.00 atm and at 400 K. The mixture reacts completely with 130 g of O 2 to produce CO 2 and H2O. Assuming ideal gas behaviour, calculate the mole fractions of C 2H4 and C2H6 in the mixture. Sol. 0.63, 0.37 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 120 EXERCISE – I STOICHIOMETRY - 1 OBJECTIVE PROBLEMS (JEE MAIN) Single correct 1. For the reaction 2x + 3y + 4z 5w Initially if 1 mole of x, 3 mole of y and 4 mole of z is taken. If 1.25 mole of w is obtained then % yield of this reaction is (A) 50% (B) 60% (C) 70% (D) 40% Sol. 5. Sol. The vapour density of a mixture of gas A (Molecular mass = 40) and gas B (Molecular mass = 80) is 25. Then mole % of gas B in the mixture would be (A) 25% (B) 50% (C) 75% (D) 10% Sol. 2. A solution of A (MM = 20) and B (MM = 10), [Mole fraction XB = 0.6] having density 0.7 gm/ ml then molarity and molality of B in this solution will be ________________ and ______________ respectively. (A) 30M,75m (B) 40M,75m (C) 30M,65m (D) 50M,55m 6. For the reaction 2A + 3B + 5C 3D Initially if 2 mole of A, 4 mole of B and 6 mole of C is taken, With 25% yield, moles of D which can be produced are _____________. Sol. (A) 0.75 (C) 0.25 (B) 0.5 (D) 0.6 Sol. 3. 125 ml of 8% w/w NaOH solution (sp. gravity 1) is added to 125 ml of 10% w/v HCl solution. The nature of resultant solution would be _________ (A) Acidic (B) Basic (C) Neutral (D) None 7. Sol. Fill in the blanks in the following table. Compound Grams Grams Molality Compd Waterof Compd Na2 CO3 ______ 250 0.0125 (A) 0.331 (B) 0.662 (C) 0.165 (D) 0.993 Sol. 4. Ratio of masses of H2 SO4 and Al2 (SO4) 3 is grams each co nt ainin g 32 grams o f S i s __________. (A) 0.86 (B) 1.72 (C) 0.43 (D) 2.15 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) STOICHIOMETRY - 1 8. Page # 121 Equal volumes of 10% (v/v) of HCl is mixed with 10% (v/v) NaOH solution. If density of pure NaOH is 1.5 times that of pure HCl then the resultant solution be : (A) basic (B) neutral (C) acidic 11. (D) can’t be predicted. Sol. 9. A definite amount of gaseous hydrocarbon was burnt with just sufficient amount of O2 . The volume of all reactants was 600 ml, after the explosion the volume of the products [CO 2(g) and H2 O(g)] was found to be 700 ml under the similar conditions. The molecular formula of the compound is : (A) C3 H8 (B) C3 H6 (C) C3 H4 (D) C4 H10 One mole mixture of CH 4 & air (containing 80% N 2 20% O 2 by volume) of a composition such that when underwent combustion gave maximum heat (assume combustion of only CH 4 ). Then which of the statements are correct, regarding composition of initial mixture. (X presents mole fraction) 1 , XO2 = 2 8 (A) X CH 4 11 (B) X CH4 3 , XO2 8 1 , X N2 8 1 2 (C) X CH4 1 , X O2 6 1 , X N2 6 2 3 11 , X N2 11 (D) Data insufficient Sol. Sol. 12. C6 H5OH(g) + O 2(g) CO 2 (g) + H2O(l) Magnitude of volume change if 30 ml of C 6H5OH (g) is burnt with excess amount of oxygen, is (A) 30 ml (B) 60 ml (C) 20 ml (D) 10 ml Sol. 10. One gram of the silver salt of an organic dibasic acid yields, on strong heating, 0.5934 g of silver. If the weight percentage of carbon in it 8 times the weight percentage of hydrogen and half the weight percentage of oxygen, determine the molecular formula of the acid. [Atomic weight of Ag = 108] (A) C4H6O 4 (B) C4 H6 O6 (C) C2H6O 2 (D) C5H10 O5 13. Sol. 10 ml of a compound containing ‘N’ and ‘O’ is mixed with 30 ml of H2 to produce H2O (l) and 10 ml of N2 (g). Molecular formula of compound if both reactants reacts completely, is (A) N2O (B) NO2 (C) N2 O3 (D) N2O 5 Sol. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 122 14. Similar to the % labelling of oleum, a mixture of H3 PO4 and P 4O 10 is labelled as (100 + x) % where x is the maximum mass of water which can react with P4 O10 present in 100 gm mixture of H 3 PO 4 and P4 O10. If such a mixture is labelled as 127% Mass of P4O 10 is 100 gm of mixture, is (A) 71 gm (B) 47 gm (C) 83gm (D) 35 gm STOICHIOMETRY - 1 17. Sol. In the quantitative determination of nitrogen using Duma’s method, N2 gas liberated from 0.42 gm of a sample of organic compound was collected over water. If the volume of N 2 gas 100 collected was ml at total pressure 860 mm 11 Hg at 250 K, % by mass of nitrogen in the organic compound is [Aq. tension at 250K is 24 mm Hg and R = 0.08 L atm mol–1 K–1 ] (A) 10 % 3 (B) 5 % 3 (C) 20 % 3 (D) 100 % 3 Sol. 15. Mass of sucrose C12H22O11 produced by mixing 84 gm of carbon, 12 gm of hydrogen and 56 lit. O2 at 1 atm & 273 K according to given reaction, is C(s) + H2(g) + O 2 (g) C12 H22O 11(s) (A) 138.5 (B) 155.5 (C) 172.5 (D) 199.5 18. Sol. 40 gm of a carbonate of an alkali metal or alkaline earth metal containing some inert impurities was made to react with excess HCl solution. The liberated CO 2 occupied 12.315 lit. at 1 atm & 300 K. The correct option is (A) Mass of impurity is 1 gm and metal is Be (B) Mass of impurity is 3 gm and metal is Li (C) Mass of impurity is 5 gm and metal is Be (D) Mass of impurity is 2 gm and metal is Mg Sol. 16. If 50 gm oleum sample rated as 118% is mixed with 18 gm water, then the correct option is (A) The resulting solution contains 18 gm of water and 118 gm H2SO 4 (B) The resulting solution contains 9 gm of water and 59 gm H2 SO4 (C) The resulting solution contains only 118 gm pure H2 SO4 (D) The resulting solution contains 68 gm of pure H2SO4 19. Sol. The percentage by mole of NO 2 in a mixture NO2(g) and NO(g) having average molecular mass 34 is : (A) 25% (B) 20% (C) 40% (D) 75% Sol. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) STOICHIOMETRY - 1 20. Page # 123 The minimum mass of mixture of A2 and B4 required to produce at least 1 kg of each product is : (Given At. mass of ‘A’ = 10; At mass of ‘B’ = 120) 5A2 + 2B4 2AB 2 + 4A2B (A) 2120 gm (C) 560 gm Sol. (B) 1060 gm (D) 1660 gm Sol. 24. 74 gm of sample on complete combustion gives 132 gm CO 2 and 54 gm of H 2 O. The molecular formula of the compound may be (A) C5H12 (B) C4 H10O (C) C3 H6O2 (D) C3 H7 O2 Sol. 21. The mass of CO2 produced from 620 gm mixture of C2H4 O2 & O2 , prepared to produce maximum energy is (A) 413.33 gm (B) 593.04 gm (C) 440 gm (D) 320 gm Sol. 25. The % by volume of C4H10 in a gaseous mixture of C4 H10, CH4 and CO is 40. When 200 ml of the mixture is burnt in excess of O 2 . Find volume (in ml) of CO 2 produced. (A) 220 (C) 440 (B) 340 (D) 560 Sol. 22. Assumi ng compl ete precipitati on of AgCl , calculate the sum of the molar concentration of all the ions if 2 lit of 2M Ag2 SO 4 is mixed with 4 lit of 1 M NaCl solution is : (A) 4M (B) 2M (C) 3M (D) 2.5 M 26. Sol. 23. 12.5 gm of fuming H2 SO4 (labelled as 112%) is mixed with 100 lit water. Molar concentration of H+ in resultant solution is : [Note : Assume that H2SO 4 dissociate completely and there is no change in volume on mixing] (A) 2 700 (B) 2 350 (C) 3 350 (D) 3 700 What volumes should you mix of 0.2 M NaCl arid 0.1 M CaCl 2 solution so that in resulting solution the concentration of positive ion is 40% lesser than concentration of negative ion. Assuming total volume of solution 1000 ml. (A) 400 ml NaCl, 600 ml CaCl2 (B) 600 ml NaCl, 400 ml CaCl2 (C) 800 ml NaCl, 200 ml CaCl2 (D) None of these Sol. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 124 27. STOICHIOMETRY - 1 An iodized salt contains 0.5% of Nal. A person consumes 3 gm of salt everyday. The number of iodide ions going into his body everyday is (A) 10–4 (B) 6.02 × 10–4 (C) 6.02 × 1019 (D) 6.02 × 1023 31. Weight of oxygen in Fe2O 3 and FeO is in the simple ratio for the same amount of iron is : (A) 3 : 1 (B) 1 : 2 (C) 2 : 1 (D) 3 : 1 Sol. Sol. 32. 28. The pair of species having same percentage (mass) of carbon is : (A) CH3COOH and C6H12O6 (B) CH3COOH and C2H5OH (C) HCOOCH3 and C12H22 O11 (D) C6 H12O 6 and C12H22O 11 Sol. Two elements X (atomic mass 16) and Y (atomic mass 14) combine to form compounds A, B and C. The ratio of different masses of Y which combines with a fixed mass of X in A, B and C is 1:3:5. If 32 parts by mass of X combines with 84 parts by mass of Y in B, then in C, 16 parts by mass of X will combine with___ parts by mass of Y. (A) 14 (B) 42 (C) 70 (D) 84 Sol. 29. 200 ml of a gaseous mixture containing CO, CO2 and N2 on complete combustion in just sufficient amount of O2 showed contraction of 40 ml. When the resulting gases were passed through KOH so l u ti on it reduces by 50 % then calculate the volume ratio 33. of VCO 2 : VCO : VN 2 in original mixture. (A) 4 : 1 : 5 (C) 1 : 4 : 5 (B) 2 : 3 : 5 (D) 1 : 3 : 5 In a textile mill, a double-effect evaporator system concentrates weak liquor containing 4% (by weight) caustic soda to produce a lye containing 25% solids (by weight). Calculate the weight of the water evaporate per 100-kg feed in the evaporator. (A) 125.0 g (C) 84.0 kg Sol. (B) 50.0 kg (D) 16.0 kg Sol. 30. Sol. Density of a gas relative to air is 1.17. Find the mol. mass of the gas. [M air = 29 g/mol] (A) 33.9 (B) 24.7 (C) 29 (D) 22.3 34. Zinc ore (zinc sulphide) is treated with sulphuric acid, leaving a solution with some undissolved bits of material and releasing hydrogen sulphide gas. If 10.8g of zinc ore is treated with 50.0 ml of sulphuric acid (density 1.153 g/ml), 65.1g of solution and undissolved material remains. In addition, hydrogen sulphide (density 1.393 g/ L) is evolved. What is the volume (in liters) of this gas? (A) 4.3 (B) 3.35 (C) 4.67 (D) 2.40 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) STOICHIOMETRY - 1 Sol. 35. Page # 125 38. A sample of an ethanol-water solution has a volume of 54.2 cm 3 and a mass of 49.6g. What is the percentage of ethanol (by mass) in the solution? (Assume that there is no change in volume when the pure compounds are mixed.) The density of ethanol is 0.80 g/cm3 and that of water is 1.00 g/cm3. (A) 18.4% (B) 37.1% (C) 33.95% (D) 31.2% A sample of clay contains 40% silica and 15% water. The sample is partially dried by which it loses 5 gm water. If the percentage of water in the partially dried clay is 8, calculate the percentage of silica in the partially dried clay. (A) 21.33% (B) 43.29% (C) 75% (D) 50% Sol. 39. Sol. The density of quartz mineral was determined by adding a weighed piece to a graduated cylinder containing 51.2ml water. After the quartz was submersed, the water level was 65.7 ml. The quartz piece weighed 38.4g. What was the density of quartz? (A) 1.71 gm/ml (B) 1.33 gm/ml (C) 2.65 gm/ml (D) 1.65gm/ml Sol. 36. A student gently drops an object weighing 15.8 g into an open vessel that is full of ethanol, so that a volume of ethanol spills out equal to the vol ume of the obj ect. The experimenter now finds that the vessel and its contents weigh 10.5 g more than the vessel full of ethanol only. The density of ethanol is 0.789 g/cm 3. What is the density of the object? (A) 6.717 gm/cm 3 (B) 4.182 gm/cm 3 3 (C) 1.563 gm/cm (D) 2.352 gm/cm 3 40. Which has maximum number of atoms of oxygen (A) 10 ml H 2O(l) (B) 0.1 mole of V 2O5 (C) 12 gm O 3(g) (D) 12.044 × 10 22 molecules of CO 2 Sol. Sol. 41. 37. A person needs on average of 2.0 mg of riboflavin (vitamin B 2) per day. How many gm of butter should be taken by the person per day if it is the only source of riboflavin? Butter contains 5.5 microgram riboflavin per gm. (A) 363.6 gm (B) 2.75 mg (C) 11 gm (D) 19.8 gm Mass of one atom of the element A is 3.9854 × 10 –23. How many atoms are contained in 1g of the element A? (A) 2.509 × 10 23 (B) 6.022 × 10 23 23 (C) 12.044 × 10 (D) None Sol. Sol. 42. The number of atoms present in 0.5 g-atoms of nitrogen is same as the atoms in (A) 12 g of C (B) 32 g of S (C) 8 g of oxygen (D) 24g of Mg : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 126 STOICHIOMETRY - 1 Sol. 43. Sol. A graph is plotted for an element, by putting its weight on X-axis and the corresponding number of number of atoms on Y-axis. Determine the atomic weight of the element for which the graph is plotted. 47. Two isotopes of an element Q are Q 97 (23.4% abundance) and Q94 (76.6% abundance). Q 97 is 8.082 times heavier than C 12 and Q 94 is 7.833 times heavier than C 12. What is the average atomic weight of the element Q? (A) 94.702 (B) 78.913 (C) 96.298 (D) 94.695 Sol. (A) infinite (C) 0.025 (B) 40 (D) 20 48. Sol. 44. The O18/O 16 ratio in some meteorites is greater than that used to calculate the average atomic mass of oxygen on earth. The average mass of an atom of oxygen in these meteorites is ___ that of a terrestrial oxygen atom? (A) equal to (B) greater than (C) less than (D) None of these The element silicon makes up 25.7% of the earth's crust by weight, and is the second most abundant element, with oxygen being the first. Three isotopes of silicon occur in nature: Si 28 (92.21%), which has an atomic mass of 27.97693 amu; Si 29 (4.70%), with an atomic mass of 28.97649 amu; and Si 30 (3.09%), with an atomic mass of 29.97379 amu. What is the atomic weight of silicon? (A) 28.0856 (B) 28.1088 (C) 28.8342 (D) 29.0012 Sol. Sol. 49. 