NEET KEY ANSWER WITH DETAILED SOLUTIONS - CODE H2 - Copy.pdf

1. The ratio of radius of gyration of a solid sphere of mass M and radius R about its own axis to the radius of gyration of the thin hollow sphere of same mass and radius about its axis is: (1) 2:5 (2) 5:2 (3) 3:5 (4) 5:3 Ans: No Options Solid sphere 𝐼 𝑠 = 𝑀 𝐾 𝑠 2 2 5 𝑀 𝑅 2 = 𝑀 𝐾 𝑠 2 𝐾 𝑠 = √ 2 5 𝑅 2 Hollow sphere 𝐼 𝐻 = 𝑀 𝐾 2 2 3 𝑀 𝑅 2 = 𝑀 𝐾 𝐻 2 𝐾 𝐻 = √ 2 3 𝑅 2 𝐾 𝑠 𝐾 𝐻 = √ 2 5 𝑅 2 3 2 𝑅 2 = √ 3 5 2. A 12 V, 60 W lamp is connected to the secondary of a steep down transformer, whose primary is connected to ac mains of 220V. Assuming the transformer to be ideal, What is the current in the primary winding? (1) 3.7A (2) 0.37A (3) 0.27A (4) 2.7A Ans: 3 For Ideal transformer 𝑃 𝑜𝑢𝑡𝑝𝑢𝑡 = 𝑃 𝑖𝑛𝑝𝑢𝑡 60 = 𝑉 𝑝 𝐼 𝑝 𝐼 𝑝 = 60 220 = 0 27 𝐴 3. If the galvanometer G does not show any deflection in the circuit shown, the value of R is given by: (1) 100  (2) 400  (3)200  (4)50  SECTIO N - A Ans : (1) Applying Kirchhoff’s loop rule ABCDEFA + 10 − 𝑖 ( 400 ) − 2 = 0 8 = i x 400 i = 1 50 A Again loop BCDEB +10 – 400i – R x i = 0 10= (400 + R) i 400 + R = 10 x 50 400 + R = 500 R = 100 Ω 4. A full wave rectifier circuit consists of two p - n junction diodes, a centre - tapped transformer, capacitor and a load resistance. Which of these components remove the ac ripple from the re ctified output? (1) Capacitor (2) Load resistance (3) A centre - tapped transformer (4) p - n junction diodes Ans: (1) Capacitor removes the A.C ripple from rectified outputs. 5. The work functions of Caesium (Cs). Potassium (K) and Sodium (Na) are 2.14eV, 2.30eV and 275eV respectively. If incident electromagnetic radiation has an incident energy of 2.20eV, which of these photosensitive surfaces may emit photoelectrons? (1) K only (2) Na only (3) Cs only (4) B oth Na and K Ans: (3) For Photo electrons Energy of Photon > Work function So, E > φ cs Emits Photoelectrons E < φ K E < φNa Does not emit photoelectrons 6. The radius of frequencies of fundamental harmonic produced by an open pipe to that of closed pipe having the same length is: (1) 1:3 (2) 3:1 (3) 1:2 (4) 2:1 Ans: (4) 𝑓 = 1 2 𝑙 𝑣 𝑓 𝑐𝑙𝑜𝑠𝑒𝑑 = 1 4 𝑙 𝑣 𝑓 𝑜𝑝𝑒𝑛 𝑓 𝑐𝑙𝑜𝑠𝑒𝑑 = 𝑣 2 𝑙 4 𝑙 𝑣 = 2 1 7. The amount of energy required to form a soap bubble of radius 2 cm from a soap solution is nearly: (surface tension of soap solution=0.03 N m - 1) (1) 3.01×10 - 4 J (2) 50.1×10 - 4 j (3) 30.16×10 - 4 J (4) 5.06×10 - 4 J An s: (1) 𝑊 = 8 𝜋 𝑅 2 𝑇 = 8 × 3 14 × ( 2 × 10 − 2 ) 2 × 0 03 = 3 01 × 10 − 4 𝐽 8. Let a wire be suspended from the ceiling (rigid support) and stretched by a weight W attached at its free end. The longitudinal stress at any point of cross - sectional area A of the wire is: (1) / 2 W A (2)Zero (3) 2 / W A (4) / W A Ans: (4) Longitudinal stress = 𝑊𝑒𝑖𝑔 ℎ 𝑡 𝐴 = 𝑤 𝐴 9. A vehicle travels half the distance with speed  and the remaining