Physics 1. A biconcave glass lens having refractive index 1.5 has both surfaces of the same radius of curvature R. On immersion in a medium of refractive index 1.75, it will behave as a (1) convergent lens of focal length 3.5 R. () convergent lens of focal length 3.0 R. (3) divergent lens of focal length 3.5 R. (4) divergent lens of focal length 3.0 R. According to lens maker formula 1 1 1 1 1 (1.5 1) f 1 R1 R R R Now, for the second material 1 1.5 0.5 0.5 1 R f 1.75 R 1.75R 1.75 f 3.5R As the focal length is positive, it will behave as a convergent lens... An electron in the hydrogen atom is in the n = 4 energy level. When this electron makes a transition to a lower energy level, the wavelength of the photon emitted is in the (1) the Lyman series only () the Balmer series only (3) the Paschen series only (4) any of the above For Lyman series, electron jumps from any higher energy state to ground state n = 1; for Balmer series, electron jumps from any higher energy state to state n = ; and for Paschen series, electron jumps from any higher energy state to state n = 3. Hence, the correct option is (4). 3. Statement 1: Static reaction is less in pulling than in pushing. Statement : (coefficient of static friction) s k (coefficient of kinetic friction). (1) Both Statements 1 and are true and Statement is correct explanation of Statement 1. () Both Statements 1 and are true and Statement is not correct explanation of Statement 1. (3) Statement 1 is true but Statement is false. (4) Statement 1 is false but Statement is true. Static friction is less in pulling than in pushing. For pulling,
For pushing, From above figures we can conclude that in pushing, normal reaction is greater than pulling. Hence, static friction increases. Statement 1 is true. (Coefficient of static friction) s k (Coefficient of kinetic friction). Statement is true. Both statements 1 and are true, and statement is not correct explanation of statement 1. Hence the correct option is (). 4. A cylindrical vessel contains a liquid of density up to height h. The liquid is closed by a piston of mass m and of cross section A. There is a small hole at the bottom of the vessel. The speed v with which the liquid comes out of the hole is mg (1) gh () gh A
mg (3) gh A mg (4) gh A We have mg PA P0 A PB P0 Applying Bernoulli s equation, 1 1 PA gh v1 PB v (1) A Now, A1 v1 A v v1 v A mg Substituting in Eq. (1), we get v gh A Hence, the correct option is () 5. Two trains, one travelling at 15 m/s and the other at 0 m/s, are heading towards one another along a straight track. Both the drivers apply brakes simultaneously when they are 500 m apart. If each train has a retardation of 1 m/s, the separation after they stop is (1) 19.5 m () 5.5 m (3) 187.5 m (4) 155.5 m For the first train, Final velocity v = 0 m/s Initial velocity u = 15 m/s Retardation a = 1 m/s Let the time be t. Now, applying the first kinematics relation, we get v = u + at 0 = 15 t t = 15 s Now applying the second kinematics relation, we get 1 15 5 s1 ut at 1515 m For the second train, Final velocity v = 0 m/s Initial velocity u = 0 m/s Retardation a = 1 m/ s Let the time be t. Now, applying the first kinematics relation we get v = u + at 0 = 0 + t t = 0 s Now applying the second kinematics relation, we get 1 1 400 s ut at 0 0 0 1
Hence, the distance between the trains will be 5 400 D 500m ( s1 s ) 500m 65 500 187.5 m Hence, the correct option is (3) 6. A H coil, a 10 μf resistor, and a 5 V battery are all connected in parallel. After a current is well established through the resistor and inductor, the battery is disconnected from the circuit by opening a switch as shown in the figure. Immediately after the battery is disconnected, Lenz s law would predict (1) The voltage across R is zero () The current through R reverses direction (3) The current through the inductor reverses direction, but its magnitude tends to remain the same (4) The magnitude and direction of the voltage across the inductor tend to remain the same When the switch is disconnected suddenly, the current in the inductor won t change abruptly, due to Lenz s law. So, the current in the loop A B C D would be i and the direction of current in the resistor would be reverse. Hence, the correct option is (3) 7. The magnifying power of telescope can be increased by (1) increasing the focal length of eyepiece () increasing the distance of object
(3) fitting eyepiece of low power (4) fitting eyepiece of high power The magnifying power of a telescope is given as o m f. f e Therefore, it can be increased by either increasing the focal length of the objective eyepiece. For a lower focal length of the eyepiece, it would have high power according to the equation 1 P e. fe Therefore, the magnifying power of telescope can be increased by fitting eyepiece of high power. Hence, the correct option is (4) 8. A filament emits electrons at a constant rate. The electrons are then subjected to a constant electric field as shown in the figure. The cross-sectional area of the beam remains constant. Statement 1: The current at cross-section B is greater than the current at A. Statement : The speed of the electrons is greater at B than at A. (1) Both Statements 1 and are true and Statement is correct explanation of Statement 1. () Both Statements 1 and are true and Statement is not correct explanation of Statement 1. (3) Statement 1 is true but Statement is false. (4) Statement 1 is false but Statement is true. The current at cross-section B is greater than the current at A. This statement is false because electrons are moving in direction opposite to the direction of electric field, thus speed of electrons at cross-section A is less than the speed of electrons at cross-section B. The speed of the electrons is greater at B than at A. This is because cross-section B is at more positive potential w.r.t. to cross-section A. Thus, Statement 1 is false but Statement is true. Hence, the correct option is (4) 9. A person bounces up and down on a trampoline, while always staying in contact with it. The motion is simple harmonic motion, and it takes 1.90 s to complete one cycle. The height of each bounce above the equilibrium position is 45.0 cm. Determine the maximum speed attained by the person? (1) 1.5 m/s ().5 m/s (3) 5. m/s (4) 3.5 m/s Since the distance from the equilibrium position to the point of maximum height is the amplitude A of the motion, we have that A = 45.0 cm = 0.45 m The angular frequency is inversely proportional to the period of the motion:
3.31 rad/s T 1.90 s The maximum speed v max attained by the person on the trampoline is vmax A (0.450 m)(3.31 rad/s) 1.49 m/s 10. In the figure, three particles of mass m = 3 g are fastened to three rods of length d = 1 cm and negligible mass. The rigid assembly rotates around point O at angular speed = 0.85 rad/s. About O, what is the rotational inertia of the assembly? (1) (3) 4.6 10 kg m 3 () 4.9 10 kg m 3 (4) 4.3 10 kg m 3 4.0 10 kg m 3 A particle contributes mr to the rotational inertia. Here, r is the distance from the origin O to the particle. The total rotational inertia is 4.610 kg m I m 3d m d m d 14md 14(.310 kg)(0.1 m) 3 11. Charged beads are placed at the corners of a square in the various configurations as shown in the Figure. Black beads have charge +Q, whereas gray beads have charge Q. Rank the configurations according to the magnitude of the electric field at the center of the square, least to greatest. (1) I > V > III > II > IV () III > V> I > IV > II (3) I = III > V > II > IV (4) V > II > I = IV > III The field at the center for the cases I V is: 1 Q E I 4 0 a, 1 4Q E II 4 0 a, 0 1 Q E, E III IV 4 0 a, 1 6Q E V 4 0 a Hence, the correct option is (4) 1. Two identical particles A and B, are attached to a string of length l, A to middle and B to one of the ends. The string is whirled in a horizontal circle, with the end O fixed (as shown in the figure). If the kinetic energy of B relative to A is E, then the absolute kinetic energies of A and B are
(1) E and E () E and 4E (3) 4E and E (4) E and 3E Since A and B are rotating with the same angular velocity, va l and vb l The kinetic energies are 1 1 E m( vb va ) m( l) 1 EA mva E 1 1 EB m( l) 4 m( l) 4E Hence, the correct option is () 13. Consider a parallel-plate capacitor originally with a charge q 0, capacitance C 0, and potential difference ΔV 0. There is an electrostatic force of magnitude F 0 between the plates, and the capacitor has a stored energy U 0. The terminals of the capacitor are connected to another capacitor of same capacitance and charge. Later the dielectric slab is removed. While the slab is being removed (1) charge on second capacitor increases () charge on second capacitor decreases (3) charge on second capacitor is constant (4) cannot be determined