45. If isotopic distribution of C 12 and C14 is 98.0% and 2.0% respectively, then the number of C 14 atoms in 12 gm of carbon is (A) 1.032 ×10 22 (B) 1.20 ×10 22 (C) 5.88 ×10 23 (D) 6.02 ×10 23 The average atomi c mass of a m ixture containing 79 mol % of 24 Mg and remaining 21 mole % of 25 Mg and 26Mg, is 24.31. % mole of 26 Mg is (A) 5 (B) 20 (C) 10 (D) 15 Sol. Sol. 50. 46. At one time there was a chemical atomic weight scale based on the assignment of the value 16.0000 to naturally occurring oxygen. What would have been the atomic weight, on such a table, of silver, if current information had been available? The atomic weights of oxygen and silver on the present table are 15.9994 and 107.868. (A) 107.908 (B) 107.864 (C) 107.868 (D) 107.872 The oxide of a metal contains 30% oxygen by weight. If the atomic ratio of metal and oxygen is 2 : 3, determine the atomic weight of metal. (A) 12 (B) 56 (C) 27 (D) 52 Sol. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) STOICHIOMETRY - 1 EXERCISE – II 1. Page # 127 OBJECTIVE PROBLEMS (JEE ADVANCED) The average mass of one gold atom in a sample of naturally occuring gold is 3.257 × 10–22 g. Use this to calculate the molar mass of gold. Sol. Sol. 5. 2. A plant virus is found to consist of uniform cylindrical particles of 150 Å in diameter and 5000 Å long. The specific volume of the virus is 0.75 cm3/g. If the virus is considered to be a single particle, find its molecular weight. [V = r2 l] Nitrogen (N), phosporus (P), and potassium (K) are the main nutri ents in pl ant fertil izers. Acco rding to an i ndustry conventi on, the numbers on the label refer to the mass % of N, P 2O 5, and K 2O, in the order. Calculate the N : P : K ratio of a 30 : 10 : 10 fertilizer in terms of moles of each elements, and express it as x : y : 1.0. Sol. Sol. EMPIRICAL & MOLECULAR FORMULA 6. MOLE 3. Calculate (a) Number of nitrogen atoms in 160 amu of NH4 NO 3 (b) Number of gram-atoms of S in 490 kg H2 SO 4 Polychlorinated biphenyls, PCBs, known to be dangerous environmental pollutants, are a group of compounds with the general empirical formula C12HmCl10–m, where m is an integer. What is the value of m, and hence the empirical formula of the PCB that contains 58.9% chlorine by mass ? Sol. (c) Grams of Al2 (SO 4) 3 containing 32 amu of S. Sol. 7. 4. A chemical compound “dioxin” has been very much in the news in the past few years. (it is the by-product of herbicide manufacture and is through to be qui te toxic.) Its formula i s C12H4 Cl4 O2 . If you have a sample of dirt (28.3 g) that contains 8.78 × 10 –8 mol es of dioxin, calculate the percentage of dioxin in the dirt sample ? Given the following empi rical formulae and molecular weights, compute the true molecular formulae : Empirical Formula Molecular weight (a) (b) CH2 CH2 O 84 150 (c) (d) HO HgCl 34 472 (e) HF 80 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 128 Sol. STOICHIOMETRY - 1 Sol. 12. 8. What is the empirical formula of a compound 0.2801 gm of which gave on complete combustion 0.9482 gm of carbon dioxide and 0.1939 gm of water ? Sol. wt of the mixture taken = 2g Loss in weight on heating = 0.11 gm Sol. 13. 9. Determine the percentage composition of a mixture of anhydrous sodium carbonate and sodium bicarbonate from the following data : A 5.5 gm sample of an organic compound gave on quantitative analysis 1.4 gm of N and 3.6 gm of C and 0.5 gm of H. If Molecular mass of the compound is 55 then calculate E.F. and M.F. Sol. A 10 g sample of a mixture of calcium chloride and sodium chloride is treated with Na2CO 3 to precipitate calcium as calcium carbonate. This CaCO 3 is heated to convert all the calcium to CaO and the final mass of CaO is 1.12gm. Calculate % by mass of NaCl in the original mixture. Sol. PROBLEMS RELATED WITH MIXTURE 10. One gram of an alloy of aluminium and magnesium when heated with excess of dil. HCl forms magnesium chloride, aluminium chloride and hydrogen. The evolved hydrogen collected over mercury at 0ºC has a volume of 1.12 liters at 1 atm pressure. Calculate the composition of the alloy. 14. A mixture of Ferric oxide (Fe2 O3 ) and Al is used as a solid rocket fuel which reacts to give Al 2 O3 and Fe. No other reactants and products are involved. On complete reaction of 1 mole of Fe2 O 3, 200 units of energy is released. (i) Write a balance reaction representing the above change. (ii) What should be the ratio of masses of Fe2 O3 and Al taken so that maximum energy per unit mass of fuel is released. (iii) What would be energy released if 16 kg of Fe2O3 reacts with 2.7 kg of Al. Sol. Sol. 11. A sample containing only CaCO3 and MaCO3 is ignited to CaO and MgO. The mixture of oxides produced weight exactly half as much as the original sample. Calculate the percentages of CaCO3 and MgCo3 in the sample. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) STOICHIOMETRY - 1 Page # 129 LIMITING REACTANT 15. 17. Titanium, which is used to make air plane engines and frames, can be obtained from titanium tetrachloride, which in turn is obtained from titanium oxide by the following process : 3 TiO2 (s) + 4C(s) + 6Cl2 (g) 2Co(g) (B) 25 ml C3H8 & 75 ml O2 3TiCl4(g) + 2CO2(g) + A vessel contains 4.32 g TiO2 , 5.76 g C and; 6.82 g Cl 2 , suppose the reactio n go es to completion as written, how many gram of TiCl 4 and be produced ? (Ti = 48) A mixture of C3H8 (g) O2 having total volume 100 ml in an Eudiometry tube is sparked & it is observed that a contraction of 45 ml is observed what can be the composition of reacting mixture. (A) 15 ml C3H8 & 85 ml O2 (C) 45 ml C3H8 & 55 ml O2 (D) 55 ml C3H8 & 45 ml O 2 Sol. Sol. 18. [Assume 100% dissociation of each salt and molecular mass of X 2– is 96] More than one correct : 16 Two gases A and B which react according to the equation aA ( g ) bB( g) cC( g) (A) [Cl –] = 20 M (B) [Na+] = 11 M dD(g ) to give two gases C and D are taken (amount not known) in an Eudiometer tube (operating at a constant Pressure and temperature) to cause the above. An aqueous solution consisting of 5 M BaCl 2, 58.8% w/v NaCl solution & 2m Na2 X has a density of 1.949 gm/ml. Mark the opti on(s) which represent correct molarity (M) of the specified ion. (C) [Total anions] = 20.5 M (D) [Total cations]=15 M Sol. If on causing the reaction there is no volume change observed then which of the following statement is/are correct. (A) (a + b) = (c + d) (B) average molecular mass may increase or decrease if either of A or B is present in limited amount. 19. (C) Vapour Density of the mixture will remain same throughout the course of reaction. (D) Total moles of all the component of mixture will change. A mixture of 100 ml of CO, CO 2 and O2 was sparked. When the resulting gaseous mixture was passed through KOH solution, contraction in volume was found to be 80 ml, the composition of initial mixture may be (in the same order) (A) 30 ml, 60 ml, 10 ml (B) 30 ml, 50 ml, 20 ml Sol. (C) 50 ml, 30 ml, 20 ml (D) 30 ml, 40 ml, 30 ml Sol. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 130 20. STOICHIOMETRY - 1 Given following series of reactions : (I) NH3 + O 2 NO + H2O (II) NO + O2 NO 2 (III) NO2 + H2O Sol. HNO3 + HNO2 (IV) HNO2 HNO3 + NO + H2O Select the correct option (s) : (A) Moles of HNO3 obtained is half of moles of Ammonia used if HNO 2 is not used to produce HNO3 by reaction (IV) 100 % more HNO3 will be produced if HNO2 is (B) 6 used to produce HNO3 by reaction (IV) than if HNO2 is not used to produce HNO 3 by reaction (IV) 23. An organic compound is burnt with excess of O2 to produce CO2 (g) and H2 O(l), which results in 25% volume contraction. Which of the following option(s) satisfy the given conditions. (A) 10 ml C3 H8 + 110 ml O2 (B) 20 ml C2H6O + 80 ml O2 (C) 10 ml C3H6O2 + 50 ml O 2 (D) 40 ml C2 H2 O4 + 60 ml O 2 Sol. 1 th (C) If HNO2 is used to produce HNO 3 then 4 of total HNO3 is produced by reaction (IV) (D) Moles of NO produced in reaction (IV) is 50% of moles of total HNO3 produced. 24. Sol. A sample of H2 O2 solution labelled as 56 volume has density of 530 gm/L. Mark the correct option(s) representing concentration of same solution in other units. (Solution contains only H2O and H2 O2) (A) MH2O 2 6 w v 17 (B) % 21. (C) Mole fraction of H 2O 2 = 0.25 Solution(s) containing 40 gm NaOH is/are (D) m H2O 2 (A) 50 gm of 80% (w/w) NaOH (B) 50 gm of 80% (w/v) NaOH [dsoln = 1.2 gm/ml] (C) 50 gm of 20 M NaOH [dsoln = 1 gm/ml] 1000 72 Sol. (D) 50 gm of 5m NaOH Sol. 25. Solution(s) containing 30 gm CH3COOH is/are (A) 50 gm of 70% (w/v) CH 3COOH [d sol = 1.4 gm/ml] (B) 50 gm of 10 M CH3COOH [dsol = 1 gm/ml] (C) 50 gm of 60% (w/w) CH3COOH (D) 50 gm of 10 m CH3 COOH Sol. 22. The incorrect statement(s) regarding 2M MgCl2 aqueous solution is/are (dsolution = 1.09 gm/ml) (A) Molality of Cl is 4.44 m (B) Mole fraction of MgCl 2 is exactly 0.035 (C) The conc. of MgCl 2 is 19% w/v (D) The conc. of MgCl2 is 19 × 104 ppm Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) STOICHIOMETRY - 1 Page # 131 26. ‘2V’ ml of 1 M Na2 SO4 is mixed with ‘V’ ml of 2M Ba(NO3 )2 solution. Sol. (A) Molarity of Na+ ion in final solution can’t be calculated as V is not known. (B) Molarity of BaSO4 in final solution is 2 M 3 4 (C) Molarity of NO3– in final solution is M 3 (D) Molarity of NO–3 in final solution is 2 M 3 Sol. 29. Column I Column II (A) 10 M MgO (dsol = 1.20 gm/ml) (P) W solvent = 120 gm per 100 ml of solution. Solute: MgO, Solvent:H2 O (B) 40% w/v NaOH (dsol. = 1.6 gm/ml) (Q) W sol = 150 gm per 100 gm solvent Solute:NaOH,Solvent:H2 O (C) 8 m CaCO3 (R) Wsolute = 120 gm per Solute:CaCO3 ,Solvent:H 2O (D) 0.6 mol fraction of ‘X’ (molecular mass = 20) 100 gm of solvent (S) W solvent = 125 gm per 100 gm of solute in ‘Y’ (molecular mass 25) Solute : X, Solvent : Y Match the Column 27. Sol. One type of artifical diamond (commonly called YAG for yttrium aluminium garnet) can be represented by the formula Y3 Al 5O 12 [Y = 89, Al = 27] Column I Column II Element Weight percentage (A) Y (P) (B) Al (Q) 32.32 % 22.73 % (C) O (R) 44.95 % Sol. 30. Bunty & Bubbly have two separate containers one having N 2 gas & other H2 gas : It is known that N 2 & H2 react to give N 2H2(l) and/or N 2 H4 (g) depending upon the ratio in which N 2 & H2 are taken & that N 2 H2 reacts with H 2 to give N 2H4 . Formation of 1 mole of N2 H4 requires 30 units of energy & formation of 1 mole of N 2 H2 (l) release 30 units of energy. From this information match Column I (representing composition of gases taken) with Column II (representing the observation) Column I 28. The recommended daily dose is 17.6 milligrams of vitamin C (ascorbic acid) having formula C6H8O 6. Match the following. Given : NA = 6 × 10 23 Column I Column II (A) O-atoms present (B) Moles of vitamin C in 1gm (Q) 5.68 × 10 –3 (C) of vitamin C Moles of vitamin C in 1gm (R) 3.6 × 1020 should be consumed daily (P) 10 –4 mole Column II (Composition of gases)(Observation) (A) 40 lit N 2 & 30 lit H2 (P) Contraction by 22.4 lit (same tem perature & pressure) (B) 11.2 lit of N2 & H2 (Q) Contraction by 20 lit. taken at 1 atm & 273 K in a ratio such that max. release of energy is observed (C) 11.2 lit of N2 & 30 lit (R) Contraction by 60 lit of H2 (same temperature & pressure) (D) 10 lit of N2 & more than 22.4 lit of H 2 (S) Contraction by 11.2 lit (same temperature & pressure) : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 132 STOICHIOMETRY - 1 Sol. COMPREHENSION 32. A 4.925 g sample of a mixture of CuCl2 and CuBr2 was dissolved in water and mixed thoroughly with a 5.74 g portion of AgCl. After the reaction the solid, a mixture of AgCl, and AgBr, was filtered, washed, and dried. Its mass was found to be 6.63 g. (a) % By mass of CuBr2 in original mixture is 31. (A) 2.24 (C) 45.3 Br2 reacts with O 2 in either of the following ways depending upon supply of O2. Br2 + 1 2 O2 Br2 O , Br2 3 2 O2 (b) % By mass of Cu in original mixture is (A) 38.68 (C) 3.86 Br2 O3 Column II (Initial reactants) (Final product) (A) 320 gm Br2 is mixed with 64 gm of O 2 (P) 1 mole Br2 O3 (B) 160 gm Br2 is mixed (Q) 1 mole (Br2 O), 2 1 with 8 gm of O 2 2 (R) 1 mole (Br2 O), 1 32 mole (Br2O3) (D) 160 gm Br2 is mixed with (S) Sol. (A) 25 (B) 50 (C) 75 (D) 60 (d) No. of moles of Cl – ion present in the solution after precipitation are (A) 0.06 (C) 0.04 (B) 0.02 (D) None Sol. mole (Br2 ) (C) 80 gm Br2 is mixed with gm of O 2 48 gm of O2 (B) 19.05 (D) None (c) % by mole of AgBr in dried precipitate is Match composition of the final mixture for initial amount of reactants. Column I (B) 74.5 (D) None 1 mole (Br2 O3), 2 1 mole (O2 ) 4 33. Na Br, u sed t o pr o duce AgBr fo r us e i n photography can be self prepared as follows : Fe + Br2 FeBr2 FeBr2 + Br2 Fe3Br 8 Fe3 Br8 + Na2CO 3 ...