distance with speed 2  . Its average speed is: (1) 4 3  (2) 3 4  (3) 3  (4) 2 3  Ans: (1) Average speed = 2 𝑉 1 𝑉 2 ( 𝑉 1 + 𝑉 2 ) = 2 × 2 𝑉 × 𝑉 ( 𝑉 + 2 𝑉 ) = 4 𝑉 2 3 𝑉 = 4 𝑉 3 10. For Young’s double slit experiment, two statements are given below: Statement I: If screen is moved away from the plane of slits, angular separation of the fringes remains constant. Statement II: If the monochromatic source is replaced by another monochromatic source of larger wavelength, the angular separation of fringes decreases. In the light of the above statements, choose the correct answer from the options given below: (1) Statement I is a true but statement II is false (2) Statement I is false but statement II is true (3) Both statement I and statement II are true. (4) Both statement I and statement II are false Ans: (1) Angular separation of the fringes is independent of the distance between the screen and the slit. Angular separation of fringes is directly proportional to wavelength. 11. Light travels distance x in time 1 t in air and 10x in time 2 t in another denser medium. What is the critical angle for this medium? (1) 1 1 2 sin 10 t t −       (2) 1 1 2 10 sin t t −       (3) 1 2 1 sin t t −       (4) 1 2 1 10 sin t t −       Ans: (2 ) Critical angle 𝑖 𝑐 = sin − 1 1 𝜇 𝜇 = 𝑐 𝑣 ⇒ 𝑐 = 𝑥 𝑡 1 𝑎𝑛𝑑 𝑣 = 10 𝑥 𝑡 2 𝜇 = 𝑥 𝑡 1 10 𝑥 𝑡 2 = 𝑡 2 10 𝑡 1 𝑖 𝑐 = sin − 1 ( 1 𝑡 2 10 𝑡 1 ) ⇒ 𝑖 𝑐 = sin − 1 ( 10 𝑡 1 𝑡 2 ) 12. An ac source is connected to a capacitor C. Due to decrease in its operating frequency: (1) Displacement current decreases (2) capacitive reactance remains constant (3) Capacitive reactance decreases (4) displacement current increa ses Ans: (1) 1 c d c i i X c    = = = i c  = If frequency decreases, i d decreases 13. T he equivalent capacitance of the system shown in the following circuit is: (1) 6 F  (2) 9 F  (3)2 F  (4) 3 F  Ans: (3) (3)(6) 18 2 3 6 9 eq c F  = = = + 14. In a plane electromagnetic wave travelling in free space, the electric field component oscillates sinusoidally at a frequency of 10 2.0 10 Hz  and amplitude 1 48 Vm − . Then the amplitude of oscillating magnetic field is : (Speed of light in free space= 8 1 3 10 ) ms −  (1) 7 1.6 10 T −  (2) 6 1.6 10 T −  (3) 9 1.6 10 T −  (4) 8 1.6 10 T −  Ans: (1) 10 2 10 f Hz =  1 0 48 E Vm − = 8 0 0 8 8 16 10 3 10 E B T C − = = =   7 1.6 10 T − =  15. In hydrogen spectrum, the shortest wavelength in the Balmer series is  . The shortest wavelength in the Bracket series is : (1) 9  (2) 16  (3) 2  (4) 4  Ans: (4) 2 2 1 1 2 RZ    =     2 1 2 1 1 4 RZ    =     2 1 4 RZ  = 2 2 1 1 1 16 4 4 RZ RZ    = =     1 4  = 1 4   = 16. A metal wire has mass ( ) 0.4 0.002 +  g, radius ( ) 0.3 0.001  mm and length ( ) 5 0.02  cm. The maximum possible percentage error in the measurement of density will nearly be: (1) 1.6% (2) 1.4% (3) 1.2% (4) 1.3% Ans: (1) (0.4 0.002) , (0.3 