Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
JEE Main 2014 PHYSICS 1. The current voltage relation of diode is given by I = (e1000V/T – 1) mA, where the applied V is in volts and the temperature T is in degree kelvin. If a student makes an error measuring ± 0.01 V while measuring the current of 5 mA at 300 K, what will be the error in the value of current in mA? (1) 0.2 mA (2) 0.02 mA (3) 0.5 mA (4) 0.05 mA Solution The given current voltage relation is I = (e1000V/T – 1) mA. Now at I = 5 mA, Eq. (1) becomes 5 = (e1000V/T – 1) or e1000V/T = 6 mA Differentiating Eq. (1) w.r.t V we get (1) dI 1000 e1000V /T dV T 1000 dI e1000V /T dV T 1000 6 mA 0.01 300 0.2 mA Hence, the correct option is (1). 2. From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle, to hit the ground, is n times that taken by it to reach the highest point of its path. The relation between H, u and n is: (1) 2gH = n2u2 (2) gH = (n – 2)2u2 2 (3) 2gH = nu (n – 2) (4) gH = (n – 2)u2 Solution Time taken by the particle to reach highest point is t1 u g 1 sy u yt ayt 2 2 1 H ut gt 2 2 t u u 2 2 gH g Using second kinematics relation we get nu u u 2 2 gH g g nu u u 2 2 gH nu u u 2 2 gH u(n 1) 2 u 2 2 gH u 2 (n 1)2 u 2 2 gH u 2 (n 2 1 2n 1) 2 gH u 2 (n 2 2n) 2 gH 2 gH n(n 2)u 2 Hence, the correct option is (3). 3. A mass m is supported by a massless string wound around a uniform hollow cylinder of mass m and radius R. If the string does not slip on the cylinder, with what acceleration will the mass fall on release? 2g 3 5g 6 (2) g 2 (4) g Solution The various forces acting on the mass m are shown below in free body diagram Let a be the acceleration with which the mass m will move downwards and be the angular acceleration produced in uniform hollow cylinder. Therefore, a R (1) Now applying condition of translational motion for mass m we have mg T ma (2) Applying the condition for rotational motion for uniform hollow cylinder we get TR mR 2 T mR T mR a R T ma (3) From (2) and (3) we get mg ma ma mg 2ma a g 2 Hence, the correct option is (2). 4. A block of mass m is placed on a surface with a vertical cross-section given by y = x3/6. If the coefficient of friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is 1 2 (1) m (2) m 6 3 1 1 (3) m (4) m 3 2 Solution At limiting equilibrium tan (1) Now dy d x3 dx dx 6 x2 2 Substituting Eq. (2) in Eq. (1) we get tan (2) x2 2 x2 0.5 2 2 x 1 x 1 Now the maximum height above the ground at which the block can be placed without slipping is x3 13 1 y m 6 6 6 Hence, the correct option is (1). 5. When a rubber-band is stretched by a distance x, it exerts a restoring force of magnitude F = ax + bx2 where a and b are constants. The work done in stretching the unstretched rubber-band by L is: 1 (1) aL2 + bL3 (2) aL2 bL3 2 2 3 1 aL2 bL3 aL bL (3) (4) 2 2 3 2 3 Solution The restoring force acting on the rubber band is F = ax + bx2 Small amount of work done can be expressed as dW = Fdx The work done in stretching the unstretched rubber-band by L will be L W (ax bx 2 )dx 0 ax 2 bx 3 2 3 L 0 aL2 bL3 2 3 Hence, the correct option is (3). 6. A bob of mass m attached to an inextensible string of length l is suspended from a vertical support. The bob rotates in a horizontal circle with an angular speed rad/s about the vertical. About the point of suspension (1) Angular momentum is conserved. (2) Angular momentum changes in magnitude but not in direction. (3) Angular momentum changes in direction but not in magnitude. (4) Angular momentum changes both in direction and magnitude. Solution The torque is given by relation mg l sin . Direction of the torque is perpendicular to the axis of rotation. As is perpendicular to L , direction of L changes but magnitude remains same. Hence, the correct option is (3). 7. Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is GM GM (1) (2) 2 2 R R GM 1 GM (3) (4) 1 2 2 1 2 2 R 2 R Solution Let the speed of each particle be v. The situation can be express with help of diagram shown below F 2 Substituting F GM F F' 2 2 R 2 GM 2 , we get 4R2 and F ' 2 GM 2 2 Mv 2 R GM 2 R 2 R 2 2 GM 2 2 R 2 2 2 GM 2 Mv 2 4R2 R GM 2 Mv 2 4R2 R GM 2 1 1 Mv 2 R 2 2 4 R 2 GM 1 1 Mv 2 R 2 4 v GM R 2 4 1 GM (1 2 2) 4 2 2 R Hence, the correct option is (4). 8. The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by 100°C is : (For steel Young's modulus is 2 × 1011 Nm–2 and coefficient of thermal expansion is 1.1 × 10–5 K–1) (1) 2.2 × 108 Pa (2) 2.2 × 109 Pa 7 (3) 2.2 × 10 Pa (4) 2.2 × 106 Pa Solution As length is constant, L Strain= T L Now pressure = stress = Y × strain = 2 × 1011 × 1.1 × 10–5 × 100 = 2.2 × 108 Pa Hence, the correct option is (1). 9. There is a circular tube in a vertical plane. Two liquids which do not mix and of densities d1 and d2 are filled in the tube. Each liquid subtends 90° angle at center. Radius joining their interface makes an angle with vertical. Ratio d1/d2 is 1 sin 1 sin 1 tan (3) 1 tan 1 cos 1 cos 1 sin (4) 1 cos (1) (2) Solution Now equating pressure at A we get R cos R sin d 2 g R cos R sin d1 g R cos R sin d 2 R cos R sin d1 R cos R sin d 1 d 2 R cos R sin cos sin cos sin Dividing numerator and denominator by cos we get cos sin d1 cos d 2 cos sin cos 1 tan 1 tan Hence, the correct option is (3). 10. On heating water, bubbles being formed at the bottom of the vessel detach and rise. Take the bubbles to be spheres of radius R and making a circular contact of radius r with the bottom of the vessel. If r << R, and the surface tension of water is T, value of r just before bubbles detach is (Density of water is w ) R2 w g (2) R 2 3T (3) R 2 w g (4) R 2 T w g 6T 3 w g T Solution When the bubble gets detached, then the surface tension will be equal to the buoyant force acting on the bubble. Buoyant force is FB T dl sin (1) Force due to surface tension is 4 FS R 3 w g 3 (2) Equating (1) and (2) we get 4 3 T dl sin 3 R w g 4 T sin dl R 3 w g 3 r 4 3 T 2 r R w g R 3 2 T 2 r 4 R3 w g R 3 2R4 w g 2w g r R2 3T 3T Hence, none of the given options is correct. r2 11. Three rods of copper, brass and steel are welded together to form a Y-shaped structure. Area of cross-section of each rod = 4 cm2. End of copper rod is maintained at 100°C whereas ends of brass and steel are kept at 0°C. Lengths of the copper, brass and steel rods are 46 cm, 13 cm and 12 cm respectively. The rods are thermally insulated from surroundings except at ends. Thermal conductivities of copper, brass and steel are 0.92, 0.26 and 0.12 CGS units respectively. Rate of heat flow through copper rod is (1) 1.2 cal/s (2) 2.4 cal/s (3) 4.8 cal/s (4) 6.0 cal/s Solution Consider the figure given below Now we know that rate of flow of heat through copper rod will be equal to sum of the rate of flow of heat through brass rod and steel rod. Therefore, we have dQc dQb dQs dt dt dt 0.92 4 100 T 0.26 4 T 0 0.12 4 T 0 46 13 12 3.68 100 T 1.04T 0.48T 46 13 12 o T 40 C Rate of heat flow through copper rod will be 0.92 4(100 40) Qc 46 0.92 4 60 4.8cal/s 46 Hence, the correct option is (3). 