Extra charge from the first capacitor will be transferred back to second capacitor, as capacitance of the first begins to decrease. 14. Green light has a wavelength of about 500 nm. Through what potential difference must an electron be accelerated to have this wavelength? (1) 6.0 μv () 5.0 μv (3) 7.0 μv (4) 5.5 μv Wavelength in terms of potential difference is given by: h mqv So, potential difference is: h V m e For electron m = m e = 9.1 10 31 kg, q = e = 1.6 10 19 C and for green light, wavelength () is 500 10 9 m. Therefore, V 6μV 15. In the modified form of Coulomb s law F 1 e I I 1 4 0 r If I 1 and I are currents because of flowing charges q 1 and q. Then the dimensional formula for 0 is (1) M L T A M L T A 1 3 () 1 3 4 (3) M L T A 1 3 (4) 1 3 4 1 M L T A We know that, Now, So, 1 I1I 1 I1I F 0 4 r 4 F r 0 F MLT I1I A r L A A 1 3 0 M L T A 3 MLT L ML T
16. A linearly polarized transverse wave is propagating in z-direction through a fixed point P in space. At time t 0, the x component E x and the y component E y of the displacement at P are 3 and 4 units, respectively. At a later time t 1, if E x at P is units, the value of E y will be (1) 5 units () 8/3 units (3) 3/8 units (4) 1/3 units For a polarized light wave its ratio of x- and y-components is always constant. Therefore, Ex E 1 x E E y1 y 3 4 E y Hence, the correct option is () E y 8 units 3 17. An elevator cab of mass m = 500 kg is descending with speed v = 4.0 m/s when its supporting cable begins to slip, allowing it to fall through a distance d = 1 m, with constant acceleration a =g/5. The net work done on the cab during the fall is (1) 1 kj () 15 kj (3) 150 J (4)10 J The figure can be represented as: From the figure, we see that the angle between the directions of Fg and the cab s displacement d is 0. W g mgd cos 0 4 5.88 10 J 59 kj (500 kg)(9.8 m/s )(1 m)(1) We can calculate work W T as W T = Td cos = m(a + g)d cos. So,
g 4 WT m g d cos mgd cos 5 5 4 (500kg)(9.8m/s )(1m)cos180 5 4 4.70 10 J 47 kj Net work = W = W g + W T = 5.88 10 4 J 4.70 10 4 J = 1.18 10 4 J 1 kj. 18. Two small satellites move in circular orbits around a planet, at distance r and r r from the center of the planet. Their time periods of rotation aret and T T respectively. ( r r, T T). Then the value of T is 3 r 3 r (1) T T () T T r r r r (3) T T (4) T T 3 r r Using Kepler s third law, 3 T a Where, a is semi-major axis circular orbit, which is special case of ellipsoidal orbit with a = b 3 T r ( T T ) ( r r) 3 r 3 r 1 1 r r T 3 r 1 1 T r T 3 r T r 3r T T r 3/ 3 3/ T T r r 1 T r 1 T r T r 19. In the figure given below a 10 g bullet moving directly upward at 1000 m/s strikes and passes through the center of mass of a 5.0 kg block initially at rest. The bullet emerges from the block moving directly upward at 400 m/s. To what maximum height does the block then rise above its initial position?
(1) 0.063 m () 0.073 m (3) 0.070 m (4) 0.065 m We think of this as having two parts: the first is the collision itself where the bullet passes through the block so quickly that the block has not had time to move through any distance yet and then the subsequent leap of the block into the air (up to height h measured from its initial position). The first part involves momentum conservation (with +y upward): 0.01kg1000m s 5.0kgv 0.01kg 400m s which yields v 1. m s. The second part involves either the free-fall equations (since we are ignoring air friction) or simple energy conservation. Choosing the latter approach, we have 1 5.0kg 1.m s 5.0kg 9.8m s h which gives the result h = 0.073 m. Hence, the correct option is () 0. A cylindrical chamber A of uniform cross-section is divided into two parts X and Y by a movable piston P which can slide without friction inside the chamber (figure below). Initially part X contains 1 mol of a monoatomic gas and Y contains mol of a diatomic gas, and the volumes of X and Y are in the ratio 1: with both parts X and Y being at the same temperature T. Assuming the gases to be ideal, the work W that will be done in moving the piston slowly to the position where the ratio of the volumes of X and Y is :1 will be (1) 5.8T () 8.3T (3) 1.3T (4) zero Total work done = Work done in chamber X + Work done in chamber Y V X V Y RT ln RT ln V1 X V1 Y 1 RT ln() RT ln RT ln(4) 5.8T 1. A uniform electric field exists in the vertically downward direction. The magnitude is 10 N/C. What is the increase in electrostatic potential as one goes up through a height of 50 cm? (1) 0 V () 15 V (3) 10 V (4) 5 V The relation between electric field and potential is dv E dr dv Edr