0.001) m g R mm =  =  (5 0.02) l cm =  m m m v A l A l   = =  = 2 m R l   = 2 100 100 100 100 m R l m R l        =  +  +  0.002 0.001 0.02 100 2 100 100 0.4 0.3 5   =  +  +      0.5 0.67 0.4 1.57% 1.6% = + + =  17. A football player is moving southward and suddenly turns eastward with the same speed to avoid an opponent. The force that acts on the player while turning is: (1) Along north - east (2) along south - west (3) Along eastward (4) along northward Ans: (1) Initially i V u j = − f V ui = Force will act along the direction of change in momentum f i p mv mv  = − ( ) m ui u j = + i.e., along North East. 18. The temperature of a gas is 50 ° C. To what temperature the gas should be heated so that the rms speed is increased by 3 times? (1) 3097 K (2) 223 K (3) 669 ° C (4) 3295 ° C Ans: (4) 0 50 273 223 T K = − + = rms V T  223 4 V V T = 0 223 16 3568 3295 T K C =  = = 19. Resistance of a carbon resistor determined from colour codes is (22000  5%)  . The colour of third band must be: (1) Orange (2) Yellow (3) Red (4) Green Ans: (1) (22000 5%) R =   3 (22 10 5%) =    Colour code: Red Red Orange Gold. Third band is orange 20. A bullet is fired from a gum at the speed of 280 1 ms − in the direction 30 ° above the horizontal. The maximum height attained by the bullet is ( ) 2 9.8 ,sin 30 0.5 g ms  − = = : (1) 1000 m (2) 3000 m (3) 2800 m (4) 2000 m Ans: (1) 2 2 max sin 2 u H g  = 2 2 0 (280) sin 30 2 9.8 =  = 1000 m 21. A Carnot engine has an efficiency of 50% when its source is at a temperature 327°. The temperature of the sink is : (1) 100°C (2) 200°C (3) 27°C (4)15°C Ans: (3) 50 1 100 2  = = 0 327 327 273 600 source T c K = = + = 1 1 600 600 2 sink sink T T  = −  = 0 sin 300 27 k T k C = = 22. In a series LCR circuit, the inductance L is 10 mH, capacitance C is 1 F  and resistance R is 100  . The frequency at which resonance occurs is: (1) 1.59 / rad s (2) 1.59 kHz (3) 15.9 / rad s (4) 15.9 kHz Ans: (2) L = 10 mH 1 C F  = 100 R =  3 6 8 1 1 1 10 10 10 10 LC  − − − = = =   4 4 1 10 / 10 rad s − = = 4 10 1.59 2 2 3.14 f kHz   = = =  23. If 0 s E dS =  over a surface, then: (1) All the changes must necessarily be inside the surface. (2) The electric field inside the surface is necessarily uniform (3) The number of flux lines entering the surface must be equal to the number of flux lines leaving it. (4) The magnitude of electric field on the surface is constant Ans: (3) If number of flux lines entering the surface is equal to number of flux lines leaving the surface. Then 0 E d s =  24. Two bodies of mass m and 9m are placed at a distance R. The gravitational potential on the line joining the bodies where the gravitational field equals zero, will be ( G =gravitational constant): (1) 16 Gm R − (2) 20 Gm R − (3) 8 Gm R − (4) 12 Gm R − Ans: (1) Let field be zero at a distance x from ‘m’ at point ‘p’ then 2 2 (9 ) ( ) Gm G m x R x = − 2 2 1 9 ( ) x R x = − 1 3 x R x = − 4 R x = (9 ) 16 3 4 4 P Gm G m Gm V R R R − = − =             25. The magnitude and direction of the current in the following circuit is (1) 5 9 A from A to B through E (2) 1.5 A from B to A through E (3) 0.2 A from B to A through E (4) 0.5 A from A to B through E Ans: (4) Equivalent emf = 10 - 5 = 5 volt 2 1 7 10 eq R = + + =  5 0.5 10 eq eq i A R  = = = Current is 0.5 A from ‘A’ to ‘B’ through ‘E’ 26. The minimum wavelength of X - rays produced by an electron accelerated through a potential differenc e of V volts is proportional to: (1) 1 V (2) 2 V (3) V (4) 1 V Ans: (4) min nC eV  = min hc eV  = min 1 V   27. The angular acceleration of a body, moving along the circumference of a circle, is: (1) Along the tangent to its position (2) along the axis of rotation (3) Along the radius, away from centre (4) along the radius towards the cent re Ans: (2) Angular velocity acceleration of a body is along the axis of the rotation. 28. The magnetic energy stored in an inductor of inductance 4 F  carrying a current of 2A is : (1) 8mJ (2) 8 J  (3) 4 J  (4) 4mJ Ans: (2) Energy stored 2 1 2 Li = 6 2 1 (4 10 )(2 ) 2 − =  6 4 10 2 8 J  − =   = 29. The half - life of a radioactive substance is 20 minutes. In how much time, the activity of substance drops to 1 16 th       of its initial value? (1) 60 minutes (2) 80 minutes (3) 20 minutes (4) 40 minutes Ans: (2) 2 1/ 20min t utes = Activity after n half lives 0 2 n A = 0 A is initial activity 0 0 4 2 2 4 n A A n = = Time taken 4 20 80min utes =  = 30. The potential energy of a long spring when stretched by 2 cm is U. If the spring is stretched by 8 cm, potential energy stored in it will be: (1) 8U (2) 16U (3) 2U (4) 4U Ans: (2) 2 1 2 U kx = 2 1 (2 ) 2 i U U k = = = 2k 2 1 (8) 2 f U k = 1 (64) 32 2 k k = = 16(2 ) 16 k U = = 31. The venturi - meter works on: (1) The principle of parallel axes (2) The principle of perpendicular axes (3) Hu y gen’s principle (4) Bernoulli’s principle Ans: (4) Venturi meter works on the principle of Bernoulli’s principle 32. The net magnetic flux through any closed surface is : (1) Infinity (2) Negative (3) Zero (4) Positive Ans: (3) Since Magnetic fiel d lines form closed loops. N et magnetic flux through a closed surface is “zero” 33. Given below are two statements: Statement I : Photovoltaic devices can convert optical radiation into electricity. Statement II : Zener diode is designed to operate under reverse bias in breakdown region. In the light of the above statemen ts, choose the most appropriate answer from the options given below: (1) Statement I is correct but Statement II is incorrect (2) Statement I is incorrect but Statement II is correct (3) Both Statement I and Statement II are correct (4) Both statement I and statement II are incorrect Ans: (3) Both statement are correct. 