12. One mole of diatomic ideal gas undergoes a cyclic process ABC as shown in figure. The process BC is adiabatic. The temperatures at A, B, and C are 400 K, 800 K and 600 K respectively. Choose the correct statement (1) The change in internal energy in whole cyclic process is 250R. (2) The change in internal energy in the process CA is 700R. (3) The change in internal energy in the process AB is –350R. (4) The change in internal energy in the process BC is – 500R. Solution The change in internal energy is given by relation U CA nCV T nCV TA TC (1) Now temperature at point A is TA = 400 K and temperature at point C is TC= 600 K. Substituting the values in Eq. (1) we get 5R U CA 1 400 600 2 5R ( 200 500 R 2 Hence, the correct option is (4). 13. An open glass tube is immersed in mercury in such a way that a length of 8 cm extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by additional 46 cm. What will be length of the air column above mercury in the tube now? (Atmospheric pressure = 76 cm of Hg) (1) 16 cm (2) 22 cm (3) 38 cm (4) 6 cm Solution The figure given below shows an open tube immersed in mercury Now we have P + x = P0 P = P0 – x [where P0 = 76 cm of Hg] P = (76 – x) Again we can write 8 A 76 = (76 – x) A (54 – x) 8 76 = (76 – x) 54 – x) 608 = 4104 – 76x – 54x + x2 x2 – 130x + 3496 = 0 x = 38 (neglecting the negative value of quadratic equation) Length of air column will, be L = 54 cm – x = 54 cm – 38 cm = 16 cm Hence, the correct option is (1). 14. A particle moves with simple harmonic motion in a straight line. In first s, after starting from rest it travels a distance a, and in next s it travels 2a in same direction then (1) Amplitude of motion is 3a (2) Time period of oscillations is 8 (3) Amplitude of motion is 4a (4) Time period of oscillations is 6 Solution As it starts from rest, we have x = A cos t. At t = 0, Eq. (1) becomes x = A cos 0 = A when t = , Eq. (1) becomes x = A cos A – a = A cos Aa cos A when t = 2, Eq. (1) becomes x = Acos2 A – 3a = Acos2 A 3a cos 2 A Now we know that cos 2= 2cos 2– 1 2 A 3a Aa 2 1 A A A 3a 2 A2 2a 2 4 Aa A2 A A2 A2 3aA A2 2a 2 4 Aa 2a 2 aA A 2a Now A a A cos 2a a 2a cos a 1 cos cos 2a 2 2 cos cos T 3 2 T 6 T 3 Hence, the correct option is (4). 15. A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s. (1) 12 (2) 8 (3) 6 (4) 4 Solution The diagram showing a pipe closed from one end is shown below Now the length of pipe l 0.85cm. Speed of sound in air v = 340 m/s. As we know in fundamental mode 4 l 4l 4 0.85 3.4cm Now the frequency can be calculated using v v 340 3.4 100 Hz Therefore, the possible frequencies are 100 Hz, 300 Hz, 500 Hz, 700 Hz, 900 Hz, and 1100 Hz. The number of possible natural oscillations of air column in the pipe is 6. Hence, the correct option is (3). 16. Assume that an electric field E 30 x 2 i exists inspace. Then the potential difference VA – VO, whereVO is the potential at the origin and VA the potentialat x = 2 m is (1) 120 J (2) –120 J (3) – 80 J (4) 80 J Solution We know that relation between electric field and potential is dV E dx dV Edx Taking definite integral both sides we get VA 2 2 dV 30 x dx VO 0 2 x3 VA VO 30 3 0 10 x 3 2 0 80 J Hence, the correct option is (3). 17. A parallel plate capacitor is made of two circular plates separated by a distance 5 mm and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is 3 × 104 V/m, the charge density of the positive plate will be close to (1) 6 × 10–7 C/m2 (2) 3 × 10–7 C/m2 4 2 (3) 3 × 10 C/m (4) 6 × 104 C/m2 Solution Dielectric constant between the two plates of parallel plate capacitor K = 2.2. Electric field in the dielectric E = 3 × 104 V/m. Let the charge density of the positive plate be =? Now using the relation for electric field E K 0 K 0 E 2.2 8.85 10 12 3 10 4 6 10 7 C/m 2 Hence, the correct option is (1). 18. In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1 kW. The voltage of the electric mains is 220 V. The minimum capacity of the main fuse of the building will be: (1) 8 A (2) 10 A (3) 12 A (4) 14 A Solution Power consumed by 15 bulbs of 40 W each = (15 40) W Power consumed by 5 bulbs of 100 W each = (5 100) W Power consumed by 5 fans of 80 W each = (5 80) W Power consumed by heater = 1 kW = 1000 W Voltage of electric mains V = 220 V. Now the total power consumed can be expressed as P total = V I 15 40 + 5 100 + 5 80 + 1000 = V I 600 + 500 + 400 + 1000 = 220 I 2500 I 220 11.36 A 12 A Hence, the correct option is (3). 19. A conductor lies along the z-axis at –1.5 z < 1.5 m and carries a fixed current of 10.0 A in a z direction (see figure). For a field B 3.0 104 e 0.2 x a y T, find the power required to move the conductor at constant speed to x = 2.0 m, y = 0 m in 5 × 10 –3 s. Assume parallel motion along the x-axis (1) 1.57 W (2) 2.97 W (3) 14.85 W (4) 29.7 W Solution Current I = 10.0 A. Time t = 5 × 10 –3 s. Length of the conductor l = 3.0 m. Magnetic field B 3.0 104 e 0.2 x a y T. The work done can be calculated as 2 W F . dx 0 2 2 IlB. dx 10 3 3.0 104 e 0.2 x dx 0 0 2 2 9 10 3 e 0.2 x 0 e0.2 x dx 9 10 0.2 0 3 3 9 10 9 103 e0.22 1 1 e 0.4 0.2 0.2 9 103 0.33 2.97 103 J Now average power can be calculated using Work Time 2.97 103 0.2 W t 5 103 2.97 W P= Hence, the correct option is (2). 20. The coercivity of a small magnet where the ferromagnet gets demagnetized is 3 103 Am–1. The current required to be passed in a solenoid of length 10 cm and number of turns 100, so that the magnet gets demagnetized when inside the solenoid, is (1) 30 mA (2) 60 mA (3) 3 A (4) 6 A Solution Coercivity of small magnet B 3 103 Am –1 . 