(i) ...(ii) (not balanced) NaBr + CO2 + Fe3O4 ...(iii) (not balanced) (a) Mass of iron required to produce 2.06 × 10 3 kg NaBr (A) 420 gm (B) 420 kg (C) 4.2 × 10 kg 5 (D) 4.2 × 10 8 gm (b) If the yield of (ii) is 60% & (iii) reaction is 70% then mass of iron required to produce 2.06 × 10 3 kg NaBr (A) 105kg (B) 105 gm (C) 10 3 kg (D) None (c) If yield of (iii) reaction is 90% then mole of CO 2 formed when 2.06 × 103 gm NaBr is formed : (A) 20 (C) 40 (B) 10 (D) None Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) STOICHIOMETRY - 1 Page # 133 Sol. 35. For a gaseous reaction, 2A(g) 3B(g) + C(g) Whose extent of disso ciati on depends o n temperature is performed in a closed container, it is known that extent of dissociation of A is different in different temperature range. With in a temperature range it is constant. (Temperature range T0 – T1, T1 – T2 , T2 – T ). A plot of P v/s T is drawn under the given condition. Given : tan 55 = 1.42, tan 50 = 1.19, tan 60 = 1.73 60° 34. N2 O5 and H2 O can react to form HNO 3, according to given reaction N2 O5 + H2O 55° 2HNO 3 50° T0 (a) If (a) Find the percentage labelling of a mixture containing 23 gm HNO3 and 27 gm N2 O5. (A) T0 – Ti T2 – T T2 Ti – Ti 1 T(k) is the degree of dissociation of A then in the temperature range Ti is lowest 1 (B) T0 – Ti (D) T2 – T Ti + 1 is highest 0 (A) 104.5% (B) 109% (C) (C) 113.5% (D) 118% (b) If initially 1 mole of A is taken in a 0.0821 l container then [R = 0.0821 atm lit/k] (b) Find the maximum and minimum value of percentage labelling : (A) 133.3% (C) 116.66%, 100% (B) 116.66%, 0% (D) None (c) Find the new labelling if 100 gm of this mixture (original) is mixed with 4.5 gm water (A) 100 + 4. 5 1 (B) 100 + (A) T0 – Ti 0. 19 (B) T0 – T1 0.095 (C) T1 – T2 0. 42 (D) T1 – T2 0.73 Sol. 4 .5 1.045 36. 4 .5 (C) 100 + 104. 5 Sol. T1 the concentration of a mi xture of HNO 3 and N2 O5 (g) can be expressed similar to oleum. Then answer the following question. 4 .5 (D) 100 + 1. 09 A 10 ml mixture of N 2, a alkane & O 2 undergo combustion in Eudiometry tube. There was contraction of 2 ml, when residual gases are passed through KOH. To the remaining mixture comprising of only one gas excess H 2 was added & after combustion the gas produced is absorbed by water, causing a reduction in volume of 8 ml. (a) Gas produced after introduction of H2 in the mixture ? (A) H2O (B) CH4 (C) CO 2 (D) NH3 (b) Volume of N 2 present in the mixture ? (A) 2 ml (B) 4 ml (C) 6 ml (D) 8 ml (c) Volume of O 2 remai ned after the fi rst combustion ? (A) 4 ml (B) 2 ml (C) 0 (D) 8 ml (d) Identify the hydrocarbon. (A) CH4 (B) C2 H6 (C) C3H8 (D) C4H10 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 134 STOICHIOMETRY - 1 Sol. Sol. 40. 37. An evacuated glass vessel weighs 50 gm when empty, 148.0 g when completely filled with liquid of density 0.98 gml–1 and 50.5 g when filled with an ideal gas at 760 mm at 300 K. Determine the molecular weight of the gas. [JEE ‘98,3] Statement-1 : 1 g-atom of sulphur contains Avogadro number of sulphur molecules Statement-2 : Atomicity of sulphur is eight. Sol. Sol. 41. 38. At 100° C and 1 atmp, if the density of liquid water is 1.0 g cm –3 and that of water vapour is 0.0006 g cm–3 , then the volume occupied by water molecules 1 L of steam at that temperature is : [JEE ‘2001 (Scr), 1] (A) 6 cm3 (B) 60 cm 3 (C) 0.6 cm3 (D) 0.06 cm3 Statement-1 : The number of O atoms in 1 gm. of O 2, 1 gm O 3 and 1 gm of atomic oxygen is same. S ta te m en t- 2 : E ach o f th e spe ci e s represents 1/16 gm-atom of oxygen. Sol. Sol. 42. Each of the questions given below consists of Statement-I and Statement-II. Use the following Key to choose the appropriate answer. (A) If both statement-1 and statement-2 are correct, and statement-2 is the correct explanation of statement-1 Statement-1 : The rati o by volume of H 2 : Cl 2 : HCl in a reaction H 2(g) + Cl 2(g) 2HCl(g) is 1 : 1 : 2. Statement-2 : Substances always react in such a way that their volume ratio i s in si mple integers. Sol. (B) If both statement-1 and statement-2 are correct, and statement-2 is not the correct explanation of statement-1 (C) If statement-1 is correct and statement-2 is incorrect 43. (D) If statement-1 is incorrect and statement-2 is correct Statement-1 : 0.2 N H 2 SO 4 solution has molarity equal to 0.2 M. Statement-2 : H2SO 4 is a diabasic acid. Sol. 39 Statement-1 : Molarity of pure water is 55.5 M. Sta tement-2 : M ol ari ty is temperature dependent parameter Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) STOICHIOMETRY - 1 Integer Type 44. 48. If one mole of ethanol (C 2H5OH) completely burns to form carbon dioxide and water, the weight of carbon dioxide formed is about - 1.5 gm of di valent metal displaced 4 gm of copper (at. wt. = 64) from a soluti on of copper sulphate. The atomic weight of the metal is- Sol. Sol. 45. Page # 135 How many grams are contai ned in 1gm-atom of Na Sol. 49. Assuming that petrol is iso-octane (C8H18) and has density 0.8 gm/ml, 1.425 litre of petrol on complete combustion will consume oxygen - Sol. 46. The mass of CaCO3 produced when carbon dioxide is passed in excess through 500 ml of 0.5 M Ca(OH) 2 wi ll be- Sol. 50. 47. The mass of oxygen that would be required to produce enough CO, which completely reduces 1.6 kg Fe2O 3 (at. mass Fe = 56) is- A mixture containing 100 gm H 2 and 100 gm O 2 is ignited so that water is formed according to the reaction, 2H 2 + O 2 2H 2O; How much water will be formed - Sol. Sol. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 136 STOICHIOMETRY - 1 SUBJECTIVE PROBLEMS (JEE ADVANCED) EXERCISE – III 1. A2 2B 2 A 2B 4 and 3 2 A2 2B 2 A 3B 4 Two substance A2 & B2 react in the above manner when A2 is limited it gives A 2 B4 , when in excess gives A3 B 4. A 2B 4 can be converted to A 3B 4 when reacted with A2 . Using this information calculate the composition of the final mixture when mentioned amount of A2 & B2 are taken 4. Chloride samples are prepared for analysis by using NaCl, KCl and NH 4 Cl separately or as mixture. What minimum volume of 5% by weight AgNO 3 solution (sp gr., 1.04 g ml–1) must be added to a sample of 0.3 g in order to ensure complete precipitation of chloride in every possible case ? Sol. (i) If 4 moles of A2 & 4 moles of B2 is taken in reaction container (ii) If 1 moles of A 2 & 2 moles of B2 is taken in 2 reaction container (iii) If 5 moles of A 2 & 2 moles of B 2 is taken 4 Sol. 5. One litre of milk weighs 1.035 kg. The butter fat is 10% (v/v) of milk has density of 875 kg/m 3. The density of fat free skimed milk is ? Sol. 2. How much minimum volume of 0.1 M aluminium sulphate solution should be added to excess calcium nitrate to obtain atleast 1 gm of each salt in the reaction. Al2 (SO 4 ) 3 3Ca(NO 3 )2 2Al( NO3 )3 3CaSO 4 6. Sol. 3. A sample of fuming sulphuric acid containing H2 SO4 , SO 3 and SO 2 weighing 1.00 g is found to require 23.47 ml of 1.00 M alkali (NaOH) for neutralisation. A separate sample shows the presence of 1.50% SO2 . Find the percentage of “free” SO3 , H2 SO4 and “combined” SO3 in the sample. 10 mL of gaseous organic compound contain C, H and O only was mixed with 100 mL of O2 and exploded under identical conditions and then cooled. The volume left after cooling was 90 mL. On treatment with KOH a contraction of 20 Ml was observed. If vapour density of compound is 23, derive molecular formula of the compound. Sol. Sol. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) STOICHIOMETRY - 1 7. Page # 137 For a hypothetical chemical reaction represented by Sol. C( g) D( g) , the following informations 3 A ( g) are known. Information (i) At t = 0, only 1 mole of A is present and the gas has V.D. = 60. (ii) At t = 30 min, the gaseous mixture consist of all three gases and has a vapour density = 75. (iii) Molecular Mass of C = 200 Calculate (a) Molecular weight of A and D. (b) Moles of each specie at t = 30 min. 10. Consider the following set reactions CH3 O || CH C O H O || CH C O H O || CH C O H Sol. n times CH3 8. PCl 3 Cl 2 PCl 3 3H 2 O AgNO3 PCl 5 H3PO 3 Silver Salt White resideu ( excess ) If 0.1 moles of silver salt is taken & wt. of residue obtained is 54 gms then what will be the molecular mass of CH CH....... CH | | CH3 | CH3 Br Br Br n 3HCl A sample containing very large amount of PCl3 was exposed to a sample of “Chlorinated water” having Cl2 dissolved in H2 O so that the above two reactions occurred. It was observed that ration of mass of PCl5 to mass of H3 PO3 was 417 : 246. From this information calculate. (i) ratio of moles of PCl 5 to moles of H3 PO 3 . Sol. (i i) ratio of mol es of Cl 2 : H 2 O present in chlorinated water (iii) Molality (m) of Chlorine in Chlorinated water. Sol. 11. 9. A mixture of H 2, N 2 & O 2 occupying 100 ml underwent reaction so as to from H2 O2 (l) and N2H2(g) as the only products, causing the volume to contract by 60 ml. The remaining mixture was passed through pyrogallol causing a contraction of 10ml. To the remaining mixture excess H2 was added and the above reaction was repeated, causing a reduction in volume of 10 ml. Identify the composition of the initial mixture in mol %. (No other products are formed) 124 gm of mixture containing NaHCO 3, AlCl 3 & KNO 3 requires 500 ml, 8% w/w NaOH solution [d NaOH = 1.8 gm/ml] for complete neutralisation. On heating same amount of mixture, it shows loss in weight of 18.6 gm. Calculate % composition of mixture by moles. Weak base formed doesn’t interfere in reaction. Assume KNO 3 does not decompose under given conditions. Sol. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 138 12. STOICHIOMETRY - 1 A mixture of three gases an alkane (general formula Cn H2n + 2), an alkene (general formula CxH2x) and O 2 was subjected to sparking to cause combustion of both the hydrocarbon at 127ºC. After the reaction three gases were present and none of the hydrocarbon remained. On passing the gases through KOH (abso rb CO 2 ), an increment in mass of KOH solution by 132 gm was observed. The remaining gases were passed over white anhydrous CuSO4 and the weight of blue hydrated CuSO4 crystals was found to be 72 gm more than that of white anhydrous CuSo4. Given that initially total 10 moles of the three gases were taken and moles of alkane and alkene were equal and if molecular mass of alkene molecular mass of alkane = 12 i.e. (Malk ene – M alkane = 12), then answer the follo wi ng questions. (Show calculations) (a) Which three gases are remained after the combustion reactions. (b) What are the number of moles of product gases. (c) What is the molecular formula of the two hydrocarbon. (d) What is the number of moles of each of the two hydrocarbons and O 2 gas taken initially. Sol. 14. The vapours of organic compound was burnt in oxygen. Equal volume of both gaseous substance were taken at same pressure and temperature. After the reaction, the system was returned to the original condition and it turn out that its vol ume has no t changed. T he pro duct of combustion contain 50% CO2 (g) and 50% H2O(g) by volume and no other gas. Find the molecular weight of organic compound (in gram/mol) in question. Sol. Sol. 15. “Prussian blue” can be prepared by the following reactions. I. Fe H 2SO 4 13. H2 O 2 H2O2 2K I 40 % yield 2KMnO 4 I2 3H2SO 4 K 2SO 4 2KOH II. FeSO 4 50% yield 2MnSO4 3O2 4H2O 100 ml of H2O sample was divided into two parts. First part was treated with KI. And KOH formed required 200 ml of M/2 H 2 SO4 for complete neutralisation. Other part was treated with just sufficient KMnO4 yielding 6.74 lit. of O2 at 1 atm & 273 K. Calculate (a) Moles of KOH produced (b) Moles of KMnO4 used (c) Total moles of H 2O 2 used in both reaction (d) Volume strength of H2 O 2 used. H2 SO 4 FeSO 4 1 O 2 2 III. Fe 2 (SO 4 ) 3 K 4 [Fe(CN) 6 ] H2 Fe 2 ( SO 4 )3 H 2O Fe 4 [Fe( CN)6 ] 3 K 2SO 4 Calculate number of moles of Fe4 [Fe(CN)6 ]3 produced, if (i) 50 moles of Fe and 30 moles of H 2 SO4 are used with sufficient amount of other reactants. (ii) 50 moles of Fe, 70 moles of H 2 SO4 and 30 moles of K 4[Fe(CN) 6 ] are used with sufficient amount of other reactants. (iii) 400 moles of Fe are used with sufficient amount of other reactants (assuming the yield of I, II & III reactions are 50%, 40% and 60% respectively). Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) STOICHIOMETRY - 1 Sol. Page # 139 20. The chief ore of Zn is the sulphide, ZnS. The ore is concentrated by froth floatation process and then heated in air to convert ZnS to ZnO. 2ZnS 3O 2 75% ZnO H2 SO 4 16. A chemist wants to prepare diborane by the reaction 6LiH + 8BF3 6LiBF4 + B2H6 2ZnSO 4 If he starts with 2.0 moles each of LiH & BF3 . How many moles of B2 H6 can be prepared. 