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MOTION SAMPLE BOOKLET CLASS XI
• ROTATIONAL
• STOICHIOMETRY - I
• TRIGONOMETRIC RATIOS & IDENTITIES
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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)
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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
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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 .......
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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?
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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
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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
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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
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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
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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.
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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
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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
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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
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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
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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.
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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.
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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
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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.
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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
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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.
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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
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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.
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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
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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
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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.
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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.
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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
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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
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(C) v/2R
(D) 3v/R
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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
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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
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(D) none
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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
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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.
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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.
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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.
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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.
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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)
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t
r
(D) data is insufficient
(B)
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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
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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
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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
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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.
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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.
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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.
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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.
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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
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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
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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.
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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.
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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.
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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.
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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.
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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
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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
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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.
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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.
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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
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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)
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3ml 2
(C) 3ml 2
(D) ml 2
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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.
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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.
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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]
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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.
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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]
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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
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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)
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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]
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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.
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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.
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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.
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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]
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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]
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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.
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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
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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
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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
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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.
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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.
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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.
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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.
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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.
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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
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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.
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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 ?
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(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 =
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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.
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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
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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
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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
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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 .
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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
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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
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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.
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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
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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
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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
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(D)
2
mol.
3
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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 %.
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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
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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
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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
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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.
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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
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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
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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
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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
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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
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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
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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.
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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.
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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.
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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
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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
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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.
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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
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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.
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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.
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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
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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)
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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
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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
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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
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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.
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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.
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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.
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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).
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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.
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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.
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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.
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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.
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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.
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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.
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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
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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
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50. NH3
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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
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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.
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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 )
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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
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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
.
.
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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
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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
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.
5
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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
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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
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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
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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.
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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
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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
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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
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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
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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)
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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
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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
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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
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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
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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
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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.
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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)
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(D)
(4
7)
3
(1
7)
4
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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
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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
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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.
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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
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(D) 1/4
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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.
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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
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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.
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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
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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.
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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.
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(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.
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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.
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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.
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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
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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.
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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.
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by cos
cos 2
sin 2
= (a2 – b2)2/3
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= 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 .
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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
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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.
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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
=
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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
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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
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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
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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.
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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
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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.
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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.
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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
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x
2
5
cm
2
1
10
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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
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