0 Number of turns of solenoid N = 100. Length of solenoid L = 10 cm = 10 102 m. Let the required current be I =? Now we know that magnetic field of solenoid is given by relation B 0 nI B 0 [ where n is the number of turns per unit length] N I L 3 103 NI 100i L 10 102 I 3A Hence, the correct option is (3). 21. In the circuit shown here, the point C is kept connected to point A till the current flowing through the circuit becomes constant. Afterward, suddenly, point C is disconnected from point A and connected to point B at time t = 0. Ratio of the voltage across resistance and the inductor at t = L/R will be equal to (1) e 1 e (3) –1 (2) 1 (4) 1 e e Solution Let the voltage drop across resistor be VR and voltage drop across inductor be VL. Applying Kirchhoff's law in closed loop, we get VR VL 0 VR VL VR 1 VL Thus the ratio of the voltage across resistance and the inductor at t = L/R will be 1:1. Hence, the correct option is (3). 22. During the propagation of electromagnetic waves in a medium (1) Electric energy density is double of the magnetic energy density (2) Electric energy density is half of the magnetic energy density (3) Electric energy density is equal to the magnetic energy density (4) Both electric and magnetic energy densities are zero Solution During the propagation of electromagnetic waves in a medium electric energy density is equal to the magnetic energy density because energy is equally divided between electric and magnetic field, that is, B2 1 0E2 2 0 2 Hence, the correct option is (3). 3 23. A thin convex lens made from crown glass has focal length f . When it is measured in two 2 different liquids having refractive indices 4/3 and 5/3, it has the focal lengths f1 and f2 respectively. The correct relation between the focal lengths is (1) f1 f2 f (2) f1 f and f2 becomes negative (3) f2 f and f1 becomes negative (4) f1 and f2 both become negative Solution 3 Refractive index of crown glass . 2 Refractive index of first liquid 1 = 4/3. Refractive index of second liquid 2 = 5/3. Applying Lens maker's formula we get 1 1 1 1 f1 1 R1 R2 1 3 / 2 1 1 1 f1 4 / 3 R1 R2 Similarly from lens maker formula, we get 1 1 1 1 f 2 2 R1 R2 1 3 / 2 1 1 1 f 2 5 / 3 R1 R2 1 3 1 1 1 f 2 R1 R2 … (1) ….. (2) …. (3) Now dividing Eq. (1) and Eq. (3) we get 1 3 / 2 1 1 9 1 1 f1 4 / 3 R1 R2 f 8 f 1 f 4 f 1 f1 3 f1 4 1 1 3 1 1 1 2 2 R1 R2 f Again dividing Eq. (2) and Eq. (3) we get 1 3 / 2 1 1 9 1 1 f 5 / 3 R R f 10 f 1 1 2 2 f 2 5 f 3 f f 5 1 1 3 1 2 2 1 1 2 2 R1 R2 f Hence, the correct option is (2). 24. A green light is incident from the water to the air–water interface at the critical angle (). Select the correct statement (1) The entire spectrum of visible light will come out of the water at an angle of 90° to the normal. (2) The spectrum of visible light whose frequency is less than that of green light will come out to the air medium. (3) The spectrum of visible light whose frequency is more than that of green light will come out to the air medium. (4) The entire spectrum of visible light will come out of the water at various angles to the normal. Solution We know that as the frequency of visible light increases refractive index increases. The critical angle are refractive index are related by relation given below 1 sin c where c is the critical angle and is the refractive index. From the above relation we conclude that refractive index is inversely related to critical angle. Therefore as the refractive index increases, the critical angle decreases. Now light having frequency greater than green light will suffer total internal reflection and light having frequency less than green light will suffer refraction and pass to air. Hence, the correct option is (2). 25. Two beams, A and B, of plane polarized light with mutually perpendicular planes of polarization are seen through a Polaroid. From the position when the beam A has maximum intensity (and beam B has zero intensity), a rotation of Polaroid through 30° makes the two beams appear equally bright. If the initial intensities of the two beams are IA and IB respectively, then IA / IB equals (1) 3 (2) 3/2 (3) 1 (4) 1/3 Solution Rotation of Polaroid is 30o. Now according to law of Malus I I 0 cos 2 For beam A, initial intensity is IA, the by Malus law the final intensity will be I A ' I A cos 2 30 2 3 I A 2 3I A (1) 4 For beam B, initial intensity is IB, the by Malus law the final intensity will be 2 1 I B I B cos 60 I B 2 I B (2) 4 Now as I A ' I B' from Eq. (1) and Eq. (2) we get 3I A I B I 1 3I A I B A 4 4 IB 3 Hence, the correct option is (4). 2 26. The radiation corresponding to 32 transition of hydrogen atoms falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of 3 × 10–4 T. If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to (1) 1.8 eV (3) 0.8 eV (2) 1.1 eV (4) 1.6 eV Solution Magnetic field B = 3 × 10–4 T. Radius of the largest circular wire r = 10.0 mm 10.0 103 m. Charge of electron e 1.6 1019 C. Mass of electron m 9.1 1031. Now the stopping potential can be calculates using relation mv r qB 2meV 1 2m V eB B e Squaring both sides we get r2 2mV B2e 2 2 4 3 19 B 2 r 2 e 3 10 10.0 10 1.6 10 V J 2m 2 9.1 10 31 0.8 eV Energy corresponding to 32 transition of hydrogen atoms will be 1 1 1 1 13.6 5 E 13.6 2 2 eV 13.6 eV eV 2 3 4 9 36 1.88eV Therefore the work function will be = 1.88eV 0.8eV=1.08eV 1.1eV Hence, the correct option is (2). 27. Hydrogen (1H1), Deuterium (1H2), singly ionized Helium (2He4)+ and doubly ionized lithium (3Li6)++ all have one electron around the nucleus. Consider an electron transition from n = 2 to n = 1. If the wave lengths of emitted radiation are 1, 2, 3 and 4 respectively then approximately which one of the following is correct? (1) 41 = 22 = 23 = 4 (2) 1 = 22 = 23 = 4 (3) 1 = 2 = 43 = 94 (4) 1 = 22 = 33 = 44 Solution The wavelength can be calculates using relation 1 1 1 RZ 2 2 2 1 2 1 3RZ 2 1 RZ 2 1 4 4 4 (1) 3RZ 2 1 For Hydrogen (1H1), Z =1 therefore the wavelength corresponding to hydrogen can be calculated using Eq. (1) as 4 4 (2) 1 2 3R 1 3R For Deuterium (1H2), Z =1 therefore the wavelength corresponding to hydrogen can be calculated using Eq. (1) as 4 4 (3) 2 2 3R 1 3R For singly ionized Helium (2He4)+, Z =2 therefore the wavelength corresponding to hydrogen can be calculated using Eq. (1) as 4 4 (4) 3 2 3R 2 12 R For doubly ionized lithium (3Li6)++ , Z =3 therefore the wavelength corresponding to hydrogen can be calculated using Eq. (1) as 4 4 (5) 4 2 3R 3 27 R Now from Eq. (2), Eq. (3), Eq. (4), and Eq. (5) we get 1 2 43 94 Hence, the correct option is (3). 28. The forward biased diode connection is (1) (2) (3) Solution (4) In forward bias, the p-side of diode should be connected to positive terminal of battery or the terminal at higher potential and n-side should be connected to negative terminal of battery (terminal to lower potential). Therefore the configuration in which diode is forward biased is Hence, the correct option is (1). 29. Match List-I (Electromagnetic wave type) with List - II (Its association/application) and select the correct option from the choices given below the lists: (a) List-I Infrared waves (i) List-II To treat muscular strain (b) Radiowaves (ii) For broadcasting (1) (2) (3) (4) (c) X-rays (iii) To detect fracture of bones (d) Ultraviolet rays (iv) Absorbed by the ozone layer of the atmosphere (a) (iv) (i) (iii) (i) (b) (iii) (ii) (ii) (ii) (c) (ii) (iv) (i) (iii) (d) (i) (iii) (iv) (iv) Solution Infrared rays are used to treat muscular strain. Thus the correct mapping is (a) (i). Radio waves are used for broadcasting. Thus the correct mapping is (b) (ii). Fractures in bones are detected using X-rays. Thus the correct mapping is (c) (iii). Ultraviolet rays are absorbed by ozone layer. Thus the correct mapping is (d) (iv). Hence, the correct option is (4). 30. A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it? (1) A meter scale (2) A Vernier caliper where the 10 divisions in Vernier scale matches with 9 division in main scale and main scale has 10 divisions in 1 cm (3) A screw gauge having 100 divisions in the circular scale and pitch as 1 mm (4) A screw gauge having 50 divisions in the circular scale and pitch as 1 mm Solution Measured value = 3.50 cm therefore least count of measuring instrument must be 0.01 cm = 0.1 mm. For Vernier caliper the least count is = 0.01 cm = 0.1 mm. Thus the student used Vernier caliper to measure the length of the rod. Hence, the correct option is (2). CHEMISTRY 31. The correct set of four quantum numbers for the valence electrons of rubidium atom (Z = 37) is (1) 5, 0, 0, +1/2 (2) 5, 1, 0, +1/2 (3) 5, 1, 1, +1/2 (4) 5, 0, 1, +1/2 Solution For Rb (Z = 37), the electronic configuration is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 5s1 As the last electron enters 5s orbital, n = 5, l = 0 and ml = 0, and since only one electron is present in the s- orbital, ms = +1/2. Hence, the correct option is (1) 32. If Z is a compressibility factor, van der Waals equation at low pressure can be written as (1) Z 1 RT/pb (2) Z = 1 a/VRT (3) Z = 1 pb/RT (4) Z = 1 + pb/RT Solution We know that compressibility factor is given by Z pV RT For one mole of gas, we have a p 2 (V b) RT V At low pressure, volume is very high, so V >> b and thus V b b. a a pV a a 1 Z 1 p 2 (V ) RT pV RT V V RT VRT VRT Hence, the correct option is (2) 33. CsCl crystallizes in body centered cubic lattice. If ‘a’ is its edge length then which of the following expressions is correct? (1) rCs rCl 3a 3a 2 3a (3) rCs rCl 2 (4) rCs rCl 3a (2) rCs rCl Solution For body centered cubic lattice of CsCl shown below, Two Cl and one Cs+ touch each other at body diagonal of cube. Thus, 3a 2rCs 2rCl 3a rCs rCl 2 Hence, the correct option is (3) 34. For the estimation of nitrogen, 1.4 g of an organic compound was digested by Kjeldahl method and the evolved ammonia was absorbed in 60 mL of M/10 sulphuric acid. The unreacted acid required 20 mL of M/10 sodium hydroxide for complete neutralization. The percentage of nitrogen in the compound is (1) 6% (2) 10% (3) 3% (4) 5% Solution Total mmol of H2SO4 taken = 60 0.1 = 6 mmol H 2SO4 2NaOH Na 2SO4 2H 2 O This implies that each mole of sulphuric acid is neutralized by two moles of NaOH. So, mmol of H2SO4 1 1 neutralized by NaOH 20 1 mmol 2 10 Therefore, mmol of H2SO4 neutralized by evolved ammonia = 6 1 = 5 mmol H 2SO 4 2NH 3 (NH 3 ) 2 SO 4 This indicates that each mole of sulphuric acid is neutralized by two moles of ammonia, so, mmol of ammonia evolved 2 5 10 mmol mmol of NH3 mmol of N atoms mass of nitrogen 10 14 mg Mass of N 140 % of N 100 100 10% Mass of organic compound 1000 1.4 Hence, the correct option is (2) 35. Resistance of 0.2 M solution of an electrolyte is 50 . The specific conductance of the solution is 1.4 S m1. The resistance of 0.5 M solution of the same electrolyte is 280 . The molar conductivity of 0.5 M solution of the electrolyte in S m2 mol1 is (1) 5 104 (2) 5 103 3 (3) 5 10 (4) 5 102 Solution Given that specific conductance of 0.2 M solution, 1.4 S m 1 1.4 102 S cm 1 1 1 l l Now, we know that 1.4 102 50 0.7 cm 1 R A A 1 l 1 0.7 2.5 103 S cm1 For 0.5 M solution, R A 280 1000 1000 2.5 103 5 S cm 2 mol1 5 104 S m 2 mol1 and its molar conductivity is M 0.5 Hence, the correct option is (1) 36. For complete combustion of ethanol, C 2 H 5 OH(l) 3O 2 (g) 2CO 2 (g) 3H 2 O(l) the amount of heat produced as measured in bomb calorimeter, is 1364.47 kJ mol1 at 25°C. Assuming ideality the enthalpy of combustion, cH, for the reaction will be (R = 8.314 J K1mol1) (1) 1366.95 kJ mol1 (2) 1361.95 kJ mol1 1 (3) 1460.50 kJ mol (4) 1350.50 kJ mol1 Solution For the given reaction, H U ng RT 1364.47 (1 8.314 298) 1366.95 kJ mol1 1000 Hence, the correct option is (1) 37. The equivalent conductance of NaCl at concentration C and at infinite dilution is C and , respectively. The correct relationship between C and is given as (where the constant B is positive) (1) C ( B )C (2) C ( B )C (3) C ( B) C (4) C ( B) C Solution According to Debye–Huckle–Onsager equation, C B C . Hence, the correct option is (3) 38. Consider separate solutions of 0.500 M C2H5OH(aq), 0.100 M Mg3(PO4)2(aq), 0.250 M KBr(aq) and 0.125 M Na3PO4(aq) at 25°C. Which statement is true about these solutions, assuming all salts to be strong electrolytes? (1) They all have the same osmotic pressure. (2) 0.100 M Mg3(PO4)2(aq) has the highest osmotic pressure. (3) 0.125 M Na3PO4 (aq) has the highest osmotic pressure. (4) 0.500 M C2H5OH (aq) has the highest osmotic pressure. Solution According to van’t Hoff equation, iCRT . For 0.500 M C2H5OH (aq), 1 0.5 RT 0.5RT For 0.100 M Mg3(PO4)2 (aq), 5 0.1 RT 0.5RT For 0.250 M KBr(aq), 2 0.25 RT 0.5RT and for 0.125 M Na3PO4 (aq), 4 0.125 RT 0.5RT So, all have the same osmotic pressure, and so they are isotonic solutions. Hence, the correct option is (1) SO3 (g) if Kp = KC(RT)x where the symbols have usual meaning then the value of x is (assuming ideality) (1) 1 (2) 1/2 (3) 1/2 (4) 1 39. For the reaction SO 2 (g) 12 O 2 (g) Solution For the reaction, x = ng = 1 (1 + 1/2) = 1/2. Hence, the correct option is (2) 40. For the non-stoichiometric reaction 2A + B C + D, the following kinetic data were obtained in three separate experiments, all at 298 K. Initial Concentration (A) Initial Concentration (B) Initial rate of formation of C (mol L1 s1) 0.1 M 0.1 M 1.2 103 0.1 M 0.2 M 1.2 103 0.2 M 0.1 M 2.4 103 The rate law for the formation of C is dC k[A][B] dt dC k[A][B]2 (3) dt (1) dC k[A]2 [B] dt dC k[A] (4) dt (2) Solution For the reaction 2A + B C + D, the rate of formation of C is d[C] k[A]x [B] y dt From the given values in the table, we have 1.2 10 3 k[0.1]x [0.1] y (1) 3 x y (2) 3 x y (3) 1.2 10 k[0.1] [0.2] 2.4 10 k[0.2] [0.1] Dividing Eq. (1) by Eq. (3), we get x 1.2 10 3 k[0.1]x [0.1] y 1 1 x 1 3 x y 2.4 10 k[0.2] [0.1] 2 2 Dividing Eq. (1) by Eq. (2), we get y 1.2 10 3 k[0.1]x [0.1] y 1 1 y 0 3 x y 1.2 10 k[0.1] [0.2] 2 dC k[A] . Therefore, rate of formation of C is dt Hence, the correct option is (4) 41. Among the following oxoacids, the correct decreasing order of acid strength is (1) HOCl > HClO2 > HClO3 > HClO4 (2) HClO4 > HOCl > HClO2 > HClO3 (3) HClO4 > HClO3 > HClO2 > HOCl (4) HClO2 > HClO4 > HClO3 > HOCl Solution Let us consider the stabilities of the conjugate bases, ClO, ClO 2 , ClO3 and ClO 4 formed from the acids, HClO, HClO2, HClO3 and HClO4, respectively. These anions are stabilized by the delocalization of the charge over oxygen atoms. More oxygen atom causes more delocalization of charge on conjugate base. So, the stability order of respective conjugate base is: ClO < ClO 2 < ClO3 < ClO 4 Now from conjugate acid base pair definition, more stable anion implies weaker is the conjugate base and stronger is the respective acid. Thus, the acidic strength decreases in the order HClO4 > HClO3 > HClO2 > HOCl Hence, the correct option is (3) 42. The metal that cannot be obtained by electrolysis of an aqueous solution of its salts is (1) Ag (2) Ca (3) Cu (4) Cr Solution The reactions at cathode for the electrolysis of aqueous salt solution of Ag, Ca, Cu and Cr salts are as follows: Ag 2e Ag H 2 O e 12 H 2 OH Cu 2 2e Cu Cr 3+ 3e Cr As Ca lies in the bottom of electrochemical series, this implies very high negative reduction potential of calcium ions. Water has less negative reduction potential, so, water is reduced instead of calcium from its aqueous solution. Thus, Ca cannot be obtained by electrolysis of its aqueous salt solution, H2 is evolved instead at the cathode. Hence, the correct option is (2) 43. The octahedral complex of a metal ion M3+ with four monodentate ligands L1, L2, L3 and L4 absorb wavelengths in the region of red, green, yellow and blue, respectively. The increasing order of ligand strength of the four ligands is (1) L4 < L3 < L2 < L1 (2) L1 < L3 < L2 < L4 (3) L3 < L2 < L4 < L1 (4) L1 < L2 < L4 < L3 Solution Strong field ligands cause higher splitting in the d orbitals. This means that the absorption of energy for d–d transition is more, and so transmitted energy is more. In case of L1, L2, L3 and L4, as the energy decreases in the order Violet > Indigo > Blue > Green > Yellow > Orange > Red so, the order ligand strength is L1 < L3 < L2 < L4. Hence, the correct option is (2) 44. Which one of the following properties is not shown by NO? (1) It is diamagnetic in gaseous state. (2) It is a neutral oxide. (3) It combines with oxygen to form nitrogen dioxide. (4) Its bond order is 2.5. Solution The MO configuration of NO is 1s 2 *1s 2 2s 2 * 2s 2 2pz2 2px2 2p 2y * 2p1x As can be seen it is an odd electron species and so paramagnetic in gaseous state with one unpaired electron. Bond order = (10 5)/2 = 2.5. Hence, the correct option is (1) 45. In which of the following reactions H2O2 acts as a reducing agent? (i) H 2 O 2 2H 2e 2H 2 O (ii) H 2 O 2 2e O 2 2H (iii) H 2 O 2 2e 2OH (iv) H 2 O 2 2OH 2e O 2 2H 2 O (1) (i), (ii) (2) (iii), (iv) (3) (i), (iii) (4) (ii), (iv) Solution The oxidation states of the species are as follows: 1 2 (i) H 2 O 2 2H 2e 2H 2 O 1 0 (ii) H 2 O 2 2e O 2 2H 1 2 (iii) H 2 O 2 2e 2O H 1 2 0 2 (iv) H 2 O 2 2 O H 2e O 2 2H 2 O In (ii) and (iv) H2O2 acts as a reducing agent as it loses electrons to reduces the other species and itself undergoes oxidation. Hence, the correct option is (4) 46. The correct statement for the molecule, CsI3, is (1) It is a covalent molecule (2) It contains Cs+ and I3 ions (3) It contains Cs3+ and I ions (4) It contains Cs+, I and lattice I2 molecule. Solution CsI3 contains Cs+ and I3 ions because Cs cannot show +3 oxidation states and it is difficult to accommodate I2 molecules into the lattice because of their large size. Hence, the correct option is (2) 47. The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1:4. The ratio of number of their molecule is (1) 1:4 (2) 7:32 (3) 1:8 (4) 3:16 Solution Let the mass of oxygen be x and that of nitrogen be 4x. So, the number of moles of oxygen is x/32 and that of nitrogen is 4x/28 = x/7 x 7 7 The required ratio is 32 x 32 Hence, the correct option is (2) 48. Given below are the half-cell reactions Mn 2 2e Mn; E o 1.18 V 2(Mn 3 e Mn 2+ ); E o 1.51 V The Eo for 3Mn2+ Mn + 2Mn3+ will be (1) 2.69 V; the reaction will not occur. (2) 2.69 V; the reaction will occur. (3) 0.33 V; the reaction will not occur. (4) 0.33 V; the reaction will occur. Solution Mn2+ + 2e Mn; Eo = 1.18V [Mn3+ + e Mn2+] 2; E° = +1.51 V [Mn2+ Mn3+ + e] 2; E° = 1.51 V Adding Eq. (1) by Eq. (3), we get (1) (2) (3) 3Mn2+ Mn + 2Mn3+ o for which E 1.18 1.51 2.69 V Negative electrode potential implies that disproportion is not spontaneous for Mn with +2 oxidation state to 0 and +3 oxidation states. Hence, the correct option is (1) 49. Which series of reactions correctly represents chemical reactions related to iron and its compound? dil. H 2SO 4 H 2SO 4 , O 2 Heat (1) Fe FeSO 4 Fe2 (SO 4 )3 Fe O 2 , heat dil. H 2SO 4 Heat (2) Fe FeO FeSO 4 Fe Cl2 , heat Heat, air Zn (3) Fe FeCl3 FeCl2 Fe O 2 , heat CO, 600 C CO, 700 C (4) Fe Fe3O 4 FeO Fe Solution The reaction in option (4) is correct. In option (1), Fe2(SO4)3 produces oxide(s) of iron, not Fe. In option (2) FeSO4 cannot be converted to Fe on heating, in this reaction, oxide(s) will be formed. In option (3), FeCl3 cannot reduced to FeCl2 in the presence of air. Hence, the correct option is (4) 50. The equation which is balanced and represents the correct product(s) is: (1) LiO 2KCl 2LiCl K 2 O (2) [CoCl(NH 3 )5 ] 5H Co 2 5NH 4 Cl excess NaOH (3) [Mg(H 2 O)6 ]2 (EDTA) 4 [Mg(EDTA)]2 6H 2 O (4) CuSO 4 4KCN K 2 [Cu(CN) 4 ] K 2SO 4 Solution The equation in option (2) is correct since both charges as well as atoms are balanced. For the rest, (1) LiO 2KCl 2LiCl K 2 O ; Li atoms are not balanced. excess NaOH (3) [Mg(H 2 O)6 ]2 (EDTA) 4 [Mg(EDTA)]2 6H 2 O ; charge is not balanced. (4) CuSO 4 4KCN K 2 [Cu(CN) 4 ] K 2SO 4 ; products are not correctly written. The correct equations for other reactions are: excess NaOH (1) [Mg(H 2 O)6 ]2 (EDTA) 4 [Mg(EDTA)]2 6H 2 O (3) LiO 2KCl 2LiCl K 2 O is unfavorable in forward reaction as LiO is stable, while K2O is unstable. (4) 2CuSO 4 10KCN 2K 3 [Cu(CN) 4 ] 2K 2SO 4 (CN) 2 Hence, the correct option is (2) 51. In SN2 reactions, the correct order of reactivity for the following compounds CH3Cl, CH3CH2Cl, (CH3)2CHCl and (CH3)3CCl is (1) CH3Cl > (CH3)2CHCl > CH3CH2Cl > (CH3)3CCl (2) CH3Cl > CH3CH2Cl > (CH3)2CHCl > (CH3)3CCl (3) CH3CH2Cl > CH3Cl > (CH3)2CHCl > (CH3)3CCl (4) (CH3)2CHCl > CH3CH2Cl > CH3Cl > (CH3)3CCl Solution Rate of SN2 reaction depends on stearic hindrance on carbon atom which bears leaving group. Increasing stearic hindrance decreases possibility of back side attack by nucleophile. This decreases the rate of SN2 reaction. Order of steric hindrance is (CH3)3CCl > (CH3)2CHCl > CH3CH2Cl > CH3Cl , so the order of reactivity towards SN2 reaction is CH3Cl > CH3CH2Cl > (CH3)2CHCl > (CH3)3CCl. Hence, the correct option is (2) 52. On heating an aliphatic primary amine with chloroform and ethanolic potassium hydroxide, the organic compound formed is (1) an alkanol (2) an alkanediol (3) an alkyl cyanide (4) an alkyl isocyanide Solution The reaction involved is RNH 2 CHCl3 3KOH(alc.) RNC 3KCl 3H 2 O Alkyl isocyanide Hence, the correct option is (4) 53. The most suitable reagent for the conversion of R–CH2–OH R–CHO is: (1) KMnO4 (3) CrO3 (2) K2Cr2O7 (4) PCC (Pyridinium Chlorochromate) Solution All others except PCC are stronger oxidizing agents and will convert R–CH2–OH R–COOH. PCC is a mild oxidizing agent and will convert R–CH2–OH R–CHO. Hence, the correct option is (4) 54. The major organic compound formed by the reaction of 1,1,1-trichloroethane with silver powder is (1) Acetylene (2) Ethene (3) 2-Butyne (4) 2-Butene Solution The reaction involved is H 3C C Cl3 6Ag Cl3 C CH3 H3C C C CH3 2-Butyne Hence, the correct option is (3) 55. Sodium phenoxide when heated with CO2 under pressure at 125°C yields a product which on acetylation produces C. The major product C is Solution The reaction is Hence, the correct option is (1) 56. Considering the basic strength of amines in aqueous solution, which one has the smallest pKb value? (1) (CH3)2NH (2) CH3NH2 (3) (CH3)3N (4) C6H5NH2 Solution C6H5NH2 is least basic as the lone pair is involved in resonance with benzene ring. In the case of three aliphatic amines in aqueous solutions, the basicity depends on two factors: +I effect of alkyl group and solvation of conjugate acid (aminium ion) of the respective base in the aqueous solution. +I effect as well as basicity due to the group decreases as (CH3)3N > (CH3)2NH > CH3NH2 > NH3. However, solvation of conjugate acids occurs by hydrogen bonding which depends on the number of hydrogen atoms attached with nitrogen atom. The order of solvation of conjugate base is (CH3)3N < (CH3)2NH < CH3NH2 < NH3. In this way, both factors are against each other and the order of resultant basic strength becomes (CH3)2NH > CH3NH2 > (CH3)3N > NH3 Hence the amine that has minimum pKb (maximum basic) is (CH3)2NH. Hence, the correct option is (1) 57. For which of the following molecule significant 0? (1) Only (a) (2) (a) and (b) (3) Only (c) (4) (c) and (d) Solution In (a) and (b), = 0 as the dipole moments cancel each other. However, in (c) and (d) more than one conformation is possible, so, the dipole moment is not zero in those cases. For example, possible conformations in (c) are shown below. Similarly for structure (d), two conformations are possible. Hence, the correct option is (4) 58. Which one is classified as a condensation polymer? (1) Dacron (2) Neoprene (3) Teflon (4) Acrylonitrile Solution Condensation polymer: Dacron is formed by condensation of ethylene glycol and terephthalic acid at 420–460 K using zinc acetate and antimony oxide as catalyst. Addition polymers: Teflon formed by addition of tetrafluoroethylene. Acrylonitrile is a monomer to form addition polymer “orlon”. Neoprene is a rubber formed by addition polymerization of chloroprene, 2-chloro-1,3-butadiene. Hence, the correct option is (1) 59. Which one of the following bases is not present in DNA? (1) Quinoline (2) Adenine (3) Cytosine (4) Thymine Solution The bases in DNA are adenine, cytosine, thymine and guanine. Quinoline is not present in DNA. Hence, the correct option is (1) LiAlH PCl alc. KOH 5 4 60. In the reaction, CH 3COOH A B C , the product C is (1) Acetaldehyde (2) Acetylene (3) Ethylene (4)Acetyl chloride Solution The reaction is Hence, the correct option is (3) MATHEMATICS 61. If X = {4n– 3n – 1: n } and Y = {9(n – 1) : n }, where N is the set of natural numbers, then X Y is equal to (1) X (3) N (2) Y (4) Y – X Solution X = {4n 3n 1: n N} 4n 3n 1 = (1 + 3)n 3n 1 nC0 30 nC1 31 nC2 32 nCn 3n 3n 1 9(n C2 n Cn 3n 2 ) 9k , where k is an integer, when n 2 When n = 1, 4n 3 n 1 = 4 3 1 = 0 = 9 k Therefore, X contains multiples of 9. Now Y = {9 (n 1): n N} = {0, 9, 18,} = all multiples of 9 Thus, X Y Y Hence, the correct option is (2) 62. If z is a complex number such that |z| 2, then the minimum value of z 1 2 5 (1) is strictly greater 2 3 5 (2) is strictly greater than but less than 2 2 5 (3) is equal to 2 (4) lies in the interval (1, 2) Solution Min z 1 is when z is at B. 2 Therefore, Min z 1 3 1 1 lies in the interval (1, 2). BC 1 . Hence, Min z 2 2 2 2 Hence, the correct option is (4) 63. If a R and the equation, –3(x – [x])2 + 2(x – [x]) + a2 = 0 (where [x] denotes the greatest integer x) has no integral solution, then all possible values of a lie in the interval: (1) (–2, –1) (2) (–, – 2) (2, ) (3) (–1, 0) (0, 1) (4) (1, 2) Solution –3(x – [x])2 + 2(x – [x]) + a2 = 0 3{x}2 + 2{x} + a2 = 0 2 2 2 2 1 1 1 1 a 2 3{x}2 2{x} 3 {x}2 {x} 3 {x} 3 3 3 3 3 Now we know that 0 {x} < 1. 2 2 1 4 1 4 1 1 2 Therefore, {x} 0 {x} 0 3 {x} 3 3 3 3 9 3 3 2 1 1 1 3 {x} 1 3 3 3 1 a2 < 1 Only possibility for non-integral solution is 0 < a2 < 1 Thus, a2 0, but when a = 0, there is integral solution for {x} = 0 Therefore, ( 1, 0) (0, 1) Hence, the correct option is (3) 64. Let and be the roots of the equation px2 + qx + r = 0, p 0. If p, q, r are in A.P. and 1 1 4, then the value of is 34 9 2 13 9 (3) 61 9 2 17 9 Solution Given that, 1 2q = p + r 1 (1) 4 (2) Since and are roots of equation px2 + qx + r = 0, we have: r p q p (3) (4) From eq (1), we have 4 From eq (2) and (3), we have 2( ) Therefore, 2 ( ) 1 1 2 4 Therefore, equation having roots 1 and 1 2q p r r 1 1 p p p 1 1 1 9 is x2 4x 9 = 0 1 4 9 0 9x2 + 4x 1 = 0 x2 x 16 16 36 52 2 13 2 1 Therefore, 4 4 81 81 9 9 9 Hence, the correct option is (2) Therefore, equation having roots and is 3 1 f (1) 1 f (2) 65. If , β 0, and f(n) = + and 1 f (1) 1 f (2) 1 f (3) K (1 )2 (1 )2 ( )2 , then K is n n 1 f (2) 1 f (3) 1 f (4) equal to (1) 1 (2) –1 1 (3) β Solution (4) 111 1 1 2 2 1 1 1 1 2 2 1 3 3 1 1 2 2 1 3 3 1 4 4 1 2 1 1 1 1 1 2 = {(1 ) (1 ) ( )2)} 2 1 2 On comparison with the given equation, we get K = 1. Hence, the correct option is (1) 66. If A is a 3 × 3 non-singular matrix such that AAAA and B A–1A, then BBequals (1) B–1 (2) (B–1) (3) I + B (4) I Solution BB = A1A(A1A) = A1AA(A1) = A1AA(A1) = IA(A1) = I(A1A) = I.I = I2 = I Hence, the correct option is (4) 67. If the coefficients of x3 and x4 in the expansion of (1 + ax + bx2)(1 – 2x)18 in powers of x are both zero, then (a, b) is equal to 14, 272 16, 272 3 3 (3) 16, 251 3 14, 251 3 Solution (1 + x + bx2)(1 2 x)18 (1 ax bx 2 ){18 C0 (2 x)0 18 C1 (2 x)1 18 C2 (2 x)2 18 C3 (2 x)3 18 C4 (2 x)4 } Coefficient of x3 18C3 (2)3 a 18C2 (2)2 b 18C1 (2) 3 2 8 18 17 16 18 17 a 4 2b 18 3 2 1 2 1 = 4 {2 3 17 16 9 17 a + a b} = 0 153a ab = 1632 51a 3b = 544 272 Only 16, satisfies it 3 Hence, the correct option is (2) 68. If (10)9 + 2(11)1 (10)8 + 3(11)2 (10)7+ … + 10(11)9 = k (10)9, then k is equal to (1) 100 (2) 110 121 441 (3) 10 (4) 100 Solution Let S = 1 109 + 2 111 108 + 3 112 107 + 9 10 118 + 10 119 (1) 11 S = 11 108 + 2 112 107 + 3 113 106 + 9 119 10 On subtracting (2) from (1), we get 1 S = 11 108 112 107 119 + 1110 109 10 = 109 11 108 112 107 119 + 1110 Therefore, 11 / 10 1 11 10 10 9 10 (11 / 10) 1 10 11 109 11 / 10 1 1 / 10 (2) 10 10 11 1 S 1010 10 1 1110 1110 1010 1110 10 10 11 9 S = 10 = 100 10 Therefore, k = 100 Hence, the correct option is (1) 69. Three positive numbers form an increasing G.P. If the middle term in this G.P. is doubled, the new numbers are in A.P. Then the common ratio of the G.P. is (1) 2 3 2 3 (3) 2 3 3 2 Solution Let the numbers are a, a r, a r2. Now 2[2ar] = a + a r2 4 a r = a + a r2 r2 4 r + 1 = 0 r 4 16 4 2 3 2 Since the numbers form an increasing G.P. so r 1 . Therefore, r 2 3 Hence, the correct option is (2) sin( cos 2 x) is equal to x 0 x2 (1) –π (2) π 70. lim (3) 2 Solution 2 sin( cos 2 x) sin (1 sin x) sin( sin 2 x) x 0 x2 x2 x2 lim x 0 sin( sin 2 x) sin( sin 2 x) sin 2 x x2 sin 2 x x2 x0 x 0 =1= Hence, the correct option is (2) 71. If g is the inverse of a function f and f '( x) (1) 1 1 {g ( x)}5 x 1 , then gʹ(x) is equal to 1 x5 1 {g ( x)}5 x Solution Given g (x) = f1 (x) Therefore, f (g(x)) = x f(g(x)) g(x) = 1 Thus, g '( x) Thus, 1 f ' g ( x) 1 1 1 {g ( x )}5 g (x) = 1+ {g(x)}5 Hence, the correct option is (2) 72. If f and g are differentiable functions in [0, 1] satisfying f(0) = 2 = g(1), g(0) = 0 and f(1) = 6, then for some c ]0, 1[ (1) f (c) = g(c) (2) f (c) = 2g(c) (3) 2f (c) = g(c) (4) 2f (c) = 3g(c) Solution Let h (x) = f (x) 2g(x) So that h (0) = f (x) 2 g(x) = 2 2 0 = 2 And h (1) = f (1) 2 g(1) = 6 2 2 = 2 Now h(x) is a differentiable function in [0, 1] and h(0) = h(1), so by Rolle’s theorem h(c) = 0 for some c (0, 1) Therefore, 0 = f (c) 2 g (c) f (c) = 2 g (c) Hence, the correct option is (2) 73. If x = –1 and x = 2 are extreme points of f (x) log |x| x2 x, then 1 1 (1) 2, 2, 2 2 (3) 6, 1 2 6, 1 2 Solution f (x) = log(x) + x2 + x 2 x 1 x Now f( 1) = 2 + 1 = 0 f '( x) 4 x 1 0 2 Thus, the equations are 2 1 0 and 8 2 0 On solving the above equations, we get 1 , 2 2 Hence, the correct option is (1) f '(2) 1 1 x 74. The integral 1 x e x dx is equal to x (1) ( x 1)e ( x 1)e x x 1 x 1 x c c xe xe x x 1 x 1 x c c Solution 1 1 1 x x 1x 1 x x 1 x x x 1 x e dx e dx x 1 e dx xe c 2 x x xf '( x ) f ( x )dx xf x c Hence, the correct option is (4) 75. The integral 1 4sin 2 0 x x 4sin dx equals 2 2 (1) 4 3 4 4 3 4 4 3 2 44 3 3 Solution 0 1 4sin 2 x x x 4sin dx 1 2sin 2 2 2 0 2 2 x x x 1 2sin dx 3 1 2sin dx 1 2sin dx 0 0 2 2 2 3 x 3 x 2 cos 2 cos 2 2 x x (1 / 2) (1 / 2) 0 3 2 3 (4) 4(0) 2 3 3 3 3 2 3 4 3 2 3 2 4 3 4 4 3 4 3 3 Hence, the correct option is (2) 76. The area of the region described by A = {(x, y): x2 + y2 1 and y2 1 – x} is 2 2 2 3 2 3 2 4 3 2 4 3 Solution y2 = 1 x x = 1 y2 1 1 y3 1 4 1 Required Area is ( 12 ) 2 (1 y 2 )dy 2 y 2 1 0 0 2 3 0 2 2 3 2 3 Hence, the correct option is (3) 77. Let the population of rabbits surviving at a time t be governed by the differential equation dp(t ) 1 p(t ) 200 If p(0) = 100, then p(t) equals dt 2 (1) 600 – 500 et/2 (2) 400 – 300 e–t/2 t/2 (3) 400 – 300 e (4) 300 – 200 e–t/2 Solution dp (t ) 1 dp (t ) p(t ) 200 dt 1 dt 2 p (t ) 200 2 On integrating we get p(t ) log 200 log p(t ) 400 1 c 2 t c1 (1) t 1 (1/ 2) 2 2 2 Using initial Conditions log 100 400 c1 (2) 2log150 c1 2 2 From eqn. (1) and eqn(2) we get p (t ) 400 1 p (t ) 400 1 log t log150 log t 2 2 300 2 p (t ) 400 et / 2 or p (t ) 400 300et / 2 300 Since initially number of rabbits is 100 and is decreasing ( p(t ) 400) 300t / 2 or p(t ) 400 300et /2 Hence, the correct option is (3) 78. Let PS be the median of the triangle with vertices P(2, 2), Q(6, –1) and R (7, 3). The equation of the line passing through (1, –1) and parallel to PS is (1) 4x + 7y + 3 = 0 (2) 2x – 9y – 11 = 0 (3) 4x – 7y – 11 = 0 (4) 2x + 9y + 7 = 0 Solution Slope of required line is 1 2 ( 9 / 2) 9 2 Equation is y 1 ( x 1) 9 y + 9 = 2 x + 2 or 2 x + 9 y + 7 = 0 9 Hence, the correct option is (4) 79. Let a, b, c and d be non-zero numbers. If the point of intersection of the lines 4ax + 2ay + c = 0 and 5bx + 2by + d = 0 lies in the fourth quadrant and is equidistant from the two axes then (1) 3bc – 2ad = 0 (2) 3bc + 2ad = 0 (3) 2bc – 3ad = 0 (4) 2bc + 3ad = 0 Solution 4ax 2ay c 0 5bx 2by d 0 x y 1 2ad abc 5bc 4ad 8ab 10ab 2 ( ad bc ) bc ad x 2 ( ab) ab 5bc 4ad y 2ab Now according to question point being in 4th quadrant and equidistance from axes, lies on y = x 5bc 4ad (bc ad ) Therefore, 2 ab ab 5bc 4ad = 2(bc ad) 5bc 4ad = 2bc 2ad 3bc 2ad = 0 Hence, the correct option is (1) 80. The locus of the foot of perpendicular drawn from the centre of the ellipse x2 + 3y2 = 6 on any tangent to it is (1) (x2 + y2)2 = 6x2 + 2y2 (2) (x2 + y2)2 = 6x2 – 2y2 (3) (x2 – y2)2 = 6x2 + 2y2 (4) (x2 – y2)2 = 6x2 – 2y2 Solution x2 y2 1 ( 2)2 x cos y sin Eqn. of tangent at P is 1 6 2 Ellipse is ( 6)2 Equation of tangent at (, ) is (1) ( x ) or y 2 = x + 2 or x + y = 2 + 2 (2) Comparing Eq (1) and Eq (2), we get cos sin 1 2 2 6 2 y cos 2 sin 2 1 1 2 2 2 2 6 2 6 2 2 2 6 2 +2 2 = (2 + 2) Therefore, locus is (x2 + y2)2 = 6 x2 + 2 y2 Hence, the correct option is (1) or 81. Let C be the circle with centre at (1, 1) and radius = 1. If T is the circle centred at (0, y), passing through origin and touching the circle C externally, then the radius of T is equal to 1 1 (1) 2 (2) 4 3 2 Solution (3) (4) 3 2 AB =1 + y BC = 1 y Now by Pythagoras theorem (1 + y) 2 = 12 + (1 y) 2 4 y = 1 Therefore, y 1 4 Hence, the correct option is (2) 82. The slope of the line touching both the parabolas y2 = 4x and x2 = –32y is 1 2 (1) (2) 8 3 1 3 (3) (4) 2 2 Solution Let a Parametric point on y2 = 4x be (t2, 2 t) Therefore, Equation of tangent is yt = x + t2 or x = yt t2 (1) Now solving x = yt t2 and x2 = 32y for points of intersection. Therefore, (yt t2)2 = 32y or y2t2 + t4 2yt3 = 32y or y2t2 + y(32 2t3) + t4 = 0 For line to be tangent, discriminate = 0 Therefore, (32 3t3)2 4t6 = 0 or 4(16 t3)2 4 (t3)2 = 0 3 3 (16 t t )(16 t 3 t 3 ) 0 d 3 8 t 2 1 1 Now slope (from 1) t 2 Hence, the correct option is (3) 83. The image of the line x3 3 x3 (3) 3 (1) y 5 1 y 5 1 z2 5 z2 5 x 1 y 3 z 4 in the plane 2x – y + z + 3 = 0 is the line 3 1 5 x3 y 5 z 2 3 1 5 x3 y 5 z 2 3 1 5 Solution Since, 3 (2) + 1 (1) + (5) (1) = 6 1 5 = 0 Therefore, Line is parallel to plane Now finding image of point (1, 3, 4) in the plane. Let the image is (x1, y1, z1) x1 1 y1 3 z1 4 2 2(1) (3) (4) 3 2 6 Therefore, 2 1 1 22 12 12 6 Therefore, x1 = 3, y1 = 5, z1 = 2 x3 y 5 z 2 Therefore, required image line is 3 1 5 Hence, the correct option is (3) 84. The angle between the lines whose direction cosines satisfy the equations l + m + n = 0 and l2 = m2 + n2 is (1) (3) 6 (2) 2 (4) 3 4 Solution Since, l=mn Therefore, ( m n)2 = m2 + n2 or 2 m n = 0 m = 0 or n = 0 If m = 0, then l2 = n2l = n 1 1 Therefore, n2 + 0 2 + n 2 = 1 n 2 n 2 2 1 1 ,0, Therefore, directions are 2 2 If n = 0, then l = m m2 + m2 + 0 = 1 1 1 Thus, m 2 m 2 2 1 1 , ,0 Therefore, directions are 2 2 1 cos 0 0 2 3 Hence, the correct option is (3) 85. If [a b b c c a ] [a b c ]2 , then is equal to (1) 0 (2) 1 (3) 2 (3) 3 Solution [a b.b cc a ] (a b).[(b c) (c a)] (a b).[(b c) c a] (a b).[(b c a)c (b c c)a ] (a b).[(b c.a )c 0] (a b.c)(a b.c) (a b.c)2 [ab.c]2 Therefore, = 1 Hence, the correct option is (2) 1 1 1 86. Let A and B be two events such that P( A B) , P( A B) and P( A) , where A stands for 6 4 4 the complement of the event A. Then the events A and B are (1) Independent but not equally likely (2) Independent and equally likely (3) Mutually exclusive and independent (4) Equally likely but not independent Solution 1 1 5 P ( A B) 1 C 6 6 1 (1) P( A B) 4 1 1 3 P ( A) P ( A) 1 4 4 4 Now, P (A B) = P (A) + P (B) P (A B) 5 3 1 5 3 1 10 9 3 4 1 P ( B ) or P ( B ) 6 4 4 6 4 4 12 12 3 3 1 1 P( A) P( B) (2) 4 3 4 Therefore, from (1) and (2), A and B are independent. Since P (A) P (B), therefore not equally likely Hence, the correct option is (1) P( A B) 87. The variance of first 50 even natural numbers is (1) 437 (3) (2) 833 4 437 4 (4) 833 Solution Variance = 2 Mean of squares Square of Mean 50 50 (2r ) (2r ) = r 1 r 1 n n 2 2 4(12 22 32 502 ) 2(1 2 3 50) 50 50 50 51 50 ( 51)(101) 4 4 6 50 2 56 = 3434 512 = 833 Hence, the correct option is (4) 2 2 1 88. Let f k ( x) (sin k x cos k x) where x R and k ≥ 1. Then, f 4 ( x) f 6 ( x) equals k 1 1 (1) (2) 4 12 1 1 (3) (4) 6 3 Solution 1 1 f 4 ( x) f 6 ( x) (sin 4 x cos 4 x) (cos 6 x sin 6 x) 4 6 1 1 [(sin 2 x cos 2 x) 2 2sin 2 x cos 2 x] [(cos 2 x sin 2 x)3 3cos 2 x sin 2 x(cot 2 x sin 2 x)] 4 6 1 1 [1 2sin 2 x cos 2 x] [1 3cos 2 x sin 2 x] 4 6 1 1 1 1 3 12 1 2 2 sin x cos x cos 2 x sin 2 x 4 2 6 2 12 12 Hence, the correct option is (2) 89. A bird is sitting on the top of a vertical pole 20 m high and its elevation from a point O on the ground is 45°. It flies off horizontally straight away from the point O. After one second, the elevation of the bird from O is reduced to 30°. Then, the speed (in m/s) of the bird is (1) 20 2 (2) 20( 3 1) (3) 40( 2 1) (4) 40( 3 2) Solution d 20 cot 30 3 d 20 3 20 20( 3 1) 20 Hence, the correct option is (2) 90. The statement ~(p ~q)is (1) A tautology (2) A fallacy (3) Equivalent to p q (4) Equivalent to ~ p q Solution p q q p q (p q) pq T T F F T T T F T T F F F T F T F F F F T F T T Bi-conditional Hence, the correct option is (3)