2H2O 2ZnO 100 % 80 % 2SO 2 ZnSO 4 H2 O 2 Zn 2H 2SO 4 O2 (a) What mass of Zn will be obtained from a sample of ore containing 291 kg of ZnS. (b) Calculate the volume of O 2 produced at 1 atm & 273 K in part (a). Sol. Sol. 17. Carbon reacts with chlorine to form CCl4 . 36 gm of carbon was mixed with 142 g of Cl2 . Calculate mass of CCl4 produced and the remaining mass of reactant. Sol. 21. MISCELLANEOUS PROBLEM 18. In a determination of P an aqueous solution of NaH2PO 4 is treated with a mixture of ammonium and magnesium ions to precipitate magnesium ammonium phosphate Mg(NH4)PO4 , 6H2O. This is h ea te d an d de co mpo sed t o ma gne si u m pyrophosphate, Mg2 P2 O 7 which is weighed. A solution of NaH 2 PO 4 yielded 1.054 g of Mg2 P2 O7. What weight of NaH2PO4 was present originally ? Sol. P4S 3 + 8O 2 P 4 O10 + 3SO 2 Calculate minimum mass of P4S 3 is required to produce atleast 1 gm of each product. Sol. 22. 19. Sol. By the reaction of carbon and oxygen, a mixture o f CO and CO 2 i s o btai ned. W hat i s the composition by mass of the mixture obtained when 20 grams of O 2 reacts with 12 grams of carbon ? A mixture of nitrogen and hydrogen. In the ratio of one mole of nitrogen to three moles of hydrogen, was partially converted into NH 3 so that the final product was a mixture of all these three gases. The mixture was to have a density of 0.497 g per litre at 25ºC and 1.00 atm. What would be the mass of gas in 22.4 liters at 1atm and 273 K ? Calculate the % composition of this gaseous mixture by volume. Sol. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 140 23. STOICHIOMETRY - 1 1gm sample of KClO3 was heated under such conditions that a part of it decomposed according to the equation (1) 2KClO3 –– 2 KCl + 3O 2 and remaining underwent change according to the equation. (2) 4KClO3 –– 3 KClO 4 + KCl 26. Density of a solution containing 13% by mass of sulphuric acid is 1.09 g/mL. Then molarity of solution will be. Sol. If the amount of O2 evolved was 112 ml at 1 atm and 273 K., calculate the % by weight of KClO4 in the residue. Sol. 27. 24. In one process for waterproofing, a fabric is exposed to (CH3)2 SiCl2 vapour. The vapour reacts with hydroxyl groups on the surface of the fabric or with traces of water to form the waterproofing film [(CH3) 2SiO] n , by the reaction n( CH3 )2 SiCl 2 2nOH 2nCl The density of a solution containing 40% by mass of HCl is 1.2 g/mL. Calculate the molarity of the solution. Sol. nH2O [( CH3 ) 2 SiO ] n wher e n stands for a l arge i nteger. Th e waterproofing film is deposited on the fabric layer upon layer. Each l ayer i s 6.0 Å thi ck [the thickness of the (CH3 )2 SiO group]. How much (CH3 )2 SiCl2 is needed to waterproof one side of a piece of fabric, 1.00 m by 3.00 m, with a film 300 layers thick ? The density of the film is 1.0 g/cm3 . 28. 15g of methyl alcohol is present in 100 mL of solution. If density of solution is 0.90 g mL–1. Calculate the mass percentage of methyl alcohol in solution Sol. Sol. CONCENTRATION TERMS 25. Calculate the molarity of the following solutions (a) 4g of caustic soda is dissolved in 200 mL of the solution. (b) 5.3 g of anhydrous sodium carbonate is dissolved in 100 mL of solution (c) 0.365 g of pure HCl gas is dissolved in 50 mL of solution. Sol. 29. Units of parts per million (ppm) or per billion (ppb) are often used to describe the concentrations of solutes in very dilute solutions. The units are defined as the number of grams of solute per million or per billion grams of solvent. Bay of Bengal has 1.9 ppm of lithium ions. What is the molality of Li + in this water ? Sol. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) STOICHIOMETRY - 1 30. A 6.90 M solution of KOH in water contains 30% by mass of KOH. What is density of solution in gm/ml. Sol. Page # 141 34. (a) Find molarity of Ca2+ and NO 3– in 2 M Ca(NO 3)2 aqueous solution of density 1.328 g/mL. (b) Also find mole fraction of solvent in solution. Sol. 35. 31. A solution of specific gravity 1.6 is 67% by weight. What will be % by weight of the solution of same acid if it is diluted to specific gravity 1.2 ? Calculate molality (m) of each ion present in the aqueous solution of 2M NH 4 Cl assuming 100% dissociation according to reaction. NH4 Cl (aq ) NH 4 (aq) Cl (aq ) Given : Density of solution = 3.107 gm/ml. Sol. Sol. 36. 32. Find out the volume of 98% w/w H2SO 4 (density = 1.8 gm/ml) must be diluted to prepare 12.5 litres of 2.5 M sulphuric acid solution. Sol. 500 ml of 2 M NaCl solution was mixed with 200 ml of 2 M NaCl solution. Calculate the final volume and molarity of NaCl in final solution if final solution has density 1.5 gm/ml. Sol. EXPERIMENTAL METHODS 33. Determine the volume of diluted nitric acid (d = 1.11 gmL–1 , 19% w/v HNO3) That can be prepared by diluting with water 50 mL of conc. HNO3 (d = 1.42 g ML–1, 69.8 % w/v). 37. Sol. What is the percentage of nitrogen in an organic compound 0.14 gm of which gave by Dumas method 82.1 c.c. of nitrogen collected over water at 27ºC and at a barometric pressure of 774.5 mm ? (aqueous tension of water at 27ºC is 14.5 mm) Sol. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com Page # 142 38. 0.20 gm of an organic compound was treated by Kjeldahl’s method and the resulting ammonia was passed into 50 cc of M/4 H2 SO4 . The residual acid was then found to require 40 cc of M/2 NaOH for neutralisation. What is the percentage of nitrogen in the compound ? STOICHIOMETRY - 1 Sol. Sol. 39. 0.252 gm of an organic compound gave on complete combustion 0.22 gm of carbon dioxide and 0.135 gm of water. 0.252 gm of the same compound gave by Carius method 0.7175 gm of silver chloride. What is the empirical formula of the compound ? Sol. 42. The molecular mass of an organi c aci d was determined by the study of its barium salt. 2.562 g of salt was quantitatively converted to free acid by the reaction 30 ml of 0.2 M H 2 SO 4, the barium salt was found to have two moles of water of hydration per Ba+2 ion and the acid is mono basic. What is molecular weight of anhydrous acid ? (At mass of Ba = 137) Sol. SOME TYPICAL CONCENTRATION TERMS 43. 40. 0.6872 gm of an organic compound gave on complete combustion 1.466 gm of carbon dioxide and 0.4283 gm of water. A given weight of the compound when heated with nitric acid and silver nitrate gave an equal weight of silver chloride. 0.3178 gm of the compound gave 26.0 cc of nitrogen at 15ºC and 765 mm pressure. Deduce the empirical formula of the compound ? Calculate composition of the final solution if 100 gm oleum labelled as 109% is added with (a) 9 gm water (b) 18 gm water (c) 120 gm water Sol. Sol. 44. For ‘44.8 V’ H2 O2 solution having d = 1.136 gm/ ml calculate (i) Molarity of H2 O2 solution. (ii) Mole fraction of H 2 O2 solution. Sol. 41. 0.80 g of the chloroplatinate of a mono acid base on ignition gave 0.262 g of Pt. Calculate the mol wt of the base. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) STOICHIOMETRY - 1 45. An oleum sample is labelled as 118%, Calculate (i) Mass of H2SO4 in 100 gm oleum sample. (ii) Maximum mass of H2 SO4 that can be obtained if 30 gm sample is taken. (iii) Composition of mixture (mass of components) if 40 gm water is added to 30 gm given oleum sample. Page # 143 Sol. Sol. 49. When a certain quantity of oxygen was ozonised in a suitable apparatus, the volume decreased by 4 ml. On addition of turpentine the volume further decreased by 8 ml. All volumes were measured at the same temperature and pressure. From these data, establish the formula of ozone. Sol. EUDIOMETRY 46. 10ml of a mixture of CO, CH 4 and N 2 exploded with excess of oxygen gave a contraction of 6.5 ml. There was a further contraction of 7 ml, when the residual gas treated with KOH. Volume of CO, CH4 and N2 respectively is Sol. 50. 47. When 100 ml of a O 2 – O 3 mixture was passed through turpentine, there was reduction of volume by 20 ml. If 100 ml of such a mixture is heated, what will be the increase in volume ? Sol. 48. 10 ml of ammonia were enclosed in an eudiometer and subjected to electric sparks. The sparks were continued till there was no further increase in volume. The volume after sparking measured 20 ml. Now 30 ml of O 2 were added and sparking was continued again. The new volume then measured 27.5 ml. All volume were measured under identical conditions of temperature and pressure. V.D. of ammonia is 8.5. Calculate the molecular formula of ammonia. Nitrogen and Hydrogen are diatomic. Sol. 60 ml of a mixture of nitrous oxide and nitric oxide was exploded with excess of hydrogen. If 38 ml of N 2 was formed, calculate the volume of each gas in the mixture. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com 1. EXERCISE – IV PREVIOUS YEARS L EVEL – I JEE MAIN The weight of 2.01 × 1023 molecules of CO is [AIEEE-2002] (A) 9.3 gm (C) 1.2 gm Sol. (B) 7.2 gm (D) 3 gm Sol. 5. 2. In an organic compound of molar mass 108 gm mol–1 C, H and N atoms are present in 9 : 1 : 3.5 by weight. Molecular formula can be – [AIEEE-2002] (A) C6H8N 2 (C) C5H6N 3 How many mol es of magnesium phosphate, Mg3(PO 4) 2 will contain 0.25 mole of oxygen atoms ? [AIEEE 2006] (A) 3.125 × 10 –2 (B) 1.25 × 10 –2 (C) 2.5 × 10 –2 (D) 0.02 Sol. (B) C7H10N (D) C4H18N3 Sol. 6. 3. Number of atoms in 560 gm of Fe (atomic mass 56 g mol–1) is – [AIEEE-2003] (A) is twice that of 70 gm N (B) is half that of 20 gm H (C) both are correct (D) None is correct (C) 67.2 L H 2(g) at STP is produced for every mole Al that reacts (D) 11.2 L H 2(g) at STP is produced for every mole HCl (aq) consumed Sol. 6.02 ×1020 molecules of urea are present in 100 ml of its solution. The concentration of urea solution is [AIEEE-2004] (A) 0.001 M (B) 0.01 M (C) 0.02 M (D) 0.1 M (Avogadro constant, N A = 6.02 ×10 23 mol –1) [AIEEE 2007] 2Al (s)+6HCl(aq) 2Al3+ (aq)+6Cl (aq)+3H2(g) , (A) 6L HCl (aq) is consumed for every 3L H2(g) produced (B) 33.6 L H 2(g) is produced regardless of temperature and pressure for every mole Al that reacts Sol. 4. In the reaction, Page # 145 STOICHIOMETRY - 1 LEVEL – II JEE ADVANCED 5. 1. How many moles of e– weigh one Kg (A) 6.023 × 1023 (C) 6. 023 10 54 9. 108 (B) 1 × 10 31 9.108 (D) 1 10 8 9.108 6.023 20% surface sites have adsorbed N2 . On heating N 2 gas evolved from sites and were collected at 0.001 atm and 298 K in a container or volume is 2.46 cm 3. Density of surface sites is 6.023 × 10 14/cm2 and surface area is 1000 cm 2 , find out the no. of surface sites occupied per molecule of N 2. [JEE 2005] Sol. [JEE ‘2002 (Scr), 1] Sol. 6. 2. Calculate the molarity of pure water using its density to be 1000 kg m–3. [JEE’ 2003] Sol. Sol. 7. 3. Dissolving 120 g of urea (mol. wt. 60) in 1000 g of water gave a solution of density 1.15 g/mL. The molarity of the solution is:[JEE 2011] (A) 1.78 M (B) 2.00 M (C) 2.05 M (D) 2.22 M One gm of charcoal absorbs 100 ml 0.5 M CH3 COOH to form a monolayer, and there by the molarity of CH3 COOH reduces to 0.49. Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid. Surface area of charcoal = 3.01 × 102 m2/gm. [JEE’ 2003] Reaction of Br2 with Na2CO3 in aqueous solution gives sodium bromide and sodium bromate with evolution of CO2 gas. The number of sodium bromide molecules involved in the balanced chemical equation is : [JEE 2011] Sol. Sol. 8. 4. Calculate the amount of Calcium oxide required when it reacts with 852 gm of P4O 10.[JEE 2005] Sol. A decapeptide (Mol. wt. 796) on complete hydrolysis gives glycine (Mol. Wt. 75), alanine and phenylalanine. Glycine contributes 47.0% to the total weight tto the hydrolysed products. The num ber of gl yci ne uni t s pre sent i n th e decapeptideis : [JEE 2011] Sol. 394,50 - Rajeev Gandhi Nag ar Kota, Ph. No. : 93 141-87482, 0744-22096 71 IVRS No : 0744-2439051, 52, 53, www. motioniitjee.com , info@motioniitjee.com Page # 146 STOICHIOMETRY - 1 ANSWER -KEY OBJECTIVE PROBLEMS (JEE MAIN) Answer Ex–I 1. A 2. A 3. A 4. 9. A 10. B 11. A 17. A 18. B 25. C 5. A 6. A 7. A 8. A 12. B 13. C 14. A 15. B 16. B 19. A 20. A 21. C 22. B 23. A 24. C 26. D 27. C 28. A 29. C 30. A 31.A 32.C 33. C 34. D 35. B 36. D 37. A 38. B 39.C 40. C 41. D 42. C 43. B 44. B 45. B 46. D 47.D 48. A 49. C 50. B Answer Ex–II 1. 196.169 gm A OBJECTIVE PROBLEMS (JEE ADVANCED) 2. 7.092 × 10 7 4. 1.0 × 10 –4% 5. 10 : 0.66 : 1 3. (a) 4 ; (b) 5000 moles; (c) 1.89 × 10 –22 gm 6. m = 4, C6H2Cl 3 7. (a) C6 H12, (b) C5 H10O5 , (c) H2O 2 , (d) Hg2 Cl 2, (e) H4 F4 10. Al = 60%; Mg = 40% 8. CH 9. C3H5N, C3H5N 11. CaCO3 = 28.4%; MgCO 3 = 71.6% 12. NaHCO3 = 14.9%; Na2 CO3 = 85.1% 13. %NaCl = 77.8% 14. (i) Fe2O 3 + 2Al Al 2 O3 + 2Fe; (ii) 80 : 27 ; (iii) 10,000 units 16. A,C 17. A, B 18. A,B,C 19. A,B 20. A,C,D 24. B,D 25. B,C 26. C 27. (A) R, (B) P, (C) Q 21. A,C 15. 9.12 22. B,D 23. A,C 28. (A) R, (B) Q, (C) P 29. (A) Q; (B) P; (C) S; (D) R 30. (A) R; (B) S; (C) P; (D) Q 31. (A) R, (B) Q, (C) S, (D) P 32. (a) C; (b) A; (c) B (d) A 34. (a) B; (b) C; (c) B 35. (a) A; (b) A 33. (a) B; (b) C; (c) B 36. (a) D; (b) B; (c) C; (d) A 37. 123 g/mol 38. C 44. 88 46. 25 40. D 41. A 42. C 43. D 47. 480 48. 24 49. 125 50. 113 45. 23 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 39. B STOICHIOMETRY - 1 Page # 147 SUBJECTIVE PROBLEMS (JEE ADVANCED) Answer Ex–III 1. (i) A2 = 1, A 3B 4 =2; (ii) B2 =1 A 2B 4=1/2; (iii) A 3 B4 = 0.5 A 2B 4 = 0.5 2. 24.51 ml 3. H2 SO4 = 35.38%, Free SO3 = 63.1%, combined SO3 = 28.89% 4. 18.38 ml 5. 1.052 gm/ml 6. C2 H6O 8. (i) 2/3 ; (ii) 2/9; (iii) m Cl2 7. (a) mA = 120, m D = 160; (b) n A 2 1000 mol / kg 9 18 2 , n C 5 1, n D 5 1 5 9. N2 = 30 ml, H2 = 40 ml 10. 495 11. AlCl 3 = 33.33 ; NaHCO3 = 50 ; KNO3 = 16.67 12.(a) CO 2, H2 O and O2 ; (b) nCO 2 = 3, n H2 O = 4 ; (c) C2H4 and CH4 are the H.C; (d) n O2 = 8 13. (a) 0.2; (b) 0.4 moles ; (c) 0.45 ; (d) 50.4 ‘V’ 14. 30 15. (a) 5, (b) 10, (c) 12 16. 0.25 mole 18. 1.1458 19. 21 : 11 17. wc = 24gm ; W CCl4 = 154 gm 20. (a) 117 kg; (b) 20.16 × 103 lit. 21. 1.14 gm 22. 12.15 gm, N 2 = 14.28% H2 = 42.86 %, NH3 = 42.86% 23. 59.72% 24. 0.9413 gram 25. (a) 0.5 M, (b) 0.5 M, (c) o.2 M 28. 16.66% 29. 2.7 × 10 –4 30. 1.288 31. 29.77 26. 1.445 27. 13.15 32. 1.736 litre33. 183.68 ml 34. (a) [Ca2+] = 2 molar [NO3 – ] = 4 molar; (b) 0.965 35. 0.6667, 0.6667 36. 2M 37. 66.67% 41. 92.70 42. 128 38. 35% 39. CH3Cl 40. C7 H10 NCl 43. (a) pure H2 SO4 (109 gm); (b) 109 gm H2 SO4, 9 gm H2O; (c) 109 gm H2SO4 , 111 gm H2O 44. (i) 4M; (ii) 0.06 45. (i) 20 gm H2 SO 4; (ii) 35.4 gm H2SO 4; (iii) H2 SO 4 = 35.4 gm, H2O = 34.6 gm 46. 5 ml, 2ml, 3ml 47. 10 ml 48. NO = 44 ml; N2O = 16 ml 49. O 3 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, 50. NH3 :info@motioniitjee.com Page # 148 STOICHIOMETRY - 1 Answer Ex–IV PREVIOUS YEARS PROBLEMS LEVEL – I 1. A 2. A JEE MAIN 3. C 4. B LEVEL – II 5. A 6. D JEE ADVANCED 1. D 2. 55.5 mol L –1 3. 5 × 10 –19 m2 4. 1008 gm 5. 0002 6. C 7. 0005 8. 0006 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) TRIGONOMETRIC RATIOS & IDENTITIES Page # 149 TRIGONOMETRIC RATIOS & IDENTITIES A. BASIC TRIGONOMETRIC IDENTITIES (a) sin² + cos² = 1 ; 1 (b) sec² ; sec ; cosec tan² = 1 (c) cosec² cot² = 1 sin 1; 1 1 cos 1 R R 1 R Important Trigonometric Ratios : (a) sin n = 0 ; (b) sin ( 2n 1) 2 cos n = (-1)n = ( 1)n & cos ; tan n = 0 where n ( 2n 1) = 0 where n 2 I I Trigonometric Functions Of Allied Angles : If is any angle , then (a) sin ( )= , 90 ± sin , 180 ± , 270 ± ; cos ( (b) sin (90°- ) = cos ; cos (90° (c) sin (90°+ ) = cos ; cos (90°+ ) = (d) sin (180° ; cos (180° ) = sin , 360 ± ; tan ( ; tan (90° ; tan (90°+ ) = cot cos ; tan (180° cos ; tan (180°+ ) = tan ; tan (270° ; tan (270°+ ) = – cot ) = sin sin )= (e) sin (180°+ ) = sin ; cos (180°+ ) = (f) sin (270° )= cos ; cos (270° (g) sin (270°+ ) = cos ; cos (270°+ ) = sin Ex.1 Express 1·2 radians in degree measure. Sol. 1·2 radians = 1·2 × = 180 degrees = 1·2 × etc. are called ALLIED ANGLES . ) = cos )= 180 22 / 7 [ sin = ) = – tan ) = cot ) = tan ) = cot 22 (approx).] 7 1·2 180 7 = 68·7272 = 68º (·7272 × 60)’ = 68º (43·63)’ 22 = 68º 43’ (·63 × 60)” = 68º 43’ 37·8” , 3 2 Calculate sin if cos Sol. For any angle belonging to the indicated interval sin =– 1 9 11 2 Ex.3 Calculate tan Sol. For any angle therefore tan =– =– 9 and 11 Ex.2 . is negative, and therefore sin = – 1 cos 2 2 10 . 11 if cos =– 5 and 5 , 3 2 . belonging to the indicated interval tan = 1 cos2 cos is positive and cos is negative, and = 2. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com TRIGONOMETRIC RATIOS & IDENTITIES Page # 150 Ex.4 Given that 5 cos2 2 sin Sol. Making a quadratic equation in sin2 (sin + 1) (5 sin = 2=0 3) = 0 sin 3 3 , = 2 2 4 cot 5 4 7 , then find the value of cot . 4 2 = 1 sin = 3 5 not possible as 5 4 7 4 3 =1 4 Ex.5 Prove that 3(sin x – cos x)4 + 4(sin6x + cos6x) + 6(sin x + cos x)2 = 13 Sol. L.H.S. = 3[(sin x – cos x)2]2 + 4[(sin2 x)3 + (cos2 x)3)] + 6(sin2 x+ cos2x + 2 sin x cos x) = 3 (sin2 x + cos2 x – 2 sin x cos x)2 + 4(sin2 x + cos2 x) (sin4x + cos4x – sin2 x cos2x)] + 6(sin2 x + cos2x + 2 sin x cos x) = 3(1 – 2 sin x cos x)2 + 4 [(sin4 x + cos4 x ) – sin2 x cos2 x] + 6 (1 + 2 sin x . cos x) = 3 (1 + 4 sin2x cos2x – 4 sin x cos x) + 4 [(sin2x + cos2x)2 – 2sin2 x cos2x – sin2 x cos 2x] + 6 + 12 sinx cos x = 3 + 12sin2x cos2x – 12 sin x cos x + 4 (1 – 3 sin2 x cos2x) + 6 + 12 sin x cos x = 3 + 12 sin2x cos2x + 4 – 12sin2x cos2x + 6 = 13 Ex.6 1 Simplify the expression b a b a sin x a . 1 Sol. b a sin x a 2 a b tan2 x where b > a > 0. After a few simple manipulations, this expression (for brevity denote it by P) can be rewritten P= sin x a a (b b tan 2 x sin x a a ) sin 2 x a cos 2 x b tan 2 x b sin 2 x Some students handle this as follows: a b tan2 x a b sin2 x cos2 x a cos2 x b sin2 x cos x and get a wrong answer: P = tan x. In this transformation what we actually have to simplify is the expression cos 2 x which is equal to |cos x|. And so the final result is P = sinx / |cos x|. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) TRIGONOMETRIC RATIOS & IDENTITIES Ex.7 If tan 1 = Let tan 2 1 =x= = 1 2 2 x 2 x2 + 2x – 1 = 0 sin 4 a 1 1 2 tan (0, 2 ), find the possible values of . 1 2 Sol. where 1 2 Page # 151 2 x= = = 2 1 cos 4 b 1 2 8 or =( 2 1) sin 8 cos 8 a3 b3 1 If Sol. Given sin 4 a or, or, or, or, or, b(a + b) sin4 b(a + b) sin4 (a + b)2 sin4 (a + b)2 sin4 [(a + b) sin2 + a(a + b) (1 – sin2 )2 = ab. + a(a + b) (1 + sin4 – 2sin2 ) = ab – 2a (a + b) sin2 + a2 + ab = ab – 2(a + b) sin2 . a + a2 = 0 – a]2 = 0 or, (a + b) sin2 –a=0 Now, sin8 a3 Ex.9 If – 2 <x< cos 4 b + 2 2 1 is not b/w (0, 2 ) 9 8 Ex.8 a b , prove that 8 b )3 (a 1 a b cos8 b3 = sin2 = a4 b4 + (a b )4 .a 3 (a b ) 4 b 3 a cos2 = a b = b a b a b a b 1 + = = ( a b )4 (a b )4 (a b )4 ( a b )3 and y = log10(tan x + sec x). Then the expression E = 10 y 10 2 y simplifies to one of the six trigonometric functions. find the trigonometric function. Sol. y = log10 (tan x + sec x), E= = 10 y 10 2 y y = log10 1 sin x = cos x cos x 1 sin x 2 1 sin x cos x = 1 sin2 x 2 sin x cos2 x 2 cos x(1 sin x) 2 sin x (1 sin x ) 2 sin2 x 2 sin x = = tan x 2 cos x (1 sin x ) 2 cos x(1 sin x ) : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com TRIGONOMETRIC RATIOS & IDENTITIES Page # 152 B. TRIGONOMETRIC FUNCTIONS OF SUM OR DIFFERENCE OF TWO ANGLES (a) sin (A ± B) = sinA cosB ± cosA sinB (b) cos (A ± B) = cosA cosB (c) sin²A (d) cos²A (e) tan (A ± B) = sin²B = cos²B sin²B = cos²B sinA sinB cos²A = sin (A+B) . sin (A B) sin²A = cos (A+B) . cos (A B) tan A tan B 1 tan A tan B (f) cot (A ± B) = cot A cot B 1 cot B cot A Factorisation Of The Sum Or Difference Of Two sines Or cosines : (a) sinC + sinD = 2 sin C D C D cos 2 2 (c) cosC + cosD = 2 cos (b) sinC C D C D cos 2 2 (d) cosC sinD = 2 cos cosD = C D C D sin 2 2 2 sin C D C D sin 2 2 Transformation Of Products Into Sum Or Difference Of sines & cosines : (a) 2 sinA cosB = sin(A+B) + sin(A B) (c) 2 cosA cosB = cos(A+B) + cos(A B) (b) 2 cosA sinB = sin(A+B) sin(A B) (d) 2 sinA sinB = cos(A B) cos(A+B) Ex.10 Suppose x and y are real numbers such that tan x + tan y = 42 and cot x + cot y = 49. Find the value of tan(x + y). Sol. tan x + tan y = 42 and cot x + cot y = 49 tan(x + y) = now, tan x tan y 1 tan x tan y tan x · tan y = tan (x + y) = (A) x + y + z = 0 x sin =y similarly 1 sin 2 x = z 1 = 49 tan y tan y tan x tan x · tan y = 49 tan x tan y 42 6 = = 49 49 7 42 42 1 ( 6 7 ) = 1 7 = 294 Ans. 2 3 Ex.11 If x sin = y sin Sol. 1 tan x cot x + cot y = 49 = z sin 4 3 then : (B) xy + yz + zx = 0 3 cos 2 3 cot 2 1 2 x 3 = cot y 2 on adding (C) xyz + x + y + z = 1 (D) none 1 2 x x + = 1 z y xy + yz + zx = 0 Ans. B Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) TRIGONOMETRIC RATIOS & IDENTITIES Page # 153 Ex.12 Find satisfying the equation, tan 15° · tan 25° · tan 35° = tan , where Sol. LHS = tan 15° · tan (30° – 5°) · tan (30° + 5°) let t = tan 30° and m = tan 5° tan = tan 15° · (0, 15°). t m t m t 2 m2 3m m3 1 3m2 · · = tan 3(5 ) · = 1 tm 1 tm 1 t 2 m2 1 3m 2 3 m 2 m (3 m 2 ) (1 3m 2 ) · = = m = tan 5°. Hence = 5° (1 3m 2 ) 3 m 2 tan 3 x 3 tan x tan3 x 1 3 tan2 x tan 30º t 2 1/ 3 t ; Ex.13 If tan A & tan B are the roots of the quadratic equation, a x2 + b x + c = 0 then evaluate a sin2 (A + B) + b sin (A + B) . cos (A + B) + c cos2 (A + B). Sol. b c ; tan A . tan B = a a tan A + tan B = b a = b c a 1 ac tan (A + B) = Now a b2 1 = b2 ( c a )2 1 = E = cos2 (A + B) [a tan2 (A + B) + b tan (A + B) + c] a )2 (c b 2 (c (c b2 a) 2 c b2 c a) 2 = a )2 (c b 2 (c b2 a) 2 c a a c a 1 c E=c c a )2 (c a c Ex.14 Show that cos2A + cos2(A + B) + 2 cosA cos(180° + B) · cos(360° + A + B) is independent of A. Hence find its value when B = 810°. Sol. cos2 A + cos2 (A + B) – [2 cosA · cosB · cos (A + B)] cos2 A + cos2(A + B) – [ {cos(A + B) + cos(A – B) } cos (A + B) ] cos2 A + cos2 (A + B) – cos2(A + B) – (cos2 A – sin2 B) = sin2 B which is independent of A now, sin2(810°) = sin2 (720° + 90°) = sin2 90° = 1 Ans. Ex.15 Simplify: cos x · sin(y – z) + cos y · sin(z – x) + cos z · sin (x – y) where x, y, z R. Sol. (1/2)[sin(y – z + x) + sin(y – z – x) + sin(z – x + y) + sin(z – x – y) + sin(x – y + z) + sin(x – y – z)] = 0 C. MULTIPLE ANGLES AND SUB-MULTIPLE ANGLES (a) sin 2A = 2 sinA cosA ; sin = 2 sin (b) cos 2A = cos²A cos = cos² 2 2 cos sin²A = 2cos²A 1 = 1 sin² 2 = 2cos² 2 1= 1 2 2 sin²A ; 2sin² 2 cos²A = 1 + cos 2A , 2sin²A = 1 cos 2A ; 2 cos² cos 2 = 1 + cos , 2 sin² 2 =1 2 . . : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com TRIGONOMETRIC RATIOS & IDENTITIES Page # 154 2 tan 2 2 tan A ; tan = 2 1 tan 2 2 1 tan A (c) tan 2A = (d) sin 2A = (f) cos 3A = 4 cos3A 2tan A 1 tan 2 A 1 tan 2 A , cos 2A = 1 tan 2 A (e) sin 3A = 3 sinA 4 sin3A 3 cosA (g) tan 3A = = cos 75° or cos 5 12 ; = sin 75° or sin 5 12 ; 3 tan A tan 3 A 1 3 tan 2 A Important Trigonometric Ratios (i) sin 15° or sin 12 cos 15° or cos sin (iii) sin Ex.16 If cot Sol. 10 = 2 2 = 3 1 2 2 2 2 2 3 = cot 75° ; ; cos or sin 18° = 8 2 2 5 1 & 4 sin 2 = 2 tan 1 tan 2 = 2·2 4 = ; 1 4 5 ; tan 8 = 2 cos 36° or cos 5 = 3 1 =2 3 1 1 ; tan 3 = 8 3 = cot 15° 2 1 5 1 4 cos 2 = 1 tan 2 1 tan2 = 1 4 3 =– 1 4 5 tan 8 = (1 + sec2 ) (1 + sec4 ) (1 + sec8 ). tan 1 cos 2 RHS = cos 2 1 cos 4 cos 4 2 cos 2 2 cos 2 2 2 cos 2 4 1 cos 8 = cos 2 cos 4 cos 8 cos 8 [ 8 cos cos 2 cos 4 ] cos = cos 8 Ex.18 If x = 7.5° then find the value of Sol. 2 = tan 75° = = 1/2, then find the values of sin2 and cos2 . Ex.17 Prove that Sol. 8 3 1 3 1 =2 3 1 tan 15° = (ii) 12 = sin 8 cos sin = cos 8 cos x cos 3x . sin 3x sin x cos x cos 3x 2 sin 2 x sin x = = tan 2x = tan (2 × 7.5) = tan 15° = 2 – 3 Ans. sin 3x sin x 2 sin x cos 2 x Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) TRIGONOMETRIC RATIOS & IDENTITIES Page # 155 Ex.19 Prove the identity, cos Sol. 3 2 4 + sin (3 8 ) sin (4 12 ) = 4 cos 2 cos 4 sin 6 . LHS : sin 4 + sin 8 + sin 12 = 2 sin 8 cos 4 + sin 8 = 2 sin 8 cos 4 + 2 sin 4 cos 4 = 2 cos 4 [sin 8 + sin 4 ] = 2 cos 4 [2 sin 6 cos 2 ] = 4cos 2 cos 4 sin 6 Ex.20 Calculate 4 sin 1 Sol. 4sin 1 6 cos 1 3 = 2 sin 2 Thus, 4 sin 1 Ex.21 If cos Sol. 1 cos = = tan2 cos 1 2 cos 2 cos 1 2 . = 2 sin 1 sin 2 3 ( 2) 2 6 1 = 3 tan2 2 1 1 cos cos tan2 2 if sin = = 2 we have cos 3 1 cos 1 cos cot2 2 1 2 = 2cos (–2) – 1 = 2 cos 2 – 1. 3 , 2 =– 2 Ans. 3 , 2 and 0 < < ) 3 . is negative for any angle belonging to the indicated interval, 3 . 5 , it follows that also negative, and therefore cos < (Componendo & dividendo) =3 4 and 5 = – 1 sin 2 3 = 2 cos 2 – 1. First of all we seek cos . Since cos Since sin 1 3 = 2 sin 6 3 1 6 2 cos 1 then find the value of tan cot (0 < 2 2 2 cos Ex.22 Calculate cos Sol. 6 cos 1 6 2 =– 3 , . For any angle belonging to this interval cos is 4 2 2 2 1 cos 2 5 . Thus cos 5 2 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, 5 . 5 :info@motioniitjee.com TRIGONOMETRIC RATIOS & IDENTITIES Page # 156 Ex.23 Calculate tan Sol. Since cos 2 7 and 32 1 cos 2 2 = , 3 4 =– 2 . belonging to the indicated interval, 39 . 8 , it follows that is negative, and therefore tan 3 4 , is negative for an angle we have cos Since if cos 2 = 2 2 1 cos 1 cos , 3 . For any angle belonging to this interval tan 8 2 2 tan 2 =– 8 39 5 . Ex.24 The figure (not drawn to scale) shows a regular octagon ABCDEFGH with diagonal AF = 1. Find the numerical value of the side of the octagon. Sol. = 22.5° ( AOB = 45º) A tan 22.5° = x 2 · 2 1 B H C O D G x = tan 22.5° = F 2 1 tan 1 cot = , find the value of . tan tan 3 3 cot cot 3 tan 1 = 3 tan = tan – tan 3 tan tan 3 3 E Ex.25 If Sol. 3 tan tan 3 2 tan + 1 3 tan 2 now, = cot cot cot 3 = 0, tan 3 tan 3 tan 2 tan 2(1 – 3 tan2 ) + 3 – tan2 = 0 3 tan (1 3 tan 2 ) + tan 3 = 0 tan2 = 5 7 tan 3 3 tan tan 3 1 3 tan 2 tan 16 3 (5 7 ) 2 3 tan 2 tan ( 3 tan 2 )(1 3 tan 2 ) = = = = Ans. 2 2 2 2 2 ·12 3 2(1 tan ) 2 1 (5 7) tan (1 3 tan )(3 tan 1 3 tan ) Alternatively: Prove that tan cot + tan tan 3 cot cot 3 = 1 now proceed Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) TRIGONOMETRIC RATIOS & IDENTITIES Page # 157 Ex.26 In a kite ABCD, AB = AD and CB = CD. If A = 108° and C = 36° then the ratio of the area of ABD to 2 the area of CBD can be written in the form Sol. a b tan 36 where a, b and c are relatively prime positive c integers. Determine the ordered triple (a, b, c). Since the triangles ABD and CBD have a common base, hence the ratio of their areas equals the ratio of their heights. h , then h = x tan 36°. x k |||ly tan 72° = then k = x tan 72°. x Since tan 36° = Hence, h x tan 36 = = k x tan 72 tan 36 2 tan 36 1 tan 2 36 = 1 tan 2 36 2 Then ordered triple (a, b, c) is (1,1, 2) Ex.27 If , , and be the roots of the equation, 2 cos 2 2 cos + 1 = 0, all lying in the interval [0, 2 ] then find the value of the product, cos . cos . cos . cos . Sol. 4 cos2 cos 2 cos 5 1 or cos 4 = = 1=0 or 5 cos 5 1 = 4 = = sin 2 10 4 8 = cos 16 = 5 10 1 5 4 10 = cos 6 10 9 3 7 ; or 5 5 5 Hence P = cos 5 cos 3 7 9 1 cos cos = 5 5 5 16 Ex.28 If sin x, sin22x and cos x · sin 4x form an increasing geometric sequence, find the numerical value of Sol. Given sin x, sin22x and cos x · sin 4x are in G.P. (r > 1 as G.P. is increasing) sin4 2x = (sin x) (cos x) (sin 4x) 16 sin4x cos4 x = sin x cos x sin 4x 3 3 16 sin x cos x = sin 4x (sin x 0, cos x 0) 16(sin x cos x)3 = 2 sin 2x · cos 2x (sin 2x)3 = sin 2x · cos 2x 2 2 sin 2x = cos 2x (sin 2x 0), 1 – cos 2x = cos 2x, y2 + y – 1 = 0 1 cos 2x = 5 2 ; cos 2x cannot be 1 1 cos 2x sin x = = 2 5 1 2 cos 2x = r= 5 1 2 · r= 5 1 hence rejected 2 cos 2x = 1 5 2 5 1 5 1 3 5 2 = = 2 2 2 2 sin 2 2 x = 4 sin x cos2x = 2 sin x(1 + cos 2x) sin x 4 5 1 = = 2 2 2 2 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com TRIGONOMETRIC RATIOS & IDENTITIES Page # 158 Ex.29 Prove using induction or otherwise that, 2 cos where R. H. S. contains n radical signs and Sol. 2 cos 2 2 cos 2 cos 2 (1 = 22 23 = = 2 1 cos 2 = 2 1 cos 2 2 cos 2 2 sin 2 2 2 n 2 2 (1 = 2 22 2n 1 = = 2 2 We have = 2 cos =2 2 cos 2 2n 1 2 ...... 2 22 = = 2 2 ...... 2 2 cos 2 2 2n Hence cos cos ) and so on. 2 (1 2 cos 1 where R. H. S. contains n radical signs 2 cos 1 2 7 . sin 2 = 2 sin cos , 15 15 15 sin 8 4 4 16 8 8 = 2 sin cos , sin = 2 sin cos . 15 15 15 15 15 15 cos cos 2 cos cos ) sin Multiplying the equalities and noting that sin Further ...... 2 (0 , ). 2 3 4 5 6 7 cos cos cos cos cos cos Ex.30 Show that cos 15 15 15 15 15 15 15 Sol. 2 cos ) In the same way Similarly 2n 15 5 15 . cos 4 2 2 = 2 sin cos , 15 15 15 16 8 7 = – sin , cos = – cos . 15 15 15 15 1 2 4 7 . cos . cos = 4 15 15 15 2 1 . and 2 sin 6 3 3 12 6 6 = 2 sin cos , sin = 2 sin cos . 15 15 15 15 15 15 1 3 6 . cos = 2 . The rest is obvious. 2 15 15 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) TRIGONOMETRIC RATIOS & IDENTITIES D. Page # 159 CONDITIONAL IDENTITIES tan A tan B tan C tan A tan B tan C 1 tan A tan B tan B tan C tan C tan A If A+B+C = then (a) tanA + tanB + tanC = tanA tanB tanC tan (A+B+C) = A B B C C A tan + tan tan + tan tan =1 2 2 2 2 2 2 (b) tan (c) sin2A + sin2B + sin2C = 4 sinA sinB sinC (d) sinA + sinB + sinC = 4 cos (e) cos 2A + cos 2B + cos 2C = –1 – 4 cos A cos B cos C (f) cos A + cos B + cos C = 1 + 4 sin Ex.31 If A + B + C = Sol. LHS= , prove that A B C cos cos 2 2 2 A B C sin sin 2 2 2 tan A = tan B . tan C tan 2 A tan 2 B tan 2 C (tan A = tan A tan B tan C tan B tan A tan A 2 cot A . tan C ) 2 2 tan A [ = 2 tan A tan B tan B tan C tan C tan A tan A tan B tan C tan A tan B tan A = = tan A] tan A 2 cot A] Ex.32 If A + B + C = and cot = cot A + cot B + cot C, show that , sin (A ) . sin (B ) . sin (C ) = sin3 . Sol. Given cot = cot A + cot B + cot C or cot cot A = cot B + cot C or sin (A ) sin (B C ) sin A = = sin sin A sin B sin C sin B sin C similarly sin (B )= or sin2 B sin sin C sin B sin (A (2) )= sin2 A sin sin B sin C sin (C )= (1) sin2 C sin sin A sin B (3) Multiplying (1) , (2) and (3) we get the result Ex.33 Find whether a triangle ABC can exists with the tangents of its interior angle satisfying, tan A = x, tan B = x + 1 and tan C = 1 – x for some real value of x. Justify your assertion with adequate reasoning. Sol. In a triangle tan A = tan A (to be proved) x + x + 1 + 1 – x = x(1 + x)(1 – x) 2 + x = x – x3; x3 = – 2; x = – 21/3 Hence tanA = x < 0 and tanB = x + 1 = 1 – 21/3 < 0 Hence A and B both are obtuse. Which is not possible in a triangle. Hence no such triangle can exist. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com TRIGONOMETRIC RATIOS & IDENTITIES Page # 160 Ex.34 Prove that (a) sin3 A cos (B – C) + sin3 B cos (C – A) + sin3 C cos (A – B) = 3 sin A sin B sin C; (b) sin3 A sin (B – C) + sin3 B sin (C – A) + sin3 C sin (A – B) = 0 if A + B + C = . Sol. (a) We have sin3 A cos (B – C) = = 1 2 sin2 A sin A cos (B – C) = sin2 A {sin (A + B – C) + sin (A – B + C)}. But since A + B + C = , we have = (b) sin3 A cos (B – C) = 1 2 sin2 A (sin 2C + sin 2B) sin2 A (sin B cos B + sin C cos C) = sin2 A sin B cos B + sin2 A sin C cos C + sin2 B sin C cos C + sin2 B sin A cos A + sin2 C sin A cos A + sin2 C sin B cos B = = sin A sin B (sin A cos B + cos A sin B) + sin A sin C (sin A cos C + cos A sin C) + sin B sin C (sin B cos C + cos B sin C) = sin A sin B sin (A + B) + sin A sin C sin (A + C) + sin B sin C sin (B + C) = 3 sin A sin B sin C. We have sin3 A sin (B – C) = = 1 2 sin2 A {cos 2C – cos 2B) = 1 sin2 C = sin2 A sin2 B sin2C × sin2 A sin A sin (B – C) = 1 sin2 C 1 sin 2 B 1 sin 2 A sin2 A sin (B + C) sin (B – C) sin2 A(sin2 B – sin2 C) 1 = sin2 A sin2 B sin2 C sin2 B 1 sin2 C 1 sin 2 B 1 sin 2 A =0 Ex.35 Given the product p of sines of the angles of a triangle & product q of their cosines, find the cubic equation, whose coefficients are functions of p & q & whose roots are the tangents of the angles of the triangle. Sol. Given sinA sinB sinC = p ; cosA cosB cosC = q Hence tanA tanB tanC = tanA + tanB + tanC = p/q Hence equation of cubic is x3 – now p p x2 + tanA tan Bx – =0 q q tan A tan B ...(i) sin A sin B cos C sin B sin C cos A sin C sin A cos B cos A cos B cos C We know that A + B + C = cos(A+B+C) = –1; cos(A+B) cosC – sin(A+B) sinC = –1 ( cosA cosB – sinA sin B) cosC – sinC (sinA cosB + cosA sinB) = –1 1+ cosA cosB cosC= sinA sinB cosC + sinB sinC cosA + sinC sinA cosB dividing by cosA cosB cosC 1 q q tan A tan B Hence (i) becomes qx3 – px 2 + (1 + q)x – p = 0 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) TRIGONOMETRIC RATIOS & IDENTITIES E. Page # 161 MAXIMUM & MINIMUM VALUES OF TRIGONOMETRIC FUNCTIONS (a) Min. value of a2 tan 2 (b) Max and Min. value of acos + bsin are a 2 (c) If f( ) = acos( – a2 (d) If b2 0, + b2 cot2 = 2ab ) + bcos( 2ab cos( 2 and ) = b 2 and – ) where a, b, a2 f( ) b2 and a2 b2 are known quantities then 2ab cos( ) (constant) then the maximum values of the expression cos cos , cos + cos , sin + sin and sin sin occurs when /2 (e) (f) (g) If 0, 2 and = (constant) then the minimum values of the expression sec + sec , tan + tan , cosec + cosec occurs when /2. If A, B, C are the anlges of a triangle then maximum value of sinA + sinB + sinC and sinA sinB sinC occurs when A = B = C = 60º In case a quadratic in sin or cos is given then the maximum or minimum values can be interpreted by making a perfect square Ex.36 Find the minimum vertical distance between the graphs of y = 2 + sin x and y = cos x. Sol. dmin = min(2 + sin x – cos x) = min[2 + 2 sin x 4 ]=2– 2 at x = 7 4 Ex.37 If a sin2x + b lies in the interval [–2, 8] for every x R then find the value of (a – b). Sol. f (x) = a sin2x + b f (x) has a maximum value of 8 which occurs when sin2x = 1 a+b=8 ....(1) |||ly f (x) has a minimum value of – 2 which occurs where sin x = 0 b=–2 ....(2) from (1) and (2) a = 10; b = – 2 a – b = 12 [Ans. 12] Ex.38 Find the greatest value of c such that system of equations x2 + y2 = 25; x + y = c has a real solution. Sol. put x = 5 cos y = 5 sin 5(cos hence, c max + sin ) = c; but (cos + sin )max = 2 and (cos + sin )min = – 2 5 2 Ans. Ex.39 Find the minimum and maximum value of f (x, y) = 7x2 + 4xy + 3y2 subjected to x 2 + y 2 = 1. Sol. Let x = cos and y = sin y = f ( ) = 7 cos2 + 4 sin cos + 3 sin2 = 3 + 2 sin 2 + 2(1 + cos 2 ) = 5 + 2(sin 2 + cos 2 ) ymax = 5 + 2 2 and but – ymin = 5 – 2 (sin 2 + cos 2 ) 2 2 2 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com TRIGONOMETRIC RATIOS & IDENTITIES Page # 162 Ex.40 If 1, (cos Sol. 2 1 , ...... , n are real numbers, show that, + cos 2 + ...... + cos n)2 + (sin 1 + ......+ sin L H S = (cos2 1 + sin2 1 ) + ....... + (cos2 + sin2 n n n )+2 )2 n2 . cos ( 1 2 ) n n+2 n ( n 1) = n2 2 Ex.41 Show that the expression cos (sin + Sol. Let y = cos (sin + or, or, F. sin 2 sin 2 sin 2 ) always lies between the values of ± 1 sin 2 ) sin 2 y – cos sin = cos ( sin 2 or, or, or, [Here or, or, or, or, C2 terms ) sin 2 (y – cos sin ) = cos (sin + sin ) y2 – 2ysin cos + cos2 = cos2 sin2 + cos2 sin2 y2 – 2ysin cos + cos2 = cos2 + cos2 . sin2 we have added cos2 on both sides to get 1 + sin2 ] y2 – 2y sin cos + cos2 = cos2 (1 + sin2 ) y2.sec2 – 2y tan + 1 = 1 + sin2 (dividing by cos2 ) 2 2 2 2 y tan – 2ytan + 1 = (1 + sin ) – y (sec2 = 1 + tan2 ) (ytan – 1)2 = (1 + sin2 ) – y2 square of a real number 0 1 + sin 2 – y2 0 2 y2 – ( 1 sin2 2 )2 2 2 0 y lies between – 1 sin 2 and 1 sin2 . SUMMATION OF TRIGONOMETRIC SERIES Sum of sines or cosines of n angles n sin + sin ( + ) + sin ( + 2 ) + ...... + sin ( sin 2 sin n 1 )= sin 2 n cos + cos ( + ) + cos ( + 2 ) + ...... + cos ( Ex.42 Find the sum of the series, cos 2n 1 + cos sin 2 cos n 1 ) = sin 2 n 1 2 n 1 2 3 5 + cos + ........ upto n terms. 2n 1 2n 1 Do not use any direct formula of summation. Sol. Let = 2n 1 S = cos + cos 3 + cos 5 + ........ cos (2n – 1) (2 sin ) S = 2 sin [cos + cos 3 + cos 5 + ........ cos (2n – 1) ] T1 = sin 2 – 0; T2 = sin 4 – sin 2 T3 = sin 6 – sin 4 Tn = sin 2n – sin 2(n – 1) 2n 2 n 1 =1 S= 2 2 sin 2n 1 sin (2 sin ) S = sin2n Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) . TRIGONOMETRIC RATIOS & IDENTITIES 35 Ex.43 Given sin 5k = tan k 1 m , where angles are measured in degrees, and m and n are relatively prime n positive integers that satisfy Sol. LHS: Page # 163 m < 90, find the value of (m + n). n S = sin 5 + sin 10 + sin 15 + .......... + sin 170 + sin 175 S 2 sin 5 5 = 2 sin [sin 5 + sin 10 + ......... + sin 175] 2 2 T1 = cos 2 sin 5 15 – cos ; 2 2 T2 = cos 15 25 – cos .........; 2 2 T35 = cos 345 355 – cos 2 2 5 5 355 180 175 175 · S = cos – cos = 2 sin · sin = 2 sin 2 2 2 2 2 2 175 2 S= 5 sin 2 sin 175 2 = 5 cos 90 2 175 175 m 2 = 175 = tan 2 = tan n cos 2 sin sin m = 175 and n = 2 m + n = 177 Ex.44 Find the sum of the series , cot 2 x . cot 3 x + cot 3 x . cot 4 x + ...... + cot (n + 1) x . cot (n + 2) x . Sol. cot x = cot [ (n + 2) x (n + 1) x ] = cot ( n 2 ) x .cot ( n 1) x 1 cot ( n 1 ) x cot ( n 2 ) x or cot x [ cot (n + 1) x cot (n + 2) x ] = cot (n + 2) x . cot (n + 1) x + 1 Hence cot (n + 1) x cot (n + 2) x = cot x [ cot (n + 1) x cot (n + 2) x ] 1 Put n = 1 , 2 , 3 , ...... , n and adding we get sum of the series = cot x [ cot 2 x cot (n + 2) x ] n 2 sin Ex.45 Let f (x) denote the sum of the infinite trigonometric series, f (x) = n 1 2x x sin n . n 3 3 Find f (x) (independent of n) also evaluate the sum of the solutions of the equation f (x) = 0 lying in the interval (0, 629). Sol. sin f (x) = n 1 2x x 1 sin n = n 3 3 2 2 sin n 1 1 2x x sin n = n 2 3 3 cos n 1 x 3n cos x 3 n 1 now substituting n = 1, 2, 3, 4........ f (x) = 1 x cos 2 3 f (x) = Lim n cos x + 1 x cos n 2 3 1 x cos 2 2 3 cos x = cos x 3 + 1 x cos 3 2 3 cos 1 [1 – cos x] now f (x) = 0 2 x 1 x cos n 2 .......... + 2 3 3 cos x = 1 cos x 3 n 1 x = 2n , n sum of the solutions in (0, 629), S = 2[ + 2 + 3 + ....... + 100 ] = 2 · 5050 = 10100 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com I TRIGONOMETRIC RATIOS & IDENTITIES Page # 164 89 1 Ex.46 Evaluate (tan n ) 2 n 11 1 Sol. S= . 1 1 (tan 1 ) 2 1 1 (tan 2 ) 2 1 (tan 3 ) 1 ........ 2 1 1 (tan 88 ) 2 1 (tan 89 ) 2 reversing the sum 1 S= 1 1 (cot 1 ) 89 2S = 2 1 (cot 2 ) 1 n 11 89 1 (tan n ) 2 1 .............................. 2 1 (cot n ) 2 = n 1 1 1 (cot 88 ) 2 1 (tan n ) 2 1 (tan n ) 2 1 (tan n ) 2 1 (cot 89 ) 2 89 1 = 1 + 1 + ....... + 1 = 89 = S = 44.5 n 1 G. ELIMINATION Ex.47 Eliminate between the equation a sec + b tan + c = 0 and p sec + q tan + r = 0. Sol. Given a sec + b tan + c = 0 ...(1) and p sec + q tan + r = 0 ...(2) Solving (1) and (2) by cross multiplication method, we have sec br qc tan pc ar br qc aq pb 2 1 aq pb sec2 – tan2 =1 2 pc ar aq pb =1 or, (br – qc)2 – (pc – ar)2 = (aq – pb)2 Ex.48 If is eliminated from the equations, a cos + b sin = c & a cos2 eliminant is, (a b)2 (a c) (b c) + 4 a2 b2 = 0 . Sol. a cos + b sin = c ..............(1) a cos2 + b sin2 = c ..............(2) sin2 From (2) Now squaring (1) a2 or b b c c + b2 a b a2 (b = c b a2 cos2 a a c2 = c) + b2 (c (a b) (b (a (a b)2 (b b)2 (b a b and cos2 = a b c) (c + b 2 sin2 2 ab a) b b c2 (b a) = 2 ab c b cos = c2 a a a) = 2 ab b c c = c, show that the c a + 2 ab sin c a + b sin2 b c c a a c)2 (c a)2 = 4 a2 b2 (b c) (c c) (c a) = 4 a2 b2 Result a) Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) TRIGONOMETRIC RATIOS & IDENTITIES Page # 165 Miscellaneous Questions Ex.49 Prove that Sol. Let = tan 7 . tan 7 = 7 or, 4 +3 = or, 4 tan 4 tan3 1 6 tan2 tan 4 4z or, 2 3 . tan = 7 7 7 1 6z or, 4z 3 2 z =– 3z 4 tan(4 ) = tan( – 3 ) 3 tan or, tan4 = –tan3 tan 3 1 3 tan 2 z3 [where tan 1 3z2 = z (suppose)] or, (4 – 4z2) (1 – 3z2) = –(3 – z2)(1 – 6z2 + z4) or 12z 4 – 16z2 + 4 = –(–z6 + 9z4 – 19z2 + 3) or, z6 – 21z4 + 35z2 – 7 = 0 ...(1) This is cubic equation in z2 i.e. in tan2 , the roots of this equation are therefore tan2 7 , tan2 2 3 and tan2 7 7 From (1), product of the roots = 7 tan2 7 . tan 2 2 3 . tan2 =7 7 7 tan 7 . tan 2 3 . tan = 7 Hence the result. 7 7 Ex.50 In triangle ABC, cos A . cos B + cos B . cos C + cos C . cos A = 1 – 2 cos A . cos B . cos C. Prove that it is possible if and only if ABC is equilateral. Sol. cos A . cos B = 1 – 2 cos A . cos B . cos C = 1 – cos C (cos (A+ B) + cos (A – B) ) = 1 – cos C (cos (A – B) – cosC) = 1 + cos (A + B) cos (A – B) +cos2 C = 1 + cos2 A – sin2 B + cos2C = cos2 A + cos2 B +cos2C = Thus we have, 2 cos A – 2 cos2A. cos A . cos B = 0 2 2 (cos A – cos B) + (cos B – cos C)2 + (cos C – cos A)2 = 0 cos A = cos B = cos C A= Thus triangle ABC is equilateral Now if =1– is equilateral 2 8 A= B= C= 3 cosA cos B = 3 and 1 – 2 cos A cos B cos C 4 3 . Hence the given expression is true if and only if ABC is equilateral. 4 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com B= C TRIGONOMETRIC RATIOS & IDENTITIES Page # 166 EXERCISE – I JEE MAIN Sol. 1. If tan +cot =a then the value of tan4 +cot4 = (A) a4 + 4a2 +2 (B) a4 – 4a2 + 2 4 2 (C) a – 4a – 2 (D) None of these Sol. 2. If a cos + b sin = 3 & a sin – b cos = 4 then a2 + b2 has the value = (A) 25 (B) 14 (C) 7 (D) None of these Sol. 5. The expression 3 4 3 sin 2 sin 4 (3 is equal to (A) 0 (B) 1 Sol. 3. The value of tan 1º tan 2º tan 3º ..... tan 89º is (A) 1 (B) 0 (C) (D) 1/2 Sol. tan x 4. 2 cos cos x 3 2 2 x . tan sin3 3 2 7 2 ) –2 sin 6 (C) 3 2 sin 6 (5 ) (D) sin 4 + sin 6 6. cos (540º – ) – sin (630º – ) is equal to (A) 0 (B) 2 cos (C) 2 sin (D) sin –cos Sol. x x when simplified reduces to : (A) sinx cosx (B) – sin2x (C) –sinx cosx (D) sin2x 7. The value of sin( (A) –1 (B) 0 ) sin ( (C) sin ) cosec2 is equal to (D) None of these Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) TRIGONOMETRIC RATIOS & IDENTITIES Page # 167 Sol. 11. In a triangle ABC if tan A < 0 then : (A) tan B . tan C > 1 (B) tan B . tan C < 1 (C) tan B . tan C = 1 (D) None of these Sol. 8. If sin sin – cos cos + 1 = 0, then the value of 1 + cot tan is (A) 1 (B) –1 (C) 2 (D) None of these Sol. 12. If tan A – tan B = x and cot B – cot A = y, then cot (A – B) is equal to (A) 1 y 1 1 (B) x x 1 1 (C) y x 1 (D) None of these y Sol. 9. The value of (A) –1 Sol. (B) 1 sin 24 º cos 6º sin 6º sin 66 º is sin 21º cos 39º cos 51º sin 69º (C) 2 (D) None of these 13. If tan 25º=x, then (A) tan 10. If 3 sin = 5 sin , then tan (A) 1 Sol. (B) 2 (C) 3 2 2 1 x2 1 x2 (B) 2x 2x tan155º tan 115º is equal to 1 tan 155º tan115º 1 x2 1 x2 (C) (D) 2 1 x 1 x2 Sol. is equal to (D) 4 14. If A + B = 225º, then the value of cot A cot B . is 1 cot A 1 cot B (A) 2 (B) 1/2 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, (C) 3 (D) 1/3 :info@motioniitjee.com TRIGONOMETRIC RATIOS & IDENTITIES Page # 168 Sol. 18. If A lies in the third quadrant and 3 tan A – 4 = 0, then 5 sin 2A + 3 sinA + 4 cosA is equal to (A) 0 (B) – 24 5 (C) 24 5 (D) 48 5 Sol. 15. The value of tan 3A – tan 2A – tan A is equal to (A) tan 3A tan 2A tan A (B) – tan 3A tan 2A tan A (C) tan A tan 2A – tan 2A tan 3A – tan 3A tan A (D) None of these Sol. 19. cos 20º 8 sin 70 º sin 50º sin10º is equal to sin2 80 º (A) 1 Sol. (B) 2 (C) 3/4 (D) None of these 16. tan 203º + tan 22º + tan 203º tan 22º = (A) –1 (B) 0 (C) 1 (D) 2 Sol. 20. If cos A = 3/4, then the value of 16cos2 (A/2) – 32 sin (A/2) sin (5A/2) is (A) –4 (B) –3 (C) 3 (D) 4 Sol. 17. The value of (A) 1 (B) 1 tan2 15º is 1 tan2 15 º 3 (C) 3 2 (D) 2 Sol. 21. The value of the expression 1 cos (A) 1/8 10 1 cos 3 10 (B) 1/16 1 cos 7 9 1 cos 10 10 (C) 1/4 (D) 0 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) is TRIGONOMETRIC RATIOS & IDENTITIES Page # 169 Sol. Sol. 24. cos0+cos +cos 7 (A) 1/2 Sol. 2 3 4 5 6 +cos +cos +cos +cos = 7 7 7 7 7 (B) –1/2 (C) 0 (D) 1 22. The numerical value of sin 12º . sin 48º . sin 54º is equal to (A) 1/2 (B) 1/4 (C) 1/16 (D) 1/8 Sol. 25. A regular hexagon & a regular dodecagon are inscribed in the same circle. If the side of the dodecagon is ( 3 –1), then the side of the hexagon is (A) 23. If (A) tan (B) tan (C) tan (D) tan Sol. = 2 , then 2 2 2 2 + tan tan 2 + tan tan 2 2 + tan + tan 2 2 + tan + tan 2 2 = tan tan 2 2 + tan = – tan tan 2 2 2 + tan tan 2 tan tan 2 2 2 tan tan 2 3 1 2 (C) 2 (D) 2 =1 tan 2 2 +1 (B) 2 =0 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com 2 TRIGONOMETRIC RATIOS & IDENTITIES Page # 170 26. In a right angled triangle the hypotenuse is 2 2 times the perpendicular drawn from the opposite vertex. Then the other acute angles of the triangle are 3 (A) & (B) & 3 6 8 8 (C) 4 & 4 3 (D) & 5 10 28. The value of cot x + cot(60º + x) + cot (120º + x) is equal to (A) cos 3x (B) tan 3x (C) 3 tan 3x (D) 3 9 tan2 x 3 tan x tan3 x Sol. Sol. , 29. If x 27. If 2 , then the value of 4 cos2 – 1 sin (A) 2 cos 2 1 sin is equal to (B) 2 sin 2 (C) 2 (D) None of these (A) 1 Sol. 4 3 2 x 2 (B) 2 then 4 sin4 x sin2 2x is always equal to (C) –2 (D) None of these Sol. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) TRIGONOMETRIC RATIOS & IDENTITIES 30. The expression Page # 171 cos 6x 6 cos 4x 15 cos 2x 10 cos 5 x 5 cos 3x 10 cos x is equal to (A) cos 2x (B) 2 cos x Sol. (C) cos2 x Sol. (D) 1 + cos x 33. For – 31. If cos (A – B) = 3/5 and tan A tan B = 2, 1 5 1 (C) cos (A + B) = – 5 (A) cosA cosB = – 2 (A) (– Sol. < < sin sin 2 lies in the interval 2 1 cos cos 2 , ) (B) (–2, 2) (C) (0, ) (D) (–1, 1) 2 5 4 (D) sin A cos B = 5 (B) sinA sinB = – Sol. 34. If 0 < x < 3 32. If A + B + C = , then cos 2A + cos 2B + cos 2C 2 is equal to (A) 1–4cos A cosB cosC (B) 4 sinA sinB sinC (C) 1+2 cosA cosB cosC (D) 1–4 sinA sinB sinC (A) (C) (4 7) 3 (1 1 , then tan x is 2 and cos x + sin x = 7) 4 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, (B) – (D) (4 7) 3 (1 7) 4 :info@motioniitjee.com TRIGONOMETRIC RATIOS & IDENTITIES Page # 172 Sol. Sol. 37. Which is correct one ? (A) sin 1° < sin 1 (B) sin 1° = sin 1 (C) sin 1° > sin 1 (D) sin 1° = sin Sol. 35. Let If sin be such that + sin (A) – 3 130 3 . 21 =– and cos 65 the value of cos (B) 180 27 , then 65 + cos =– 6 65 (D) – is 2 3 130 (C) 6 65 Sol. 38. The value of cos 10° – sin 10° is (A) Positive (B) Negative (C) 0 Sol. (D) 1 36. The value of the expression cos 1° cos 2° ......... cos 179° equals (A) 0 (B) 1 (C) 1/ 2 (D) – 1 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) TRIGONOMETRIC RATIOS & IDENTITIES Page # 173 EXERCISE – II JEE ADVANCED (OBJECTIVE ) LEVEL – I SINGLE CORRECT 1. If tan A and tan B are the roots of the quadratic equation x2 – ax + b = 0, then the value of sin2 (A + B) (A) a 2 a2 (1 b )2 (B) a2 (C) (b c )2 Sol. a2 a 2 b2 a2 (D) 2 b (1 a )2 Sol. 4. If A + B + C = 2. If A = tan 6º tan 42º and B = cot 66º cot 78º, then (A) A = 2B (B) A = 1/3 B (C) A = B (D) 3A = 2B Sol. & sin A A B tan = 2 2 k 1 k 1 (A) (B) k 1 k 1 C C 2 = k sin 2 , then tan (C) k k 1 (D) k 1 k Sol. 3. 1 cos 290 º 2 3 (A) 3 1 3 sin 250 º 4 3 (B) 3 (C) = 5. In any triangle ABC, which is not right angled 3 (D) None of these cos A . cosec B . cosec C is equal to (A) 1 (B) 2 (C) 3 (D) None of these : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com TRIGONOMETRIC RATIOS & IDENTITIES Page # 174 Sol. 9. The value of tan (A) cot 8 (B) cot 16 16 + 2 tan (C) cot 16 8 + 4 is equal to –4 (D) None of these Sol. 6. If 3 cos x + 2 cos 3x = cos y, 3 sin x + 2 sin 3x = sin y, then the value of cos 2x is (A) –1 (B) 1/8 (C) –1/8 (D) 7/8 Sol. 10. The value of cos is equal to (A) 1/2 (B) 0 Sol. 7. If cos then + cos cos 3 cos = a, sin + sin = b and 19 +cos 3 5 17 +cos +...+ cos 19 19 19 (C) 1 (D) None of these =2 , = (A) a2 + b2 – 2 (C) 3 – a2 – b2 Sol. (B) a2 + b2 – 3 (D) (a2 + b2) /4 11. If 3 < 4 (A) 1 +cot (C) 1 – cot Sol. , then 2 cot 1 sin 2 is equal to (B) –1 – cot (D) –1 + cot 8. If A + B + C = & cos A = cos B . cos C then tan B . tanC has the value equal to (A) 1 (B) 1/2 (C) 2 (D) 3 Sol. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) TRIGONOMETRIC RATIOS & IDENTITIES 12. If f( ) = sin4 1 ,1 (A) 2 + cos2 , then range of f( ) is 1 3 3 , ,1 (B) (C) 2 4 4 (D) None of these 13. If 2 cos x + sin x = 1, then value of 7 cos x + 6 sin x is equal to (A) 2 or 6 (B) 1 or 3 (C) 2 or 3 (D) None of these Sol. 14. If cosec A + cot A = Sol. 21 22 (B) 15 16 3 15. If 0° < x < 90° & cos x = Sol. (A) Page # 175 11 , then tan A is 2 44 117 (C) (D) 117 43 10 , then the value of log10 sin x + log10 cos x + log10 tan x is (A) 0 (B) 1 (C) –1 (D) None of these Sol. 16. If cot + tan = m and 1 cos – cos = n, then (A) m (mn2)1/3 – n(nm2 )1/3 = 1 (B) m(m2n)1/3 – n(nm2 )1/3 = 1 (C) n (mn2)1/3 – m(nm2)1/3 = 1 (D) n(m2 n)1/3 – m(mn2 )1/3 = 1 Sol. 17. If 2 sec2 – sec4 – 2 cosec2 4, then tan is equal to (A) 1/ 2 Sol. (B) 1/2 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, (C) 1/2 2 + cosec4 (D) 1/4 :info@motioniitjee.com = 15/ TRIGONOMETRIC RATIOS & IDENTITIES Page # 176 20. If sin 2 = k, then the value of tan3 1 tan 2 cot 3 1 cot 2 is equal to (A) 1 k2 2 k2 (B) k k (C) k 2 + 1 (D) 2 – k2 Sol. 18. If sin A sin B 3 cos A and 2 cosB 5 , 0 < A, B < /2, 2 then tan A + tan B is equal to (A) 3 / 5 (B) 5 / 3 (C) 1 (D) ( 5 3)/ 5 Sol. 21. If f( ) = sin2 then f (A) 2 3 15 2 3 + sin2 4 , 3 + sin2 is equal to (B) 3 2 (C) 1 3 (D) Sol. 19. If 3 sin x + 4 cos x = 5 then 4 sin x – 3 cos x is equal to (A) 0 (B) 1 (C) 5 (D) None of these Sol. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 1 2 TRIGONOMETRIC RATIOS & IDENTITIES LEVEL – II MULTIPLE CORRECT sin x cos x 1. The value of = cos3 x (A) 1+tanx + tan2x –tan3x (C) 1–tanx + tan2x +tan3x Sol. Page # 177 (B) 1+tan x+tan2x+tan3x (D) (1 + tan x) sec2x 2. If (sec A + tan A) (sec B + tan B) (sec C + tan C) = (sec A – tan A) (sec B – tan B) (sec C – tan C) then each side is equal to (A) 1 (B) –1 (C) 0 (D) None of these Sol. 4. If tan2 = 2 tan2 cos 2 + sin2 is (A) 1 (B) 2 + 1, then the value of (C) –1 (D) Independent of Sol. 5. The value of cos (A) (C) 10 10 cos 2 4 8 16 cos cos cos is 10 10 10 10 2 5 64 (B) – cos( / 10 ) 16 (D) – cos( / 10) 16 10 2 5 64 Sol. 3. The value of (cos 11º sin 11º ) is (cos 11º sin11º ) (A) –tan 304º (B) tan 56º (C) cot 214º (D) cot 34º Sol. 6. If x + y = z, then cos2 x + cos2 y + cos 2 z – 2 cos x cos y cos z is equal to (A) cos2 z (B) sin2 z (C) cos (x + y – z) (D) 1 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com TRIGONOMETRIC RATIOS & IDENTITIES Page # 178 Sol. 9. An extreme value of 1 + 4 sin + 3 cos is (A) – 3 (B) – 4 (C) 5 (D) 6 Sol. 7. If tan A + tan B + tan C = tan A . tan B . tan C, then (A) A, B, C may be angles of a triangle (B) A + B + C is an integral multiple of (C) sum of any two of A, B, C is equal to third (D) None of these Sol. 8. In a triangle tan A + tan B + tan C = 6 and tan A tan B = 2, then the values of tanA, tan B and tan C are (A) 1, 2, 3 (B) 2, 1, 3 (C) 1, 2, 0 (D) None of these Sol. 10. If the sides of a right angled triangle are {cos2 + cos2 + 2cos( )} and {sin2 + sin2 + 2sin( )},then the length of the hypotneuse is (A) 2 [1 + cos( )] (B) 2 [1 – cos( )] (C) 4 cos2 2 (D) 4 sin2 2 Sol. 11. For 0 < < /2, tan + tan 2 + tan 3 = 0 if (A) tan = 0 (B) tan 2 = 0 (C) tan 3 = 0 (D) tan tan 2 = 2 Sol. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) TRIGONOMETRIC RATIOS & IDENTITIES 12. (a+2) sin (A) 3/4 + (2a – 1) cos (B) 4/3 (C) Page # 179 = (2a+1) if tan 2a (D) a2 1 = Sol. 2a a2 1 Sol. 15. The equation sin6x + cos6x = a2 has real solution if (A) a (C) a Sol. 13. If tan x = 2b , (a a c (–1,1) 11 22 (B) a (–1, –1/2) (D) a (1/2, 1) c) y = a cos2x + 2b sin x cos x + c sin2x z = a sin2x – 2b sin x cos x + c cos2x, then (A) y = z (B) y + z = a + c (C) y – z = a – c (D) y – z = (a – c)2 + 4b2 Sol. 16. If 3 sin =sin (2 + ), then tan ( ) – 2 tan (A) independent of (B) independent of (C) dependent of both and (D) independent of but dependent of Sol. 14. cos A sin A (A) 2 tann cos B sin B A B 2 (C) 0 : n is odd n sin A cos A sin B cosB (B) 2 cotn n A B : n is even 2 (D) None of these : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com is TRIGONOMETRIC RATIOS & IDENTITIES Page # 180 EXERCISE – III JEE ADVANCED Comprehension # 1 Comprehension # 2 If cos + cos = a and sin + sin = b and is arithmetic mean between and , then sin 2 + cos 2 = 1 + nb(a b) . a2 b2 4. In this triangle tan A + tan B + tan C is equal to where n is some integer then answer the following questions : 1. The value of n is (A) 0 (B) 1 Sol. Let p be the product of the sines of the angles of triangle ABC and q is the product of the cosines of the angles. (A) p + q (B) p – q (C) p q (D) None of these Sol. (C) 2 (D) – 2 n 2. If for n obtained in above question, sin A = x, then sin A sin 2A sin 3A sin 4A is a polynomial in x, of degree (A) 5 (B) 6 (C) 7 (D) 8 Sol. 5. tan A tan B + tan B tan C + tan C tan A is equal to 3. If degree of polynoimal obtained in previous question is p and (p – 5) + sin x, cos x, tan x are in 9 6 5 G.P., then cos x + cos x + 3 cos x – 1 = (A) –1 (B) 0 (C) 1 (D) None of these Sol. 6. The value of tan A + tan B + tan C is (A) 1 + q (B) 1 q q (C) 1 + p (D) 1 p p Sol. 3 (A) p3 3pq 2 q3 (B) 3 3 q p 3 (C) 3 p3 q3 (D) Sol. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) p3 3pq q3 TRIGONOMETRIC RATIOS & IDENTITIES Page # 181 Matrix Match Type 1. Column - I (A) sin 420° cos 390° + cos (–660°) sin (–330°) (B) tan 315° cot (–405°) + cot 495° tan (–585°) (C) The value of (D) Value of 4 Column - II (P) 0 (1 tan 8 )(1 tan 37 ) = (1 tan 22 )(1 tan 23 ) 1 2 sin x is 3 (Q) 1 (R) 2 (S) 5 (where [.] represents greatest integer function) Sol. 2. Column - I (A) If for some real x, then equation x+ Column - II (P) 2 1 = 2 cos holds x then cos is equal to (B) If sin + cosec = 2, 2008 2008 then sin + cosec is equal to 4 4 (C) Maximum value of sin + cos is 2 2 (D) Least value of 2 sin + 3 cos is Sol. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com (Q) 1 (R) 0 (S) –1 TRIGONOMETRIC RATIOS & IDENTITIES Page # 182 SUBJECTIVE TYPE 1. Eliminate from the relations a sec a2 sec2 = 5 + b2 tan2 Sol. 2. If tan = –5/12, then show that = 1 – b tan , is not in the second quadrant, sin( 360º ) tan( 90 º ) sec(270 º ) cos ec ( ) 181 338 Sol. 3. Prove that 1 cot 2 1 cot 2 4 cos 2 cot 4 sec 9 = cosec 4 . 2 4 Sol. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) TRIGONOMETRIC RATIOS & IDENTITIES Page # 183 6. If 0 < 2 < /4, then show that 2(1 cos 4 ) = 2 cos . 4. Prove than sec 8 A 1 (i) sec 4 A 1 Sol. tan 8A tan 2A Sol. 7. Prove that tan tan (60º + ) tan (60º – ) = tan 3 and hence deduce that tan 20º tan 40º tan 60º tan 80º = 3. Sol. (ii) cos A sin A cos A sin A – = 2 tan 2A cos A sin A cos A sin A Sol. 5. If A+B =45º, prove that (1+tan A) (1+tan B) = 2 and hence deduce that tan 22 1º 2 8. Prove that 4(cos3 20º+cos3 40º)=3(cos 20º+cos 40º) Sol. 2 1 Sol. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com TRIGONOMETRIC RATIOS & IDENTITIES Page # 184 9. Prove that sin2 +sin22 +sin23 +....+sin2 n = n sin n cos(n 1) – 2 2 sin Sol. 11. If x + y + z = 2 show that, sin 2x + sin 2y + sin 2z = 4 cosx cosy cosz. Sol. 12. If x + y = + z, then prove that sin2x + sin2y – sin2z = 2 sin x sin y cos z. Sol. 10. If is the exterior angle of a regular polygon of n sides and is any constant, then prove that sin + sin ( ) + ....... up to n terms = 0 Sol. 13. If A + B + C = 2S then prove that cos (S – A) + cos(S – B) + cos (S – C) + cos S = 4 cos A B C cos cos 2 2 2 Sol. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) TRIGONOMETRIC RATIOS & IDENTITIES 14. If A + B + C = 0º then prove that sin 2A + sin 2B + sin 2C = –4 sin A sin B sin C. Sol. Page # 185 17. Prove that, sin3x . sin3 x + cos 3 x . cos3 x = cos3 2x. Sol. 15. Find the extreme values of 2 3 cos x cos x cos 2 3 x Sol. 18. If tan sin 2 = = tan tan , prove than 1 tan . tan sin 2 sin 2 . 1 sin 2 . sin 2 Sol. 16. Find the maximum and minimum values of (i) cos 2x + cos2 x Sol. (ii) cos 2 Sol. 4 x (sin x – cos x)2 19. Show that : (i) cot 7 or 2 1º 1º or tan 82 = ( 3 2 2 3 4 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, 2 )( 2 1) 6 :info@motioniitjee.com TRIGONOMETRIC RATIOS & IDENTITIES Page # 186 Sol. (ii) tan 142 1º =2+ 2 2 3 6. Sol. 21. Calculate the following without using trigonometric tables : (i) tan 9º – tan 27º – tan 63º + tan 81º Sol. 20. If sin x + sin y = a & cos x + cos y = b, show that, sin (x + y) = 2ab x y 4 a 2 b2 =± . 2 and tan a b 2 a 2 b2 2 Sol. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) TRIGONOMETRIC RATIOS & IDENTITIES (ii) cosec 10º – Page # 187 3 sec 10º Sol. (iii) 2 2 sin10º sec 5º 2 cos 40º sin5º 2 sin35º Sol. 22. If cos ( prove that cos + cos Sol. ) + cos ( + cos = 0, sin 3 , 2 ) + cos ( )= + sin + sin = 0 (iv) cot 70º + 4 cos 70º Sol. 23. If (v) tan 10º – tan 50º + tan 70º Sol. ax cos by sin = a2 – b2, Show that (ax)2/3 + (by)2/3 Sol. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, ax sin by cos cos 2 sin 2 = (a2 – b2)2/3 :info@motioniitjee.com = 0. TRIGONOMETRIC RATIOS & IDENTITIES Page # 188 24. If Pn = cosn + sinn and Qn = cosn – sinn , then show that Pn – Pn – 2 = – sin2 cos2 Pn – 4 Qn – Qn–2 = –sin2 cos2 Qn – 4 and hence show that P4 = 1 – 2 sin2 cos2 Q4 = cos2 – sin2 Sol. 26. If A + B + C = , Prove that tanB tanC+tanC tanA+tanA tanB=1+secA . sec B .secC. Sol. 25. If sin ( ) = a & sin ( then find the value of cos2 ( Sol. ) = b (0 < ) – 4 ab cos ( /2) ) 27. If tan2 +2 tan . tan 2 =tan2 +2 tan . tan2 , then prove that each side is equal to 1 or tan = ± tan . Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) TRIGONOMETRIC RATIOS & IDENTITIES Page # 189 their radii. Sol. Sol. 3 3 , <x< , find the value of 4 2 x x sin and cos . 2 2 31. If tan x = Sol. 28. For all ) Sol. in 0, 2 show that cos (sin ) > sin (cos 32. Prove that : (i) sec4 A (1 – sin4 A) – 2 tan2 A = 1 Sol. (ii) Sol. cot2 (sec 1 sin 1) = sec2 . 1 sin 1 sec 29. Find the length of an arc of a circle of radius 10 cm which subtends an angle of 45° at the centre. Sol. 30. If the arcs of the same length in two circles subtend angles 75° and 120° at the centre, find the ratio of 33. In a ABC, prove that : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com TRIGONOMETRIC RATIOS & IDENTITIES Page # 190 sin A B C +sin +sin =1+4sin 2 2 2 A 4 sin B 4 sin C 4 Sol. (b) sin 4 16 sin4 3 5 7 3 sin4 sin4 16 16 16 2 Sol. 37. If X = s i n 34. Prove that : cos² + cos² ( + ) cos ( + ) = sin² Sol. 2cos 3 , Y = cos 12 cos 3 12 7 + s in 12 7 + cos 12 then prove that X Y 12 12 + s in + cos Y = 2 tan 2 . X Sol. 35. Prove that : cos 2 = 2 sin² + 4cos ( + ) sin sin + cos 2( + ) Sol. 36. Prove that : (a) tan 20° . tan 40° . tan 60° . tan 80° = 3 Sol. 38. If m tan ( - 30°) = n tan ( + 120°) , that cos 2 = m n 2(m n) . Sol. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) show TRIGONOMETRIC RATIOS & IDENTITIES Page # 191 41. If + = , prove that cos² + cos² + cos² = 1 + 2 cos Sol. cos cos . 39. If A + B + C = , prove that tan A tan B tan C = (tan A ) 2 (cot A ) . Sol. 40. Show that 1 sin A cos A 1 cos B sinB 42. In A, B, C denote the angles of a triangle ABC then prove that the triangle is right angled if and only if sin4A + sin4B + sin4C = 0 Sol. 2sin A 2sinB . sin(A B) cos A cosB Sol. 2 , prove that 7 tan . tan 2 + tan 2 . tan 4 + tan 4 . tan = 7. Sol. 43. If = : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com TRIGONOMETRIC RATIOS & IDENTITIES Page # 192 46. Prove that the average of the numbers nsinnº, n = 2, 4, 6,...180, is cot 1º. Sol. 44. Let k = 1º, then prove that 88 n 1 . cos(n 1)k cos nk 0 cos k sin 2 k . Sol. 47. Show that elliminating x & y from the equations, sin x + sin y = a ; cos x + cos y = b & tan x + tan y = c gives 45. If cos A = tan B, cos B = tan C and cos C = tan A, then prove that sin A = sin B = sin C = 2 sin 18º. Sol. Sol. 8ab = c. (a 2 b2 )2 4a2 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) TRIGONOMETRIC RATIOS & IDENTITIES Page # 193 EXERCISE – IV PREVIOUS YEARS LEVEL – I JEE MAIN 1. If cos x + cos y + cos x y = 2 (B) cos (C) cot = 0, then cot (A) sin Sol. = 0 and sin x + sin y + sin [AIEEE-2002] (D) 2 sin 3. Let If sin be s uc h tha t + sin 21 =– and cos 65 the value of cos (A) – 3 130 (B) < – + cos is- 2 3 130 < 3 . 27 =– , then 65 [AIEEE-2004] (C) 6 65 (D) 6 65 Sol. 2. cos 1°. cos 2°. cos 3°.... cos 179° = [AIEEE-2002] (A) 0 (B) 1 (C) 2 (D) 3 Sol. 4. Let A and B denote the statements A : cos + cos + cos = 0 [AIEEE-2009] B : sin + sin + sin = 0 If cos +cos +cos = 3 ,then : 2 (A) A is false and B is true (B) both A and B are true (C) both A and B are false (D) A is true and B is false : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com TRIGONOMETRIC RATIOS & IDENTITIES Page # 194 Sol. Sol. 5. Let cos ( + where 0 (A) ) = 4 25 16 (B) 4 and let sin ( 5 . Then tan 2 = 56 33 (C) – ) = 5 , 13 [AIEEE-2010] 19 12 (D) 20 7 7. In a PQR i f 3 si n P + 4 cos Q = 6 and 4 sin Q + 3 cos P = 1, then the angle R is equal to : [AIEEE-2012] (A) 4 (B) 3 4 (C) 5 6 (D) Sol. Sol. 6. If A = sin2x + cos4x, then for all real x : [AIEEE-2011] (A) 3 4 (C) 1 A A 1 2 (B) 13 16 (D) 3 4 A A 1 13 16 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 6 TRIGONOMETRIC RATIOS & IDENTITIES LEVEL – II 1. (a) Let f( ) = sin (sin (A) 0 only when (C) 0 for all real Sol. 0 Page # 195 JEE ADVANCED + sin 3 ). Then f( ) : [JEE 2000 (Scr.), 1] (B) 0 for all real (D) 0 only when 0. (b) Find the smallest positive values of x & y satisfying, x–y= 4 , cot x+cot y=2. [REE 2000, 3] Sol. (b) In any triangle ABC, prove that, cot A B C A B C + cot + cot = cot cot cot . 2 2 2 2 2 2 [JEE 2000 (Mains), 3] Sol. 3. If + = 2 and (A) 2(tan + tan ) (C) tan + 2tan Sol. + = then tan equals [JEE 2001 (Scr.), 1] (B) tan + tan (D) 2tan + tan 2. (a) Find the maximum and minimum values of cos 2x sin 2x 27 . 81 . Sol. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com TRIGONOMETRIC RATIOS & IDENTITIES Page # 196 Sol. 4. If and then + (A) are acute angles sin = 1/2, cos = 1/3, [JEE 2004 (Scr.)] 2 , , 3 2 (B) 2 3 (C) 2 5 , 3 6 (D) 5 , 6 Sol. tan 5. In an equilateral triangle, 3 coins of radii 1 unit each are kept so that they touch each other and also the sides of the triangle. Area of the triangle is [JEE 2005 (Scr.)] A (A) 4 2 3 B C (C) 12 7 3 4 (B) 6 4 3 (D) 3 cot 6. Let (0, /4) and t1 =(tan ) , t2 =(tan ) , tan cot t3 = (cot ) , t4 = (cot ) , then [JEE 2006, 3] (A) t1 > t2 > t3 > t4 (B) t4 > t3 > t1 > t2 (C) t3 > t1 > t2 > t4 (D) t2 > t3 > t1 > t4 Sol. 7 3 4 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) TRIGONOMETRIC RATIOS & IDENTITIES Page # 197 (b) For 0 < 6 cosec m 1 (A) /4 < /2, the solution(s) of (m 1) 4 cosec (B) /6 m 4 (C) /12 = 4 2 is (are) (D) 5 /12 Sol. One or more than one is/are correct : [Q.7 (a) & (b)] 1 sin4 x cos4 x 7. (a) If + = , then [JEE 2009, 4+4] 5 2 3 2 1 sin8 x cos8 x 2 (A) tan x = (B) + = 3 125 8 27 1 2 sin8 x cos8 x 2 (C) tan x = (D) + = 3 125 8 27 Sol. 8. The maximum value of the expression 1 sin2 3 sin cos 5 cos2 is [JEE 2010] Sol. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :info@motioniitjee.com TRIGONOMETRIC RATIOS & IDENTITIES Page # 198 9. The positive integer value of n > 3 satisfying the 1 equation sin n 1 2 sin n 1 3 sin n 10. Let , [JEE 2012] [0,2 ] be such that is [JEE 2011] 2 cos (1 sin ) sin2 tan 2 cot 2 Sol. tan(2 Then (A) 0 (C) 4 3 ) 0and 1 3 2 sin cannot satisfy (B) 2 3 2 (D) 2 3 2 4 3 2 Sol. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) cos 1 TRIGONOMETRIC RATIOS & IDENTITIES Page # 199 Answer Ex–I JEE MAIN 1. B 2. A 3. A 4. D 5. B 6. A 7. A 8. D 9. A 10. D 11. B 12. C 13. A 14. B 15. A 16. C 17. C 18. A 19. B 20. C 21. B 22. D 23. A 24. D 25. D 26. B 27. A 28. D 29. B 30. B 31. C 32. D 33. A 34. B 35. A 36. A 37. A 38. A Answer Ex–II JEE ADVANCED (OBJECTIVE) LEVEL – I SINGLE CORRECT 1. A 2. C 3. B 4. A 5. B 6. A 7. B 8. C 9. B 10. A 11. B 12. C 13. A 14. C 15. C 16. A 17. A 18. D 19. A 20. B 21. B LEVEL – II MULTIPLE CORRECT 1. BD 2. AB 3. ABCD 4. D 5. BD 6. CD 7. AB 8. AB 9. BD 10. AC 11. CD 12. BD 13. BC 14. BC 15. BD 16. AB Answer Ex–III Comprehension # 1 JEE ADVANCED 1. C 2. A Comprehension # 2 3. B 4. C 5. B 6. D Matrix Match Type 1. (A)–(Q) ; (B)–(R) ; (C)–(Q) ; (D)–(P) Subjective Type 21. (i) 4 (ii) 4 (iii) 4 (iv) 30. r1 : r2 = 8 : 5 2. (A)–(Q, S) ; (B)–(P) ; (C)–(Q) ; (D)–(P) 1. a 2b2 + 4a 2 = 9b2 3 (v) 3 15. – 1 1 , 4 4 16. (i) 2, –1 (ii) 2, 0 25. 1 – 2a2 – 2b2 31. sin x 2 3 10 29. and cos : 0744-2209671, 08003899588 | url : www.motioniitjee.com, x 2 5 cm 2 1 10 :info@motioniitjee.com TRIGONOMETRIC RATIOS & IDENTITIES Page # 200 Answer Ex–IV PREVIOUS YEARS LEVEL – I 1. C 2. A JEE MAIN 3. A 4. B 5. B LEVEL – II 6. A 7. D JEE ADVANCED 5 –5 1. (a) C 2. (a) max. = 3 & min. = 3 5. B 6. B ; (b) x = 7. (a) A, B ; (b) C, D 5 ;y= 12 6 8. 2 3. C 4. B 9. 7 10. A,C,D Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)

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