34. The errors in the measurement which arise due to unpredictable fluctuations in temperature and voltage supply are: (1) Least count errors (2) Random errors (3) Instrumental errors (4) person al errors Ans: (2) Fluctuations in temperature and voltage supply correspond to Random errors 35. An electric dipole is placed at an angle of 30  with an electric field of intensity 5 1 2 10 NC −  . It experiences a torque equal to 4 Nm. Calculate the magnitude of charge on the dipole, if the length is 2 cm. (1) 4 mC (2) 2mC (3) 8mC (4) 6mC Ans: (2) sin E    = 5 0 4 (2 10 ) sin30 p =  5 5 4 2 4 10 2 10 p C m −  = =   P Qd Q d  = → = 5 3 2 4 10 2 10 2 2 10 C mC − − −  = =  =  36. The x t − graph of a particle performing simple harmonic motion is shown in the figure. The acceleration of the particle at 2 t = s is: (1) 2 2 16 ms  − (2) 2 2 16 ms  − − (3) 2 2 8 ms  − (4) 2 2 8 ms  − − Ans: (2) T= 8 sec A=1m 𝜔 = 2 𝜋 𝑇 = 2 𝜋 8 = 𝜋 4 rad/s x = A sin 𝜔 t = 1.sin ( 𝜋 4 t ) = sin ( 𝜋 4 t ) V= 𝑑 𝑥 𝑑 𝑡 = 𝜋 4 cos ( 𝜋 4 t ) a = 𝑑 𝑉 𝑑 𝑡 = - 𝜋 2 16 sin ( 𝜋 4 t ) (t = 2 sec) a = - 𝜋 2 16 sin ( 𝜋 × 2 4 ) = - 𝜋 2 16 sin 𝜋 2 = - 𝜋 2 16 ×1 = - 𝜋 2 16 m/s - 2 SECTIO N - B 37. A horizontal bridge is built across a river. A student standing on the bridge throws a small ball vertically upwards with a velocity 4 ms - 1 . The ball strikes the water surface after 4s. The height of bridge above water surface is ( ) 2 10 Take g ms − = : (1) 64 m (2) 68 m (3) 56 m (4) 60 m Ans: (1) u= 4 m/s t= 4 sec g= - 10 m/s 2 s = ut + 1 2 a t 2 = + 4(4) + 1 2 ( - 10 ) (4 2 ) = - 16 + 80 = - 64 m 38. A wire carrying a current I along the positive x - axis has length L. It is kept in a magnetic field (2 3 4 ) B i j k T = + − . The magnitude of the magnetic force acting on the wire is: (1) 5 IL (2) 3 IL (3) 3 IL (4) 5 IL Ans: (1) 𝐿 ⃗ = 𝐿 𝑖 ̂ 𝐵 ⃗ = ( 2 𝑖 ̂ +3 𝑗 ̂ - 4 𝑘 ̂ ) T 𝐹 = I( 𝐿 ⃗ × 𝐵 ⃗ ) = 𝐼 [ 𝐿 𝑖 ̂ × ( 2 𝑖 ̂ +3 𝑗 ̂ - 4 𝑘 ̂ ) ] = 𝐼𝐿 ( 3 𝑘 ̂ + 4 𝑗 ̂ ) F = IL √ 3 2 + 4 2 = 5IL 39. The resistance of platinum wire at 0 ° C is 2  at 80 ° C. The temperature coefficient of resistance of the wire is: (1) 2 1 3 10 C  − −  (2) 1 1 3 10 C  − −  (3) 4 1 3 10 C  − −  (4) 3 1 3 10 C  − −  Ans: (1) R 0 = 2 𝛺 R 80 = 6.8 𝛺 R 80 = R 0 (1+ 𝛼𝑡 ) 6.8= 2( 1+ 𝛼 ×80) 6.8 = 2+ 160 𝛼 6.8 – 2 = 160 𝛼 4.8 = 160 𝛼 𝛼 = 48 1600 𝛼 = 3× 10 - 2 0 C - 1 40. The radius of inner most orbit of hydrogen atom is 11 5.3 10 −  m. What is the radius of third allowed orbit of hydrogen atom? (1) 0 1.59 A (2) 4.77 0 A (3) 0.53 0 A (4) 1.06 0 A Ans: (2) 𝑟 0 = 5.3 × 10 - 11 m 𝑟 𝑛 = n 2 𝑟 0 𝑟 3 = (3) 2 ×5.3 × 10 - 11 = 47.7 × 10 - 11 m = 4.77× 10 - 10 m = 4 77Å 41. 