Taking the magnitude only we get dv Hence, the correct option is (4) Edr (10 N/C) (0.5m) 5V. A small but measurable current of 1. 10 10 A exists in a copper wire whose diameter is.5 mm. The number of charge carriers per unit volume is 8.49 10 8 m 3. Assuming the current is uniform, calculate the current density and electron drift speed. 5 15 5 15 (1).410 A/m, 1.8 10 m/s ().0 10 A/m, 1.8 10 m/s (3) 5 15.4 10 A/m, 1.4 10 m/s (4) 5 15.0 10 A/m, 1.4 10 m/s : (a) The magnitude of the current density vector is 10 i i 4(1.10 A) 5 J.410 A/m 3 A d / 4 (.510 m) (b) The drift speed of the current-carrying electrons is 5 J.410 A/m 15 vd 1.810 m/s 8 3 19 ne (8.47 10 m )(1.6010 C) 3. A planet is at average distance d from the sun, and its average surface temperature is T. Assume that the planet receives energy only from the sun, and loses energy only through radiation from its surface. n Neglect atmospheric effects. If T d,the value of n is (1) () 1 (3) 1/ (4) 1/4 According to Stefan Boltzmann law, u AT Hence the correct option is (3) 4 1 1 T T T d 1/4 1/4 ( A) ( d ) 4. Statement 1: When a current is passed through a helically coiled spring, the spring contracts as if it is compressed. Statement : Two parallel current-carrying wires attract each other. (1) Both Statements 1 and are true and Statement is the correct explanation of Statement 1. () Both Statements 1 and are true but Statement is not the correct explanation of Statement 1. (3) Statement 1 is true but Statement is false. (4) Statement 1 is false but Statement is true. Consider the following figure, 1/
As a helical coil has many parallel coils carrying current, it is compressed due to attractive force between these coils. 5. A film of water is formed between two straight parallel wires of length 10 cm each separated by 0.5 cm. If their separation is increased by 1 cm while still maintaining their parallelism, how much work will have to be done (Surface tension of water = 7. 10 N/m) (1) 7. 10 6 J () 1.44 10 4 J (3).88 10 6 J (4) 5.76 10 5 J Surface tension of water ( s) 7. 10 N m Length of parallel wires 10cm 0.10m The parallel wires are separated by 0.5cm 0.005m And, also their separation is increased by1cm. Hence, we know that Work done Surface tensionincrease in area Increase in area [0.10 0.015 0.10 0.005] 0.00m So, Work done Surface tension Increase in area 7.10 0.00 Hence, the correct option is () 4 1.4410 J 6. The diagram of a logic circuit is given below. The output F of the circuit is represented by (1) W (X Y) () W (X Y)
(3) W (X Y) (4) W (X Y) The out of the upper OR gate =W+X The out of the lower OR gate =W+Y = W+X W+X The out of the AND gate So, when we simplify it, we get W+X W+X =W W+W X+Y +X Y (as W W W, and 1 X Y 1) Hence, the correct option is (3) =W 1+X+Y +X Y =W+X Y 7. An open pipe is suddenly closed at one end, as a result of which the frequency of the third harmonic of the closed pipe is found to be higher by 100 Hz. Then the fundamental frequency of the open pipe is (1) 00 Hz () 300 Hz (3) 40 Hz (4) 480 Hz Let f0 be fundamental frequency of open pipe having frequencyvand length l v f0 l Third harmonic of closed pipe: nv f3 4l Here n = 3, so, 3v f3, 4l 3v v 100 4l l v 00Hz l 8. Statement 1: The photon s kinetic energy becomes four times greater, when the wavelength of radiation reduces to half. Statement : The photon s momentum doubles, when the wavelength of radiation reduces to half. (1) Both Statements 1 and are true and Statement is correct explanation of Statement 1. () Both Statements 1 and are true and Statement is not correct explanation of Statement 1. (3) Statement 1 is true but Statement is false. (4) Statement 1 is false but Statement is true. The expressions for photon energy and wavelength are as follows: hc h E p Hence, the correct option is (4)
9. The figure shows the circular path taken by an electron in a uniform magnetic field of magnitude 6 B 4.00 10 T. The electron s constant speed is v 00m/ s. The frequency of the motion f is 5 5 (1) 1.5 10 Hz ().15 10 Hz (3) 5 1.1 10 Hz (4) 5.1 10 Hz The period is the time taken to make one full circle: Circumference r T Speed v r r v mv q B m q B m 31 (9.1110 kg) q B 19 6 (1.6010 C)(4.010 T) T (since q e) 6 8.94410 s 8.94 μs The frequency is the number of circles per second: 1 5 1 5 f 1.11810 s 1.1 10 Hz T Hence, the correct option is (3) 30. A transmitting antenna at the top of a tower has a height of 36 m, and the height of the receiving antenna is 49 m. The maximum distance between them for satisfactory communication in the line-of-sight mode is (Use radius of the Earth = 6400 km.) (1) 40.5 km () 45.6 km (3) 46.5 km (4) 45.0 km For maximum distance between them for satisfactory communication in the line-of-sight mode is: d d d Rh Rh M T R T R Hence, the correct option is (3) d M 6 6 6.4 10 36 6.4 10 49 46500m 46.5km