10 resistors, each of resistance R are connected in series to a battery of emf E and negligible internal resistance. Then those are connected in parallel to the same battery, the current is increased n times. The value of n is: (1) 1 (2) 1000 (3) 10 (4) 100 Ans: (4) R 1 = 10 R (in series) R 2 = 𝑅 10 ( In parallel) I 2 = nI 1 E= I 1 R 1 = I 2 R 2 I 1 × 10 R = n I 1 × 𝑅 10 n = 100 42. A satellite is orbiting just above the surface of the earth with period T. If d is the density of the earth with period T. If d is the density of the earth and G is the universal constant of gravitation, the quantity 3 Gd  represents: (1) T 3 (2) T (3) T (4) T 2 Ans: (4) M = 4 3 𝜋 R 3 d T = 2 𝜋 √ 𝑅 3 𝐺𝑀 = 2 𝜋 √ 𝑅 3 𝐺 × 4 3 𝜋 𝑅 3 𝑑 T = √ 3 𝜋 𝐺𝑑 3 𝜋 𝐺𝑑 = T 2 43. Two thin lenses are of same focal lengths (f), but one is convex and the other one is concave. When they are place in contact with each other, the equivalent focal length of the combination will be: (1) / 2 f (2) Infinite (3) Zero (4) / 4 f Ans: (2) f 1 = f f 2 = - f 1 𝐹 = 1 f 1 + 1 f 2 1 𝐹 = 1 𝑓 - 1 𝑓 1 𝐹 = 0 F = ∞ 44. For the following logic circuit, the truth table is : (1) 0 0 1 0 1 0 1 0 1 1 1 0 A B Y (2) 0 0 0 0 1 0 1 0 0 1 1 1 A B Y (3) 0 0 1 0 1 1 1 0 1 1 1 0 A B Y (4) 0 0 0 0 1 1 1 0 1 1 1 1 A B Y Ans: (4) A B A B A B A B 0 0 1 1 0 1 0 1 1 1 0 0 1 0 1 0 1 0 0 0 0 1 1 1 45. A bullet from gun is fired on a rectangular wooden block with velocity u. When bullet travels 24 cm through the block along its length horizontally, velocity of bullet becomes 3 u . Then it further penetrates into the block in the same direction before coming to rest exactly at the other end of the block. The total length of the block is: (1) 28 cm (2) 30 cm (3) 27 cm (4) 24 cm Ans: (3) Case 1: A to B 2 2 2 v u as − = 2 2 2 24 9 u u a − =   2 8 48 9 u a − = 2 54 u a = − Case 2 : B to C 2 2 2 v u ax − = 2 2 0 2 3 54 u u x   − − =       2 2 2 9 54 u u x =   x = 3cm Total length = 24 + 3 = 27cm 46. In the figure shown here, what is the equivalent focal length of the combination of lenses (Assume that all layers are thin)? (1) - 100 cm (2) - 50 cm (3) 40 cm (4) - 40 cm Ans: (1) By Len’s make r’s formula 1 2 1 1 1 ( 1) f R R    = − −     1 1 1 1 (1.6 1) 20 f   = − −      1 1 6 1 3 10 20 100 cm f − − =  = ......... (1) 2 1 1 1 (1.5 1) 20 20 f   = − +     2 1 1 2 1 2 20 20 cm f =  = .......... (2) 3 1 1 1 (1.6 1) 20 f −   = − −      3 1 6 1 3 10 20 100 cm f − − =  = ........ (3) 1 2 3 1 1 1 1 f f f f = + + 1 3 1 3 100 20 100 f − = + − 1 3 5 3 1 100 100 f − + − − = = f = - 100 cm 47. An electric dipole is placed as shown in the figure. The electric potential (in 10 2 V) at point P due to the dipole is 0 (  = permittivity of free space and = 0 1 4 K  =  ): (1) 8 5 qK       (2) 8 3 qK       (3) 3 8 qK       (4) 5 8 qK       Ans: (3) P q q V V V + − = + 2 2 ( ) (2 10 ) (8 10 ) kq k q − − − = +   2 3 3 10 8 8 kq kq =  = 48. A very long conducting wire is bent in a semi - circular shape form A to B as shown in figure. The magnetic field at point P for steady current configuration is given by: