Volume 25 Managing Editor Mahabir Singh Editor Anil Ahlawat
No. 4
April 2017
Corporate Offi ce: Plot 99, Sector 44 Institutional area, Gurgaon -122 003 (HR). Tel : 0124-6601200 e-mail :
[email protected] website : www.mtg.in Regd. Offi ce: 406, Taj Apartment, Near Safdarjung Hospital, New Delhi - 110029.
CONTENTS
Competition Edge Physics Musing Problem Set 45
8
NEET Practice Paper
12
JEE Advanced Practice Paper
21
BITSAT Practice Paper
31
AIIMS Practice Paper
40
AMU Practice Paper
54
Physics Musing Solution Set 44
63
Crossword
89
Class 11 Brain Map
52
MPP
65
Class 12 Brain Map
53
CBSE Board - Solved Paper 2017
69
Key Concept
77
MPP
82
Subscribe online at www.mtg.in Individual Subscription Rates
Combined Subscription Rates
1 yr.
2 yrs.
3 yrs.
1 yr.
2 yrs.
3 yrs.
Mathematics Today
330
600
775
PCM
900
1500
1900
Chemistry Today
330
600
775
PCB
900
1500
1900
Physics For You
330
600
775
PCMB
1000
1800
2300
Biology Today
330
600
775
Send D.D/M.O in favour of MTG Learning Media (P) Ltd. Payments should be made directly to : MTG Learning Media (P) Ltd, Plot No. 99, Sector 44, Gurgaon - 122003 (Haryana) We have not appointed any subscription agent. Owned, Printed and Published by MTG Learning Media Pvt. Ltd. 406, Taj Apartment, New Delhi - 29 and printed by HT Media Ltd., B-2, Sector-63, Noida, UP-201307. Readers are adviced to make appropriate thorough enquiries before acting upon any advertisements published in this magazine. Focus/Infocus features are marketing incentives. MTG does not vouch or subscribe to the claims and representations made by advertisers. All disputes are subject to Delhi jurisdiction only. Editor : Anil Ahlawat Copyright© MTG Learning Media (P) Ltd. All rights reserved. Reproduction in any form is prohibited.
PHYSICS FOR YOU | APRIL ‘17
7
P
PHYSICS
MUSING
hysics Musing was started in August 2013 issue of Physics For You. The aim of Physics Musing is to augment the chances of bright students preparing for JEE (Main and Advanced) / AIIMS / NEET / Other PMTs with additional study material. In every issue of Physics For You, 10 challenging problems are proposed in various topics of JEE (Main and Advanced) / various PMTs. The detailed solutions of these problems will be published in next issue of Physics For You. The readers who have solved five or more problems may send their detailed solutions with their names and complete address. The names of those who send atleast five correct solutions will be published in the next issue. We hope that our readers will enrich their problem solving skills through “Physics Musing” and stand in better stead while facing the competitive exams.
SINGLE OPTION CORRECT TYPE
1. PQ is an infinite current carrying conductor. AB and CD are smooth conducting rods on which a conductor EF moves with constant velocity v as shown in figure. The force needed to maintain constant speed of EF is (No gravity is there.) 2
1 µ0 Iv b ln a vR 2π 2 1 µ0 Iv b ln (b) a R 2π 2 v µ0 Iv b ln (c) a R 2π 2 v µ Iv b (d) 0 ln a R 2 π (a)
2mgAε0
(b)
4mgAε0 k
mgAε0
2mgAε0 k
(d)
3. A ring carries uniform charge Q and has radius R. The electric field at a small distance x from the centre along the radius will be Qx Qx (a) (b) 3 8πε0R3 4πε0R (c)
2. The plates S and T of an uncharged parallel plate capacitor are connected across a battery. The battery is then disconnected and the charged plates are now connected in a system as shown in the figure. The system shown is in equilibrium. All the strings are insulating and massless and spring constant of massless spring is k. The magnitude of charge on one of the capacitor plates is (Area of each plate is A.)
(a)
(c)
Qx 12πε0R
3
(d)
16πε0R3
4. Consider a cubical container containing only photons. Assuming that photons behave like ideal gas then the relation between the pressure (P) and average energy density (u) will be (Assuming only photons exists.) u 2u (a) P = (b) P = 3 3 3 (c) P = u (d) P = u 2 5. Two ends of an inductor of inductance L are connected to two parallel conducting wires. A rod of length l and mass m is given velocity v0 as shown in figure. The whole system is placed in perpendicular magnetic field B. Find the maximum current in the inductor. (Neglect gravity and friction) mv0 m (a) (b) v L L 0 (c)
mv02 L
(d) None of these
By Akhil Tewari, author Foundation of physics for JEE Main & advanced, professor, iiTians paCE, Mumbai.
8
physics for you | april ‘17
Qx
MULTIPLE OPTION CORRECT TYPE
6. The diagram shows the P-V diagram of a cyclic process ABCA.
3V 5 12V (c) Volume of compartment II is 5 (b) Volume of compartment I is
(d) Final pressure in compartment I is
5P 3
INTEGER TYPE
(a) (b) (c) (d)
Work done in process A → B is 0.036 J. Work done in process B → C is –0.024 J. Work done in process C → A is zero. Work done in cycle ABCA is 0.06 J.
7. An ideal gas undergoes an expansion from a state with temperature T1 and volume V1 to V2 through three different polytropic processes A, B and C as shown in the P-V diagram. If |DUA|, |DUB| and |DUC| be the magnitude of changes in internal energy along the three paths respectively, then (a) |DUA| < |DUB| < |DUC| if temperature in every process decreases (b) |DUA| > |DUB| > |DUC| if temperature in every process decreases (c) |DUA| > |DUB| > |DUC| if temperature in every process increases (d) |DUA| < |DUB| < |DUC| if temperature in every process increases 8. A partition divides a container having insulated walls into two compartments I and II. The same gas fills the two compartments whose initial parameters are given. The partition is a conducting wall which can move freely without friction. Which of the following statements is/are correct, with reference to the final equilibrium position? P, V, T I
2P, 2V, T II
(a) The pressure in the two compartments are equal. 10
physics for you | april ‘17
9. A conducting tube is passing through a bath. A liquid at temperature 90°C and specific heat s is entering at one end of tube. Rate of flow of liquid is 1 kg s–1 and exit temperature is 50° C. In bath another liquid having specific heat 2s and inlet temperature 20° C is entering at a rate of 2 kg s–1. If the exit temperature of liquid coming out of the bath is 10x. Find x (in °C). (Assume steady state condition) 10. Heat is generated uniformly per unit volume inside a spherical volume of radius 1 m at rate of 20 W m–3. The thermal conductivity of the spherical volume 1 is W m −1 °C −1 and the temperature of outer 6 surface of the sphere is 20°C. If the temperature of the centre of the sphere is x × 10 (in °C), find x.
1. The measurement of two quantities is given as A = 1.0 m ± 0.2 m and B = 2.0 m ± 0.2 m. The correct value for AB is (a) 1.4 m ± 0.4 m (b) 1.41 m ± 0.15 m (c) 1.4 m ± 0.3 m (d) 1.4 m ± 0.2 m 2. A particle moving along x-axis has acceleration a, t at time t, given by a = a0 1 − , where a0 and T T are constants. The particle has zero velocity at t = 0. In the time interval between t = 0 and the instant when a = 0, the particle’s velocity vx is 1 (a) a0T 2 (b) a0T2 2 1 (c) a0T (d) a0T 2 3. One coolie takes 1 minute to raise a suitcase through a height of 2 m but the second coolie takes 30 s to raise the same suitcase to the same height. The powers of two coolies are in the ratio (a) 1 : 3 (b) 2 : 1 (c) 3 : 1 (d) 1 : 2 4. A body of mass m is resting on a wedge of angle q as shown in figure. The wedge is given an acceleration a. The value of a so that the mass m just falls freely is A
m
a
(a) g
B
C
(b) g sinq (c) g tanq (d) g cotq
5. When the angle of projection is 75°, a ball falls 10 m short of the target. When the angle of projection is 45°, it falls 10 m ahead of the target. Both are projected from the same point with the 12
Physics For you | April ‘17
same speed in the same direction. The distance of the target from the point of projection is (a) 15 m (b) 30 m (c) 45 m (d) 10 m 6. A ball of mass m is thrown upwards with a velocity v. If air exerts an average resisting force F, the velocity with which the ball returns to the thrower is (a) v
mg mg + F
(b) v
m mg + F
(c) v
mg − F mg + F
(d) v
mg + F mg
7. A 2 kg block is connected with two springs of force constants k1 = 100 N m–1 and k2 = 300 N m–1 as shown in figure. The block is released from rest with the springs unstretched. The acceleration of the block in its k1 lowest position is (g = 10 m s–2) (a) zero 2 kg (b) 10 m s–2 upwards k2 (c) 10 m s–2 downwards (d) 5 m s–2 upwards 8. A ball is released from the top of an inclined plane of inclination q. It reaches the bottom with velocity v. If the angle of inclination is doubled, then the velocity of ball on reaching the ground is (a) v (b) 2v (c) v 2 cos θ (d) v 2sin θ 9. If a sphere is rolling, the ratio of the translational energy to total kinetic energy is given by (a) 7 : 10 (b) 2 : 5 (c) 10 : 7 (d) 5 : 7 10. The radii of circular orbits of two satellites A and B of the earth are 4R and R respectively. If the speed of satellite A is 3v, then the speed of satellite B will be
3v 3v (b) 6v (c) 12v (d) 4 2 11. A body of mass m is taken from the earth’s surface to the height equal to twice the radius R of the earth. The change in potential energy of body will be 1 2 (a) 3mgR (b) mgR (c) 2mgR (d) mgR 3 3 12. The density of water at 20 °C is 998 kg m–3 and at 40 °C, 992 kg m–3. The coefficient of volume expansion of water is (a) 3 × 10–4 °C–1 (b) 2 × 10–4 °C–1 –4 –1 (c) 6 × 10 °C (d) 10–4 °C–1 (a)
13. A mild steel wire of length 2L 2L and cross-sectional area A is x stretched, well within elastic m limit, horizontally between two pillars as shown in the figure. A mass m is suspended from the mid-point of the wire. Strain in the wire is x2
x x2 x2 (a) (b) (c) (d) L 2L L 2L2 14. Two mercury droplets of radii 0.1 cm and 0.2 cm collapse into one single drop. The amount of energy released is (The surface tension of mercury T = 435.5 × 10–3 N m–1 ) (a) 32 × 10–7 J (b) 30 × 10–7 J –7 (c) 28 × 10 J (d) 31 × 10–7 J 15. A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500 °C and then placed on a large ice block. The maximum amount of ice that can melt is (Specific heat of copper = 0.39 J g–1 °C–1, heat of fusion of water = 335 J g–1). (a) 1.4 kg (b) 1.5 kg (c) 1.1 kg (d) 1.3 kg 16. The coefficient of volume expansion of glycerine is 49 × 10–5 °C–1. The fractional change in its density for a 30 °C rise in temperature is (a) 0.0145 (b) 0.0156 (c) 0.0167 (d) 0.0247 17. The efficiency of Carnot engine is 50% and temperature of sink is 500 K. If temperature of source is kept constant and its efficiency raised to 60%, then the required temperature of sink will be (a) 100 K (b) 600 K (c) 400 K (d) 500 K 18. A monatomic gas at pressure P1 and volume V1 is 1 th compressed adiabatically to of its original 8 volume. The final pressure of the gas is 14
Physics For you | April ‘17
(a) 64P1
(b) P1
(c) 16P1
(d) 32P1
19. The driver of a car approaching a vertical wall notices that the frequency of the horn of his car changes from 400 Hz to 450 Hz after being reflected from the wall. Assuming speed of sound to be 340 m s–1, the speed of approach of car towards the wall is (a) 10 m s–1 (b) 20 m s–1 –1 (c) 30 m s (d) 40 m s–1 20. A siren placed at a railway platform is emitting sound of frequency 5 kHz. A passenger sitting in a moving train A records a frequency of 5.5 kHz while the train approaches the siren. During his return journey in a different train B he records a frequency of 6.0 kHz while approaching the same siren. The ratio of velocity of train B to that of train A is (a) 242/252 (b) 2 (c) 5/6 (d) 11/6 21. The potential energy of a simple harmonic oscillator of mass 2 kg in its mean position is 5 J. If its total energy is 9 J and its amplitude is 0.01 m, its time period would be (a) p/10 s (b) p/20 s (c) p/50 s (d) p/100 s 22. If potential (in volts) in a region is expressed as V(x, y, z) = 6xy – y + 2yz, the electric field (in N C–1) at point (1, 1, 0) is
(a) −(2i + 3j + k ) (c) −(3i + 5 j + 3k )
(b) −(6i + 9 j + k ) (d) −(6i + 5 j + 2k )
23. Capacity of an isolated sphere is increased by n times when it is enclosed by an earthed concentric sphere. The ratio of their radii is
n n2 2n 2n + 1 (b) (c) (d) n −1 n −1 n +1 n +1 24. Two point charges + 3 mC and + 8 mC repel each other with a force of 40 N. If a charge of – 5 mC is added to each of them, then the force between them will become (a) – 10 N (b) + 10 N (c) + 20 N (d) – 20 N (a)
25. The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of 10 W is (a) 0.8 W (b) 1.0 W (c) 0.2 W (d) 0.5 W 26. A wire of resistance 4 W is stretched to twice of its original length. The resistance of stretched wire would be (a) 8 W (b) 16 W (c) 2 W (d) 4 W
27. The intensity of magnetic field at a point X on the axis of a small magnet is equal to the field intensity at another point Y on its equatorial axis. The ratio of distance of X and Y from the centre of the magnet will be (a) 2–3 (b) 2–1/3 (c) 23 (d) 21/3 28. A proton, a deuteron and an a-particle enter a region of uniform magnetic field B perpendicularly, with the same kinetic energy. The ratio of the radius of their circular paths is (a) 1 : 2 : 1 (b) 1 : 2 : 2 (c)
2 :1:1
(d)
2 : 2 :1
29. In an inductor of self inductance L = 2 mH, current changes with time according to the relation I = t2 e–t. The time at which the emf becomes zero is (a) 4 s (b) 2 s (c) 1 s (d) 0.5 s 30. A coil is placed in a transverse magnetic field of 0.02 T. If this coil starts shrinking at a rate of 1 mm s–1, while its radius remains 4 cm, then the value of induced emf is (a) 2 mV (b) 5 mV (c) 8 mV (d) 50 mV 31. A resistor of resistance 30 W, inductor of reactance 10 W and capacitor of reactance 10 W are connected in series to an ac voltage source e = 300 2 sin wt. The current in the circuit is (a) 10 2 A (b) 10 A (c) 30 11 A
(d) 30/ 11 A
32. A current of 2 A is increasing at a rate of 4 A s–1 through a coil of inductance 2 H. The energy stored in the inductor per unit time is (a) 16 J s–1 (b) 4 J s–1 (c) 2 J s–1 (d) 1 J s–1 33. The magnetic field in a plane electromagnetic wave is given by, B = (200 mT) sin [(4.0 × 1015 s–1)(t – x/c)]. The average energy density corresponding to the electric field is (a) 1.6 × 10–2 J m–3 (b) 8 × 10–3 J m–3 (c) 3.2 × 10–2 J m–3 (d) 8 × 10–4 J m–3 34. Light from a point source in air falls on a convex spherical glass surface of refractive index m = 1.5 and radius of curvature R = 20 cm. The distance of light source from the glass surface is 100 cm. The distance of the image formed is (a) 100 cm (b) 200 cm (c) 300 cm (d) 120 cm
35. The radius of curvature of each surface of a convex lens of refractive index 1.5 is 40 cm. The power of the lens is (a) 1.5 D (b) 2.0 D (c) 2.5 D (d) 2.1 D 36. A ray of light passes through an equilateral glass prism, such that the angle of incidence is equal to the angle of emergence. If the angle of emergence is 3/4 times the angle of the prism. The refractive index of the glass prism is (a) 1.414 (b) 1.214 (c) 1.523 (d) 1.423 37. A ray of light incident on an equilateral glass prism (mglass = 3 ) moves parallel to the base of the prism inside it. The angle of incidence for this ray is (a) 60° (b) 90° (c) 30° (d) 45° 38. For photoelectric emission from certain metal the cut off frequency is u. If radiation of frequency 2u impinges on the metal plate, the maximum possible velocity of the emitted electron will be (m is the mass of electron) hυ 2hυ hυ hυ (a) (b) 2 (c) (d) 2m m m m 39. When the energy of the incident radiation is increased by 20%, the kinetic energy of the photoelectrons emitted from a metal surface increased from 0.5 eV to 0.8 eV. The work function of the metal is (a) 0.65 eV (b) 1.0 eV (c) 1.3 eV (d) 1.5 eV 40. Two radiations of photon energies 1 eV and 2.5 eV, successively illuminate a photosensitive metallic surface of work function 0.5 eV. The ratio of the maximum speeds of the emitted electrons is (a) 1 : 4 (b) 1 : 2 (c) 1 : 1 (d) 1 : 5 41. Ratio of longest wavelengths corresponding to Lyman and Balmer series in hydrogen spectrum is 9 5 3 7 (a) (b) (c) (d) 31 27 23 29 42. A radioactive material decays by simultaneous emission of two particles with respective half lives 1620 and 810 years. The time in years, after which one fourth of the material remains is (a) 540 years (b) 1080 years (c) 118 years (d) 24 years 43. The capacitance of a spherical conductor of radius 1 m is (a) 9 × 10–9 F (b) 10 mF (c) 1.1 × 10–10 F (d) 1 mF Physics For you | April ‘17
15
44. According to Bohr model of hydrogen atom, only those orbits are permissible which satisfy the condition mv 2 h (a) mv = nh (b) = n 2π r h (c) mvr = n 2π
h (d) mvr = n 2π 2
45. When a coil carrying a steady current is short circuit, the current in it decreases h times in time t0. The time constant of the circuit is t t (a) 0 (b) 0 η −1 ln η t (d) 0 η
(c) t0 ln h
solutions
1. (d) 2. (c) : Given, At time t = 0, velocity, v = 0 t Acceleration, a = a0 1 − T When a = 0,
t 0 = a0 1 − T
(a0 = constant)
t = 0 or t = T . T dv Also, acceleration a = dt \
\
1−
vx
∫ dv = 0
t =T
T
t ∫ adt = ∫ a0 1 − T dt t =0 0 T
\
a T2 1 a t2 v x = a0t − 0 = a0T − 0 = a0T 2T 2T 2 0
3. (d) : Power, P =
Work done
Time taken Here, work done (= mgh) is same in both cases. \
30 s 30 s 1 P1 t2 = = = = P2 t1 1 minute 60 s 2
4. (d) : The horizontal acceleration a of the wedge should be such that in time the wedge moves the horizontal distance BC, the body must fall through a vertical distance AB under gravity. Hence, 1 1 BC = at2 and AB = gt2 2 2 16
Physics For you | April ‘17
tanq =
g AB g or a = = g cotq = tanθ BC a
5. (b) : Since, range R = \ and
u2 sin 2 × 75° g
u2 sin 2θ g
= R – 10
...(i)
u2 sin 2 × 45° u2 = R + 10 or = R + 10 g g
(R + 10) sin 150° = R – 10 or (R + 10)1/2 = R – 10 or R = 30 m
(from eqn. (i))
6. (c) : For upward motion, retarding force = mg + F mg + F \ Retardation, a = m v2 v 2m = Distance, s = 2a 2(mg + F ) For downward motion, net force = mg – F mg − F \ Acceleration, a′ = m 2 2 v′ v′ m = Distance, s′ = 2a′ 2(mg − F ) As s = s′ ∴ \
v′ = v
v 2m v′2 m = 2(mg + F ) 2(mg − F )
mg − F mg + F
7. (b) : Here, m = 2 kg, k1 = 100 N m–1, k2 = 300 N m–1 When block is released from rest, let its maximum downward displacement be x. \ Decrease in potential energy of the block = Increase in potential energy of two springs 1 mgx = (k1 + k2)x2 2 2mg 2 × 2 × 10 = x= = 0.1 m k1 + k2 100 + 300 Acceleration of block in this position is (k + k )x − mg a= 1 2 m [(100 + 300)0.1] − [2 × 10] = 10 m s–2, upwards a= 2 8. (c) : When inclination of plane is q, the downward acceleration of body along the plane is a = g sinq
Using the relation, v2 = u2 + 2as, we have v2 = 0 + 2 × g sin q × s
or
v=
...(i) 2g ssin θ When inclination of plane is 2q, the downward acceleration of body along the plane is a1 = g sin2q Then v12 = 0 + 2 g sin2q × s or
v1 =
\
v1/v =
12. (a) : Here, T1 = 20 °C, T2 = 40 °C, r20 = 998 kg m–3, r40 = 992 kg m–3 ρT ρT 1 1 = As ρT2 = (1 + γ∆T ) 1 + γ(T2 − T1 ) \ or
2 g s × 2 sin θ cos θ
or
2 cosθ or v1 = v 2 cos θ
9. (d) 10. (b) : Orbital speed of the satellite around the earth GM r For satellite A, rA = 4R, vA = 3v
998 998 or 992 = 1 + γ(40 − 20) 1 + 20 γ 998 998 6 or 20 γ = 1 + 20 γ = −1 = 992 992 992 6 1 –4 –1 = 3 × 10 °C γ= × 992 20 992 =
13. (a) :
is v =
m
GM rA For satellite B, rB = R, vB = ?
...(i)
GM rB Dividing eqn. (ii) by eqn. (i), we get r v r ∴ B = A or vB = vA A rB vA rB
...(ii)
∴ vA =
∴ vB =
4R or vB = 6v R
11. (d) : Gravitational potential energy at any point at a distance r from the centre of the earth is GMm U =− r At the surface of the earth, r = R GMm \ Ui = − R At a height h from the surface, r = R + h = R + 2R = 3R (h = 2R (Given)) GMm \ Uf =− 3R Change in potential energy, GMm GMm −− DU = Uf – Ui = − 3R R GM GMm 1 2 GMm 2 = mgR Q g = 2 = 1− = 3 R 3 3 R R
1/2
x2 ∆L = 2 (L + x ) − 2 L = 2 L 1 + − 2L L2 1 x2 = 2 L 1 + − 2 L (By binomial theorem) 2 L2 ∆L x2 x2 = \ Strain = = L 2 L 2 L2 14. (a) : Let V1 and V2 be the volumes of two mercury droplets of radii r1 and r2 respectively. r1 = 0.1 cm = 10–3 m, r2 = 0.2 cm = 2 × 10–3 m Surface tension, T = 435.5 × 10–3 N m–1. V = Volume of the resulting drop of radius r V = V1 + V2 4 3 4 3 4 3 or πr = πr1 + πr2 3 3 3 2
Substituting the given values, we get v B = 3v
From the figure, Increase in length, DL = (PR + RQ) – PQ = 2PR – PQ
or or
2 1/2
r3 = r13 + r23 = (10–3)3 + (2 × 10–3)3 = 9 × 10–9 r = 2.08 × 10–3 m ≈ 2.1 × 10–3 m
Decrease in surface area, DA = 4p[(r12 + r22) – r2]
or
DA = 4 × 3.14 [10–6 + 4 × 10–6 – 4.41 × 10–6]
or
DA = 7.4 × 10–6 m2
Energy released, E = T × DA = 435.5 × 10–3 × 7.4 × 10–6 = 32 × 10–7 J 15. (b) : Here, mass of copper block, m1 = 2.5 kg Specific heat of copper, s = 0.39 J g–1 °C–1 = 0.39 × 103 J kg–1 °C–1 Physics For you | April ‘17
17
Temperature of furnace, DT = 500 °C Latent heat of fusion, L = 335 J g–1 = 335 × 103 J kg–1 If Q be the heat absorbed by the copper block, then Q = m1sDT ...(i) Let m2 (kg) be the mass of ice melted when copper block is placed on it, then Q = m2L ...(ii) From eqns. (i) and (ii), we get m1sDT = m2L or or
m s ∆T m2 = 1 L 2.5 × 0.39 × 103 × 500 m2 = = 1.455 kg = 1.5 kg 335 × 103
16. (a) : Here, g = 49 × 10–5 °C–1, DT = 30 °C Let there be m grams of glycerine and its initial volume be V. Suppose that the volume of the glycerine becomes V′ after a rise of temperature of 30 °C then, V′ = V(1 + gDT) = V(1 + 49 × 10–5 × 30) or V′ = 1.0147 V m Initial density of the glycerine, ρ = V Final density of the glycerine, m m ρ ρ′ = = = = 0.9855 ρ V ′ 1.0147V 1.0147 Therefore, fractional change in the value of density of glycerine, ρ − ρ′ ρ − 0.9855 ρ = = 0.0145 ρ ρ 17. (c) : Efficiency h of a Carnot engine is given by T η = 1 − 2 , where T 1 is the temperature of the T1 source and T2 is the temperature of the sink. Here, T2 = 500 K 500 \ 0. 5 = 1 − or T1 = 1000 K T1 T′ Now, 0.6 = 1 − 2 1000 or 18
T2′ = 400 K
(T2′ is the new sink temperature)
Physics For you | April ‘17
18. (d) : Ideal gas equation, for an adiabatic process is g g g PV = constant or P1V1 = P2V2 For monatomic gas γ = \
P1V15/3
V = P2 1 8
5 3
5/3
⇒ P2 = P1 × (2)5 = 32P1 340 + v s 19. (b) : Here, 450 = 400 340 − v s 340 + v s 9 or = 340 − v s 8 or 9(340) – 9vs = 8(340) + 8vs or 17vs = 340 or vs = 20 m s–1 v + vA 5 20. (b) : (uapproach)A = 5.5 = v v + vB (uapproach)B = 6 = 5 v Where v is the velocity of sound. vA v Now, 5.5 = 1 + 5 ⇒ A = 0.1 v v vB v 1 Similarly, 6 = 1 + 5 ⇒ B = v v 5 v ∴ B =2 vA 21. (d) : or Q
\ or or
1 2 mv max = 9 J – 5 J = 4 J 2 mv2max = 8 or 2v2max = 8 ⇒ vmax = 2 m s–1 vmax = 2 m s–1 = Aw v 2 w = max = = 200 rad s–1 A 0.01 2π = 200 T 2π π T= = s 200 100
22. (d) 23. (a) : From the figure C1 = 4pe0R1 RR C2 = 4pe0 1 2 R2 − R1 As C2 = nC1
R1 C1
R1 R2 C2
\ or \
24. (a) : Here, F = 40 = F′ =
28. (a) : In a magnetic field,
4 πε0 R1R2 = n4pe0R1 R2 − R1 R2 = n or R2(n – 1) = R1n R2 − R1 R2 n = R1 n − 1
k(3 − 5)(8 − 5) r2
=
mv mv 2 1 = Bqv or r = and K = mv2 Bq r 2 mv =
r2 −6k
rp : rd : ra =
R
27. (d) : If d1 is distance of point X on axial line and d2 is distance of point Y on equatorial line µ 2M µ M then B1 = 0 , B2 = 0 3 4 π d1 4 π d23
or
M
d23
Bq
⇒ r∝
m q
mp qp
:
md
qd
:
mα qα
m 2m 4m : : = 1: 2 :1 e e 2e
=
I I or IR + Ir = e Here, R = 10 W, r = ?, r e = 2.1 V, I = 0.2 A \ 0.2 × 10 + 0.2 × r = 2.1 2 + 0.2r = 2.1 1 0.2r = 0.1 or r = = 0.5 Ω 2 26. (b) : Resistance of a wire, l ...(i) R=ρ =4 Ω A When wire is stretched twice, its new length will be l ′. Then l ′ = 2l \ lA = l′A′ where A′ is the new cross sectional area or A′ = l A = l A = A l′ 2l 2 \ Resistance of the stretched wire is l′ 2l l R′ = ρ = ρ = 4ρ A′ ( A / 2) A = 4(4 W) = 16 W (Using eqn. (i))
µ0 2M µ0 = 4 π d13 4π d d13 = 2d23 or 1 = 21/ 3 d2
2mK
For same values of K and B
k(3)(8)
r2 F 40 F′ 1 = – 10 N = − ; F′ = − = − 4 4 F 4 ε 25. (d) : I = R+r
As B1 = B2 ∴
2mK so r =
29. (b) dr = 1 mm s–1 = 10–3 m s–1 dt r = 4 cm = 4 × 10–2 m, e = ? Q f = BA = B(pr2) dφ d dr \ e = = (Bpr2) = Bp2r dt dt dt 22 = 0.02 × × 2 × 4 × 10–2 × 10–3 7
30. (b) : Here, B = 0.02 T,
= 5 × 10–6 V = 5 mV
31. (b) : Here, R = 30 W, XL = 10 W, XC = 10 W e = 300 2 sin wt, e0 = 300 2 V 300 ε ε0 / 2 I = rms = = 2 2 2 Z 30 + (10 − 10)2 R + ( X L − XC ) I = 10 A 32. (a) 33. (b) : Here, B0 = 200 mT = 200 × 10–6 T = 2 × 10–4 T In electromagnetic wave, average energy density corresponding to electric field is equal to average energy density corresponding to magnetic field, i.e., uav = =
2 1 1 B0 ε0 E02 = 4 4 µ0
(2 × 10 −4 )2
−7
4 × (4 π × 10 )
= 8 × 10–3 J m–3
34. (a) : Here, m1 = 1, m2 = 1.5, u = – 100 cm R = + 20 cm, [R is +ve for a convex refracting surface] Physics For you | April ‘17
19
As
Dividing eqn. (i) by eqn. (ii), we get
µ2 µ1 µ2 − µ1 − = v u R
2 vmax 2 vmax
\
1. 5 1 1. 5 − 1 1 + = = v 100 20 40
or
3 1 1 5−2 3 = − = = 2v 40 100 200 200
\
v = + 100 cm
1 1 1 = (m – 1) − f R1 R2
37. (a) : Here A = r + r′ = r + r = 2r A 60° \ r= = = 30° 2 2 (Here r = r′ Q Light moves parallel to the base of prism.) mglass =
sin i sin r
sin i =
1 3 or ∠i = 60° 3× = 2 2
sin i sin 30°
38. (a) 39. (b) : The kinetic energy of emitted photoelectrons, K = hu – f0 As per question, 0.5 eV = hu – f0 ...(i) 0.8 eV = 1.2hu – f0 ...(ii) On solving eqns. (i) and (ii), we get, f0 = 1.0 eV 40. (b) : According to Einstein’s photoelectric equation
1 2 mv = 1 eV − 0.5 eV = 0.5 eV 2 max1 1 2 and mvmax = 2.5 eV − 0.5 eV = 2 eV 2 2 20
Physics For you | April ‘17
...(i)
1 1 1 5R ...(ii) = R − = 22 32 36 λ Balmer λ Dividing eqn. (ii) by eqn. (i), Lyman = 5R × 4 = 5 λ Balmer 36 3R 27
n
As
N 1 1 = = N0 2 4
or
t = 2T =
43. (c)
...(i) ...(ii)
⇒ n = 2 or
t = 2 T
2 × 1620 3240 = 1080 years = 3 3 44. (c)
45. (a) : The decay of current in a coil is given by I = I0e–t/t
I In time t0, I = 0 η I0 \ = I0e–t0/t or e–t0/t = h–1 η Taking log of both sides, t − 0 loge e = – loge h τ t ⇒ 0 = logeh τ t = t0/loge h = t0/ln h
MPP CLASS XII
1 2 mvmax = hυ − φ0 2 ∴
1 1 3R = R − = 12 22 4 λ Lyman
T ×T 1620 × 810 1620 T= 1 2 = = years T1 + T2 1620 + 810 3
36. (a)
or
1 1 1 = R − 2 2 λ n1 n2
42. (b) : Here, T1 = 1620 years, T2 = 810 years
1 1 + = (1.5 – 1) = 2.5 D 0.40 0.40
3=
vmax 0. 5 0. 5 1 1 ⇒ = = 2 vmax 2 2 2
1
35. (c) : Here, m = 1.5, R1 = + 40 cm = 0.40 m, R2 = – 40 cm = – 0.40 m
or
=
2
41. (b) : From
Thus, the image is formed at a distance of 100 cm from the glass surface, in the direction of incident light.
P=
1
1. (b) 6. (a) 11. (a) 16. (c) 21. (b, c) 26. (9)
2. 7. 12. 17. 22. 27.
(d) (b) (c) (b) (a, c, d) (b)
ANSWER KEY
3. (a) 8. (b) 13. (a) 18. (c) 23. (a,c) 28. (c)
4. (d) 9 . (a) 14. (c) 19. (c) 24. (5) 29 . (a)
5. 10. 15. 20. 25. 30.
(d) (c) (a) (a, b, c) (7) (c)
on Exam st ay 21 M
PAPER-I
It would fall onto the Sun in a time
SECTION 1 (Maximum Marks : 15) • • • •
This section contains FIVE questions. Each question has FOUR options (a), (b), (c) and (d). ONLY ONE of these four options is correct. For each question, darken the bubble corresponding to the correct option in the ORS. For each question, marks will be awarded in one of the following categories :
Full Marks : +3 If only the bubble corresponding to the correct option is darkened. Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : –1 In all other cases.
1. A parallel plate air capacitor is connected to a battery. The quantities charge, voltage, electric field and energy associated with this capacitor are given by Q0, V0, E0 and U0 respectively. A dielectric slab is now introduced to fill the space between the plates with battery still in connection. The corresponding quantities now given by Q, V, E and U are related to the previous one as (a) Q > Q0 (b) V > V0 (c) E > E0 (d) U = U0 M 2. A heavy container containing an ideal gas is kept on a horizontal surface. A L smooth piston of mass M is at rest as shown in the figure. The natural length of the spring is L. Now the piston is given a small downward push. Assuming the temperature of the gas to be constant, and there is vacuum over the piston, the time period for small oscillations is
(a) 2π
M k
(b) 2π
L g
(c) 2π
ML Mg + kL
(d) None of these
3. If a planet revolving around the Sun with time period T, is suddenly stopped in its orbit supposed to be circular.
2 (a) t = T 7
2 (b) t = T 8
2π (c) t = T 7
π (d) t = T 2
4. A metallic ring of radius r with a uniform metallic spoke of negligible mass and length r is rotated about its axis with angular velocity w in a perpendicular uniform magnetic field B as shown in figure. The central end of the spoke is connected to the rim of the wheel through a resistor R as shown. The resistor does not rotate, its one end is always at the center of the ring and the other end is always in contact with the ring. A force F as shown is needed to maintain constant angular velocity of the wheel. F is equal to (the ring and the spoke has zero resistance) (a)
B 2 ωr 2 8R
(b)
B 2 ωr 3 (c) 2R
B 2 ωr 2 2R
ω
Fb
R
B 2 ωr 3 (d) 4R
F
5. Find the pressure at which temperature attains its maximum value if the relation between pressure and volume for an ideal gas is P = P0 + (1 – a)V2; x > 1 (a)
2P0 3
(b)
P0 3
(c) P0
(d)
4 P0 3
SECTION 2 (Maximum Marks : 32) • • • •
This section contains EIGHT questions. Each question has FOUR options (a), (b), (c) and (d). ONE OR MORE THAN ONE of these four option(s) is(are) correct. For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS. For each question, marks will be awarded in one of the following categories : PHYSICS FOR YOU | APRIL ‘17
21
Full Marks : +4 If only the bubble(s) corresponding to the correct option(s) is(are) darkened.
Partial Marks : +1 For darkening a bubble corresponding to each correct option, provided NO incorrect option is darkened.
Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : –2 In all other cases. •
For example, if (a), (c) and (d) are all the correct options for a question, darkening all these three will result in +4 marks; darkening only (a) and (d) will result in +2 marks; and darkening (a) and (b) will result in –2 marks, as a wrong option is also darkened.
6. If P, Q, R are physical quantities, having different dimensions, which of the following combinations can never be a meaningful quantity? (a) (P – Q)/R (b) PQ – R (c) (R + Q)/P (d) (PR – Q2)/R 7. Mercury of density (rHg) is poured into cylindrical communicating vessels of cross-sectional area A1 and A2 (A1 > A2). A solid iron cube of volume V0 and density riron is dropped into the broad vessel, and as a result the level of the mercury in it rises. Then liquid of density rliq poured into the broader vessel until the mercury reaches the previous level in it. The height of liquid column h is V0ρiron (a) if the liquid does not submerge ρliq ( A2 + V02/3 ) the block Vρ (b) 0 iron if the liquid does not submerge the ρliq A2 block V (c) 0 if the liquid submerge the block A2 (d)
V0 ρiron − ρliq ρHg if the liquid submerge A2 ρHg − ρliq ρliq the block
8. Two parallel plates A and B are joined together to form 100°C A B 0°C a compound plate as shown in figure. The thicknesses of the plates are 4.0 cm and 2.5 cm respectively and the area of cross section is 100 cm2 for each plate. The thermal conductivities are KA = 200 W m–1°C–1 for the plate A and KB = 400 W m–1°C–1 for the plate B. The outer surface of the plate A is maintained at 100°C and the outer surface of the plate B is maintained at 0°C. (a) the rate of heat flow through any cross section is 3810 W. 22
PHYSICS FOR YOU | APRIL ‘17
(b) the temperature at the interface is 24°C (c) the equivalent thermal conductivity of the compound plate is 548 W m–1°C–1 (d) ratio of thermal resistance of plate A to the equivalent thermal resistance is 0.76. 9. Two rays of light A and B with wavelength 5000 Å travel parallel to each other in air. Ray A encounters a 1 mm thick layer of glass with refractive index m = 1.5. Then, (a) ray B will complete more oscillation than ray A (b) both rays will complete same number of oscillations (c) ray A will complete more oscillations than ray B (d) the actual difference in number of oscillation made by two waves over the 1 mm distance is 1000. 10. An external magnetic field is decreased to zero, due to which a current is induced in a circular wire loop of radius r and resistance R placed in the field. This current will not become zero, (a) at the instant when external magnetic field stops changing (t = 0), the current in the loop as a function of time for t > 0 is given by i0e–2Rt/m0pr (b) at the instant when B stops changing (t = 0), the current in the loop as a function of time µ IR t > 0 is given by 0 2r (c) the time in which current in loop decreases to 10 –3 I 0 (from t = 0) for R = 100 W and r = 5 cm is given by
3π2 ln 10
s 1010 (d) the time in which current in loop decreases to 10–3 I0 (from t = 0) for R = 100 W and r = 5 cm is given by
3π 2
s 106 11. Two thin convex lenses of focal lengths f1 and f2 are separated by a horizontal distance d (where d < f1 , d < f2) and their centres are displaced by a vertical separation D as shown in the figure. y
O
d
x
Taking the origin of coordinates O, at the centre of the first lens, the x and y coordinates of the focal point of this lens system, for a parallel beam of rays coming from the left, are given by f f (a) x = 1 2 , y = ∆ f1 + f2 (b) x =
f1 ( f1 + d ) ∆ ,y= f1 + f2 − d f1 + f2
(c) x =
∆( f1 − d ) f1 f2 + d( f1 − d ) ,y= f1 + f2 − d f1 + f 2 − d
(d) x =
f1 f2 + d( f1 − d ) , y = 0. f1 + f2 − d
12. A small ball starts moving from A over a fixed track as shown in the figure. Surface AB has friction. 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 respectively. Then (a) hA > hC ; KB > KC A C (b) hA > hC ; KC > KA hA hC (c) hA = hC ; KB = KC (d) hA < hC ; KB > KC B
13. In the arrangement shown in figure, gas is thermally insulated. An ideal gas k is filled in the cylinder having pressure m, S P0 (> atmospheric pressure Pa). Spring of force constant k is initially unstretched. P0 Piston of mass m and area S is frictionless. In equilibrium piston rises up a distance x0, then kx mg (a) Final pressure of the gas is Pa + 0 + S S 1 2 (b) Work done by the gas is kx0 + mgx0 2 (c) Decrease in internal energy of the gas is 1 2 kx + mgx0 + Pa Sx0 2 0 12 2 (d) Work done by the gas is kx 2 0 SECTION 3 (Maximum Marks : 15) • • • •
This section contains FIVE questions. The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive. For each question, darken the bubble corresponding to the correct integer in the ORS. For each question, marks will be awarded in one of the following categories :
Full Marks : +3 If only the bubble corresponding to the correct answer is darkened. Zero Marks : 0 In all other cases.
14. Two cylindrical rollers of diameters D and d respectively rest on a horizontal plane as shown in figure. The diameter of the larger roller is four times that of smaller one. The larger F roller wound round with a string D is pulled with a horizontal force d F. Assuming that the coefficient of friction is m for all surfaces of contact, find the larger value of m (in 10–1) as the larger roller can be pulled over the smaller one. 15. A simple pendulum is suspended from a peg on a vertical wall. The pendulum is pulled away from the wall to a horizontal position and released. The bob hits the wall, the restitution
Peg O
coefficient being 2/ 5 . What is the minimum number of impacts with the wall after which the maximum angular displacement becomes less than 60°? 16. In the circuit shown in figure, the internal resistance of the cell is negligible. For the value of R = (40/x) W, no current flows through the galvanometer. Find x. R
2V G
0.5 A 40
0.5 A 40
17. A column of mercury of length 10 cm is contained in the middle of a horizontal tube of length 1 m which is closed at both the ends. The two equal lengths contain air at standard atmospheric pressure of 76 cm of mercury. The tube is now turned to vertical position. Then the column of mercurcy will be displaced by x cm. The value of x is 18. Two identical small equally charged conducting balls are suspended from long threads secured at one point. The charges and masses of the balls are such that they are in equilibrium when the distance between them is a (the length of the thread L >> a). One of the balls is then discharged. Again for the certain value of distance b (b << L) between the balls, the equilibrium is restored, the value of (a3/b3) PHYSICS FOR YOU | APRIL ‘17
23
PAPER-II SECTION 1 (Maximum Marks : 18) • • • •
This section contains SIX questions. Each question has FOUR options (a), (b), (c) and (d). ONLY ONE of these four options is correct. For each question, darken the bubble corresponding to the correct option in the ORS. For each question, marks will be awarded in one of the following categories : Full Marks : +3 If only the bubble corresponding to the correct option is darkened. Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : –1 In all other cases.
1. An alpha particle with kinetic energy K = 0.50 MeV is deflected through an angle of q = 90° by the Coulomb field of a stationary Hg nucleus. The least curvature radius of its trajectory and the minimum approach distance between the particle and the nucleus respectively are. (a) 0.56 pm and 0.23 pm (b) 0.34 pm and 0.45 pm (c) 0.23 pm and 0.56 pm (d) 0.45 pm and 0.34 pm 2. In the following R-C circuit, the capacitor is in the steady state. The initial separation of the capacitor plates is x0. If at t = 0, the separation between the plates starts changing so that a constant current flows through R, find the velocity of the moving plates as a function of time. The plate area is A. 2 I / ε0 A I / ε0 A 1 (a) t+ x0 IR − E E − IR
I / ε0 A I / ε0 A 1 (b) t + IR − E E − IR x0 I / ε0 A I / ε0 A 1 (c) t + − IR IR x E E − 0
−2
−2
2
24
PHYSICS FOR YOU | APRIL ‘17
4. A closed container of volume 0.02 m3 contains a mixture of neon and argon gases, at a temperature of 27°C and pressure of 1 × 105 N m–2. The total mass of the mixture is 28 g. If the molar masses of neon and argon are 20 and 40 g mol–1 respectively, the masses of the individual gases in the container assuming them to be ideal is (Universal gas constant R = 8.314 J mol–1 K–1). (a) Mneon = 4 g and Margon = 8 g (b) Mneon = 4 g and Margon = 24 g (c) Mneon = 2 g and Margon = 16 g (d) Mneon = 8 g and Margon = 16 g 5. A screw gauge having 100 equal divisions and a pitch of length 1 mm is used to measure the diameter of a wire of length 5.6 cm. The main scale reading is 1 mm and 47th circular division coincides with the main scale. Find the curved surface area of wire (in cm2) to appropriate significant figure. (a) 2.0 cm2 (b) 4.6 cm2 2 (c) 2.6 cm (d) 4.0 cm2 6. A smooth semicircular wire-track of radius R is fixed in a vertical plane. One end of a massless spring of natural length 3R/4 is attached to the lowest point O of the wire-track. A small ring of mass m, which can slide on the track, is attached to the other end of the spring. The ring is held stationary at point P such that the spring makes an angle of 60° with the vertical. The spring constant k = mg/R. Consider the instant when the ring is released, and determine the normal reaction of the ring.
2
−1 I / ε0 A I / ε0 A 1 (d) t+ x0 IR − E E − IR 3. A conducting bar with B mass m and length L slides over horizontal L rails that are connected I to a voltage source. The voltage source maintains a constant current I in the rails and bar, and a constant, uniform, vertical
magnetic field B fills the region between the rails (as shown in figure). If the bar has mass m, find the distance d that the bar must move along the rails from rest to attain speed v. 2 2 v 2m 3v 2m (a) (b) 5v m (c) v m (d) 2 ILB 2 ILB ILB 2 ILB
(a) (c) • • •
3mg 8
(b)
mg 4
C
60° 3mg mg (d) O 4 2 SECTION 2 (Maximum Marks : 32)
P
This section contains EIGHT questions. Each question has FOUR options (a), (b), (c) and (d). ONE OR MORE THAN ONE of these four option(s) is(are) correct. For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS.
•
•
For each question, marks will be awarded in one of the following categories : Full Marks : +4 If only the bubble(s) corresponding to all the correct option(s) is(are) darkened. Partial Marks : +1 For darkening a bubble corresponding to each correct option, provided NO incorrect option is darkened. Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : –2 In all other cases. For example, if (a), (c) and (d) are all the correct options for a question, darkening all these three will result in +4 marks; darkening only (a) and (d) will result in +2 marks; and darkening (a) and (b) will result in –2 marks, as a wrong option is also darkened.
7. Mariner 4 was designed to travel from earth to Mars in an elliptical orbit with its perihelion at earth and its aphelion at Mars. Assume that the orbits of earth and Mars are circular with radii RE and RM respectively. Neglect the gravitational effects of the planets on Mariner 4. (a) The velocity of Mariner 4 relative to earth by which it may leave the earth is vr =
2GMRM GM − RE ( RM + RE ) RE
(b) Its velocity relative to Mars by which it will reach the orbit of mars is vr ′ =
2GMRE GM − R M ( RM + R E ) RM
(c) The time period of revolution of mariner 4 3/2 R around the Sun is T = TE (RE + RM ) E 2
−3/2
(d) The time taken for Mariner 4 to reach Mars is 3/2
1 R + RM t= E 2 2 RE
years
8. A tube of length l and radius R carries a steady flow of fluid whose density is r and viscosity is h. The fluid flow velocity depends on the distance r from r2 the axis of the tube as v = v0 1 − . Choose the R2 correct option(s) from the following. (a) the volume of the fluid flowing across the 1 section of the tube per unit time is π2v0 R2 2 (b) the kinetic energy of the fluid within the tube’s 1 2 2 volume is πlR ρv0 6
(c) the friction force exerted on the tube by the fluid is 4phlv0 (d) the pressure difference at the ends of the tube 4ηlv0 is R2 9. A point charge +q is projected from point A towards an infinitely long line charge having linear charge density l with kinetic energy K0. The distance of the closest approach will be +
a
+
+
x
+
A
B
+
(a)
2 πε0 K 0 − qλ ae
(b) ae
2 πε0 K 0 (c) a 1 − qλ
πε K − 0 0 qλ
2 πε0 K 0 (d) qλ
10. The coordinates of a particle moving in a plane are given by x(t) = a cos(pt) and y(t) = b sin(pt) where a, b and p are positive constants of appropriate dimensions. Then, (a) the path of the particle is an ellipse (b) the velocity and acceleration of the particle are normal to each other at t = p/(2p) (c) the acceleration of the particle is always directed towards a origin (d) the distance travelled by the particle in time interval t = 0 to t = p/(2p) is a 11. A neutron of kinetic energy 65 eV collides inelastically with a singly ionized helium atom at rest. It is scattered at an angle of 90° with respect to its original direction. [Given : Mass of He atom = 4 × (mass of neutron), Ionization energy of H atom = 13.6 eV]. (a) the allowed values of energy of neutron after collision are 6.36 eV and 0.312 eV (b) the allowed values of energy of helium atom after collision are 17.84 eV and 16.328 eV (c) Frequencies of emitted radiations during de-excitation from n3 to n2 is 1.82 × 1015 Hz (d) Frequencies of emitted radiations during de-excitation from n3 to n1 is 9.85 × 1014 Hz PHYSICS FOR YOU | APRIL ‘17
25
12. A triode has plate characteristics in the form of parallel lines. At a grid voltage of –1 V, the anode current I (in mA) is given in terms of plate voltage V (in V) by the algebraic relation: I = (0.125V – 7.5). For grid voltage of – 3 V, the current at anode voltage of 300 V is 5 mA. (a) the plate resistance is r p = 4 × 103 W (b) the transconductance is gm = 12.5 × 10–3A V–1 (c) the amplification factor of the triod is m = 100 (d) None of these 13. In a Young’s experiment the light source is at distance l1 = 20 mm and l2 = 40 mm from the slits. The light of wavelength l = 500 nm is incident on slits separated at a distance 10 mm. A screen is placed at a distance D = 2 m away from the slits as shown in figure. P l1
S
S1 l2 d
C
S2 D
(a) The value of q relative to the central line where maxima appear on the screen is n sin −1 2 − 1 . 40 (b) Number of maxima appear on the screen will be 40. (c) Minimum thickness of a slab of refractive index 1.5 be placed on the path of one of the ray to obtain minima at C is 500 nm. (d) Phase difference at C will be 30 p. 14. A solid cone and a solid sphere are arranged as shown in the figure. The centre of mass is (a) at 3R if m1 = m2 = m and R1 = R2 (b) at 2R from centre of mass of solid cone if r1 = r2 and R1 = R2 11R if r1 = 2r2 and R1 = R2 3 (d) Both (a) and (b) (c) at
26
PHYSICS FOR YOU | APRIL ‘17
SECTION 3 (Maximum Marks : 12) • • • • •
This section contains TWO paragraphs. Based on each paragraph, there are TWO questions. Each question has FOUR options (a), (b), (c) and (d). ONLY ONE of these four options is correct. For each question, darken the bubble corresponding to the correct option in the ORS. For each question, marks will be awarded in one of the following categories :
Full Marks : +3 If only the bubble corresponding to the correct option is darkened.
Zero Marks : 0 In all other cases.
PARAGRAPH 1 A variable capacitor with a range from 10 to 365 pF is used with a coil to form a variable frequency LC circuit to tune the input to a radio. Ratio of maximum to minimum frequencies may be tuned with such a capacitor is 6.04. If this capacitor is to tune from 0.54 to 1.60 MHz, the ratio computed in is too large. By adding a capacitor in parallel to the variable capacitor this range may be adjusted. 15. Which of the following is required capacitor? (a) 20 pF (b) 10 mF (c) 30 mF (d) 36 pF 16. What inductance should be chosen in order to tune the desired range of frequencies? (a) 5 H (b) 0.22 mH (c) 550 mH (d) 2 H PARAGRAPH 2
3 has a 2 small object O on its inner surface. It is observed from a point outside the sphere on the side opposite to the object. The inner surface A and outer surface B are concentric and uniformly thick. A hollow sphere of glass of refractive index
µ = 3/2 R O X 2R A
B
R = 0.25 m
17. As light from the object refracts in surface A, the image is formed at a distance (from X) of (a) 0.25 m (b) 0.3 m (c) 0.375 m (d) 0.4 m 18. The distance of the final image from the point object is (a) 0.375 m (c) 0.6 m
(b) 0.75 m (d) 0.036 m
SOLUTIONS
Paper-I 1. (a) : Let, initial capacity = C, final capacity = KC \ Initial charge Q0 = CV0, Final charge Q = (KC)V0 Q Now, =K , \ Q > Q0 (QK > 1) Q0 V = V0 = Potential difference across the battery V \ E = E0 = 0 d 1 Energy, U 0 = CV02 2 1 U 2 U = (KC )V0 or =K (Q K > 1) 2 U0 \ U > U0. 2. (c) : Since the temperature of the gas is constant PDV + VDP = 0 P P ( Ax ) ⇒ DP = − ∆V = − V AL Mg PA ⇒ (DP)A = − x=− x L L Mg Net restoring force = − + k x L ML \ T = 2π Mg + kL 3. (b) : If the mass of the Sun is M and radius of the planet’s orbit is r, then as v0 = (GM /r ), T=
2πr r = 2πr v0 GM
4π2r 3 ...(i) GM Now, if the planet (when stopped in the orbit) has velocity v when it is at a distance x from the sun, by conservation of mechanical energy, i.e., T 2 =
−∫
0
r
= −r ∫
2
2GM r − x dx , − = dt r x
i.e., −
dx 2GM (r − x ) = dt r x
0 r x or ∫ dt = − dx ∫ 0 2GM r (r − x ) Substituting x = r sin2 q and solving the RHS. t
0
π/2
(1 − cos 2θ)dθ 0
πr 1 = −r θ − sin 2θ = 2 π /2 2 r πr × 2GM 2 From eqn. (i) and (ii), t=
...(ii)
2 t = T 8 4 2 4. (d) : Induced emf in the spoke is shown in figure. t=
1
T
i.e.,
D
I/2 C
I O
ε
E
I/2
A B
1 ε Bωr 2 ε = Bωr 2 , I = = 2 R 2R There will be no induced emf separately in parts ADC or AEC. B 2 ωr 3 Fb = IrB = 2R Balancing torque about E, F r = Fbr/2 F B 2 ωr 3 ⇒ F= b = 2 4R (Force due to current I/2 will act on circular parts also, but their torque about E will be zero.) 5. (a) : Given, P = P0 + (1 – a)V2 Q PV = nRT \ (P0 + (1 – a)V2)V = nRT (P0 + (1 − α)V 2 )V nR dT P0 3V 2 (1 − α) = + = 0 ⇒ P0 = 3(a – 1)V2 dV nR nR P0 V2 = 3(α − 1) 2P P0 = 0 P = P0 + (1 − α) 3 3(α − 1)
⇒ T=
1 2 GMm GMm mv + − = 0 − 2 x r or
1/2
2 0 r sin θ x dx = − ∫ r ⋅ 2 ⋅ sin θ cos θ d θ π /2 r − r sin2θ (r − x )
or \
6. (a, c) : Since, sum and difference of physical quantities having different dimensions, are not meaningful. So, (P – Q) and (R + Q) will not be valid in (b, d), [PQ] = [R], is possible in (d), [PR] = [Q2] is possible. PHYSICS FOR YOU | APRIL ‘17
27
7. (a, d) 8. (a, b, d) 9. (c, d) : Time taken by the ray A to travel through the glass, t1 =
d vg
[d = 1 mm and vg = velocity of light in glass = c/m] µd 1.5 × 10 −3 \ t1 = = = 5 × 10 −12 s 8 c 3 × 10
d 10 −3 10 = = × 10 −12 s 8 c 3 × 10 3 5 10 Dt = t1 – t2 = 5 − × 10 −12 = × 10 −12 s 3 3 The frequency of the light (oscillation per second) does not change inside the glass. t1 > t2 ⇒ ray A will make more oscillations. n1 = Number of oscillations for ray A = ut1 Similarly, n2 = Number of oscillations for ray B = ut2 c Dn = n1 – n2 = u(t1 – t2) = (∆t ) λ 8 3 × 10 5 = × 10 −12 = 1000 −10 3 5000 × 10 For ray B, t2 =
10. (a, c) : Flux linked with loop due to its own magnetic field, µ I µ πrI f = 0 (πr 2 ) = 0 2r 2 emf induced = − I= I
∫
I0
Now,
µ πr dI dφ =ε=− 0 dt 2 dt
µ πr dI ε =− 0 R 2R dt t
dI 2R −2 Rt /µ0 πr =−∫ ⋅ dt ; I = I 0e µ π I r 0 0
10–3I
0
= I 0e
which gives t =
−
2 Rt µ0 πr
3π2 ln10 1010
s.
11. (c) 12. (a, b) : At point A, potential energy of the ball = mghA At point B, potential energy of the ball = 0 At point C, potential energy of the ball = mghC Total energy at point A, EA = KA + mghA Total energy at point B, EB = KB Total energy at point C, EC = KC + mghC According to the law of conservation of energy, EA = EB = EC E A = E B ⇒ K B > KA ...(i) 28
PHYSICS FOR YOU | APRIL ‘17
EB = EC ⇒ KB > KC ...(ii) EA = EC KA + mghA = KC + mghC K − KA or hA − hC = C ...(iii) mg ⇒ hA > hC ; KC > KA ...(iv) Option (b) is correct. From eqns. (i), (ii) and (iv), we get hA > hC ; KB > KC Option (a) is correct. 13. (a, c) 14. (5) : Here C1 and C2 are the centres of the rollers. OC1 is the perpendicular difference between the heights of centers of cylindrical roller from the ground. 1 \ C1C2 = R + r = (D + d) 2 1 OC1 = R – r = (D – d) 2 F
C1 D–d 2
D+d 2
O
N1 N1 C2
90 – N2
C2 N2 W
D −d 2 dD ...(i) or cos a = D+d D+d Let N1 be the normal reaction between the two rollers and N2 between the smaller cylinder and the horizontal plane. The larger cylinder can be pulled over the smaller one provided the latter neither rolls nor slides. There will be no rolling provided the sum of the moments of forces about C2 is zero mN1d = mN2d; N1 = N2 There will be no sliding provided the su m of resolved parts of the forces in horizontal direction is zero mN2 + mN1 cos (90 – a) = N1 cos a mN1 + mN1 sin a = N1 cos a (As N1 = N2) cos α m= ...(ii) 1+ sin α ⇒ sin a =
d (using eqns. (i) and (ii)) D Hence the necessary condition is that m=
m≥
d D
or m ≥
1 ⇒ m ≥ 5 × 10–1 4
15. (4) : When the bob is at the O l A horizontal position A, its height n above B = length of the string = l. l Therefore, potential energy at A = D C h mgl. When the bob is released, it B hits the wall at B and the entire potential energy is converted into kinetic energy. If v is the velocity with which the bob hits the wall, then 1 2 mv = mgl or v = 2gl ...(i) 2 Speed of the bob after the first rebound, v1 = ev Speed of the bob after the second rebound, v2 = ev1 = e2v Speed of the bob after n rebounds, vn = env If the ball rises to a position C at a height h = BD after n rebounds, then from the principle of conservation of energy, we have 1 2 mv = mgh 2 n v 2 (env )2 e 2nv 2 h= n = = ...(ii) 2g 2g 2g Using eqn. (i) in (ii), we get 2 gle 2n h= = le 2n ...(iii) 2g If qn is the angle the string subtends with vertical after n rebounds, it follows from figure that h = OB – OD = l – l cos qn = l (1 – cos qn) ...(iv) From Eqs. (iii) and (iv), we have le2n = l(1 – cos qn) or e2n = (1 – cos qn) 1 For qn to be less than 60°, i.e., cos qn is greater than , 2 1 1 (1 – cos qn) must be less than . Thus, e 2n < 2 2 2 Given, e = 5 \
2 5
2n
<
n
1 1 4 or < 2 5 2
or
1 (0.8)n < or n ln (0.8) < ln (0.5) 2
or
–(0.0969) n < – 0.3010 or n >
or V = 4
\ \ or
2 R = 0.05 A 40 I3 I3 = 0.5 – 0.05 0.5 A = 0.45 A 40 V −2 = 0.45 I1 R 4−2 2 40 R= Ω = = 0.45 0.45 9
17. (3) 18. (4) : Here A and B are two identical balls suspended at a point O as shown in figure.
G 40 2 I2
0
0.5 A
(Q R = 40/x W)
O T F
A
L
a
B
F
mg
1 q2 T sin q = F = 4 πε0 a2 T cos q = mg From eqn. (i) and (ii), tan q =
2V
2
...(i) q
2
...(ii)
4πε0a2mg
or
q2 a a/2 = (Since for small q, tan q = ) 2 2L 4 πε0a mg L
or
q2 a3 = L 2πε0mg
...(iii)
When one of the balls is discharged, the balls come closer and touch each other and again separate due to repulsion. The charge on each ball after touching each other is q/2. Replacing q with q/2 in (iii), we get (q /2)2 b3 ...(iv) = L 2πε0mg a3 From eqns. (iii) and (iv), 3 = 4 b Paper-II 1. (c) : We shall ignore the recoil of Hg nucleus.
0.3010 0.0969
or n > 3.1 The least integer greater than 3.1 is 4. Hence n = 4. Thus, after 4 impacts, the amplitude will become less than 60°. 16. (9) : Since no current is flowing through galvanometer \ I1 = I2 V −2 2−0 = 40 40
I1 =
Let A be the point of closest approach to the centre C, AC = rmin. At A the motion is instantaneously circular because the radial velocity vanishes. Then if v0 is the speed of the particle at A, the following equations hold PHYSICS FOR YOU | APRIL ‘17
29
z1z2e 2 1 K = mv02 + (4 πε0 ) rmin 2
...(i)
E
mv0 rmin = 2mK b
...(ii) (r = rmin is the radius of curvature of the path at A) mv02 z z e2 ...(iii) = 1 2 2 ρmin (4 πε0 ) rmin 2
z1 z2 e θ cot ...(iv) (4 πε0 ) 2K 2 here K is kinetic energy of incident alpha particle. From eqn. (ii) and (iii) 2 z z e2 2kb2 z1 z2 e θ or ρmin = 1 2 cot 2 = 2 (4 πε0 ) 2k ρmin (4 πε0 ) b=
with z1 = 2, z2 = 80, k = 0.50 MeV and q = 90°, we get rmin = 0.23 pm From eqns.(ii) and (iv) we write
z1 z2 e 2
(cot θ / 2) , v0 (4 πε0 ) 2mL Substituting in eqn. (i) rmin =
1 θ mv02 + 2mK v0 tan 2 2 Solving for v0 we get, 2K θ θ v0 = sec − tan m 2 2 K=
θ cot z1z2e 2 2 Then rmin = θ (4 πε0 )2K θ sec − tan 2 2
=
z1z2e 2 θ 1 + cosec (4 πε0 )2K 2
Substitution gives rmin = 0.56 pm 2. (b) : Let q be the instantaneous charge on the capacitor when a steady current I flows through the circuit. Applying KVL on the circuit, we have ε A qx q + IR Q C = 0 ...(i) E = + IR or E = ε0 A x C Differentiating eqn. (i) with respect to time, we get 0=
q ε0 A
Ix dx +0 + ε0 A dt
or q = −
From eqn (ii) substituting the value of q in eqn (i), we have
(Q q = It)
Ix dx = velocity ...(ii) where (dx / dt ) dt
= −I
or
x2 + IR ε0 A(dx / dt )
− I / ε0 A 2 dx =v= x E − IR dt
...(iii)
I / ε0 A ...(iv) = dt x 2 E − IR Integrating the above expression w.r.t. time, we get 1 1 I / ε0 A ...(v) − = t x x0 E − IR or −
dx
From eqn (iii) and (v), we get I / ε0 A v= IR − E
I / ε0 A 1 t + x0 E − IR
−2
v 2 v 2m = 2a 2 ILB 4. (b) : Temperature of mixture = 27°C = 300 K Let M = mass of neon gas in the mixture \ Mass of Argon in the mixture = (28 – M) M \ Number of gram moles of neon, n1 = 20 \ Number of gram moles of argon 28 − M n2 = 40 Using Dalton’s law of partial pressures, n RT n RT RT P= 1 + 2 or P = ( n1 + n2 ) V V V 28 − M RT M or P = + 40 V 20 40PV or (M + 28) = RT 40 × (1 × 105 ) × 0.02 or (M + 28) = 32g 8.314 × 300 or M = 4 g \ Mneon = 4 g and Margon = 24 g 5. (c) : Least count of screw gauge pitch 1 mm = = = 0.01 mm N 100 Diameter = D = 1 + (47 × 0.01) = 1.47 mm = 0.147 cm Area of curved surface, S = pDl 22 ∴ S = × 0.147 × 5.6 = 2.5872 cm2 7 Round off to two significant digits = 2.6 cm2. 3. (d) : v 2 = 2ad ⇒ d =
Contd. on Page no. 86
30
PHYSICS FOR YOU | APRIL ‘17
1. A stone of mass 2 kg is projected upward with kinetic energy of 98 J. The height at which the kinetic energy of the stone becomes half of its original value, is given by (a) 5 m (b) 2.5 m (c) 1.5 m (d) 0.5 m 5 2. If one mole of a monatomic gas g = is mixed 3 7 with one mole of a diatomic gas g = , the value 5 of g for the mixture is (a) 1.40 (b) 1.50 (c) 2.91 (d) 3.07 3. A block of mass m1 lies on a smooth horizontal table and is connected to another freely hanging block of mass m2 by a light inextensible string passing over a smooth fixed pulley situated at the edge of the table as shown in the figure. Initially the system is at rest with m1 at a distance d from the pulley. The time taken for m1 to reach the pulley is m1 m2
(a) (c)
m2 g m1 + m2 2m2 d (m1 + m2 ) g
(b)
2d(m1 + m2 ) m2 g
(d) None of these
4. An ideal gas heat engine operates in Carnot cycle between 227 °C and 127 °C. It absorbs 6 × 104 cal of heat at higher temperature. Amount of heat converted to work is (a) 4.8 × 104 cal (b) 6 × 104 cal 4 (c) 2.4 × 10 cal (d) 1.2 × 104 cal ^ ^ 5. A projectile is given an initial velocity of ( i + 2 j ). The
cartesian equation of its path is (Take g = 10 m s–2).
(a) y = 2x – 5x2 (c) 4y = 2x – 5x2
(b) y = x – 5x2 (d) y = 2x – 25x2
6. Of the following quantities which one has the dimensions different from the remaining three? (i) Energy density (ii) Force per unit area (iii) Product of charge per unit volume and voltage (iv) Angular momentum per unit mass (a) (i) (b) (ii) (c) (iii) (d) (iv) 7. In an adiabatic change, the pressure and temperature of a monatomic gas are related as P ∝ TC, where C equals 5 3 5 2 (a) (b) (c) (d) 2 5 3 5 8. A block of mass m is attached with a massless spring of force constant k. The block is placed over a rough inclined surface for which the coefficient of friction 3 is m = . The minimum value of M required to 4 move the block up the plane is (Neglect mass of string and pulley and friction in 3 pulley) sin 37° = 5 4 3 3 6 (a) m (b) m (c) m (d) m 5 2 5 5 9. Three charges, each + q, are placed C(q) at the corners of an isosceles E D(Q) 2a 2a triangle ABC of equal sides BC A ( q ) B(q) and AC = 2a, as shown in figure. D and E are the mid points of BC and CA. The work done in taking a charge Q from D to E is qQ qQ (a) (b) 8 p ε0 a 4 p ε0 a 3qQ (c) Zero (d) 4 p ε0 a Physics For you | April ‘17
31
10. The volume of a metal sphere increases by 0.24% when its temperature is raised by 40°C. The coefficient of linear expansion of the metal is 10–5
°C–1
(a) 2 × (c) 18 × 10–5 °C–1
10–5
°C–1
(b) 6 × (d) 1.2 × 10–5 °C–1
11. The change in the gravitational potential energy when a body of mass m is raised to a height nR above the surface of the earth is (R is the radius of the earth) n n mgR mgR (a) (b) n − 1 n + 1 mgR (c) nmgR (d) n 12. Two pith balls carrying equal charges are y suspended from a y/2 common point by strings r r′ length, the equilibrium separation between them is r, as shown in figure. Now the strings are rigidly clamped at half the height. The equilibrium separation between the balls, now becomes 2r (a) 3
1 (b) 2
r (c) 3 2
2r (d) 3
2
13. The mass of a rocket is 500 kg and the relative velocity of gases ejecting from it is 250 m s–1 with respect to the rocket. Find the rate of burning of the fuel to give the rocket an initial acceleration of 20 m s–2 in the vertically upward direction. (Take g = 10 m s–2) (a) 300 kg s–1 (c) 90 kg s–1
(b) 60 kg s–1 (d) 30 kg s–1
14. A body is released from a point distance r from the centre of the earth. If R is the radius of the earth and r > R, then the velocity of the body at the time of striking the earth will be (a)
gR
(b)
32
(c) t1/4
Physics For you | April ‘17
(a) 1/2
(b) 1/4
(c) 2
(d) 1
18. A man running has half the kinetic energy of a boy of half his mass. The man speeds up by 1 m s–1 and then has Kinetic energy as that of the boy. What were the original speeds of the man and the boy? (a) (b) (c)
( (
(
)
2 m s −1 ; 2 2 − 1 m s −1
) ( 2 + 1) m s ; 2 ( −1
2 −1 m s , 2 −1
) 2 + 1) m s
2 − 1 m s −1 −1
(d) None of these 19. When a body is placed in surroundings at a constant temperature of 20°C and heated by a 10 W heater, its temperature remains constant at 40°C. If the temperature of the body is now raised from 20°C to 80°C in 5 min at a uniform rate, the total heat it will lose to the surroundings will be (a) 3000 J (b) 3600 J (c) 4500 J (d) 5400 J 20. A square conducting loop of side length L carries a current I. The magnetic field at the centre of the loop is (a) independent of L (b) linearly proportional to L2 (c) inversely proportional to L (d) linearly proportional to L 21. Find the position of the image formed by the lens combination given in figure. f = + 10, –10, + 30 cm
2gR (r − R) 2gRr (d) r r−R 15. A body is moved along a straight line by a machine delivering a constant power. The distance moved by the body in time t is proportional to (b) t3/2
(b) P ∝ r2 1 (d) P ∝ 2 r 17. The work done in turning a magnet of magnetic moment M by an angle of 90° from the meridian, is n times the corresponding work done to turn it through an angle of 60°. The value of n is given by (a) P ∝ r (c) P ∝ r3
2gR
(c)
(a) t3/4
16. A spherical body of radius r radiates power P and its rate of cooling is R. Then
(d) t1/2
O 30 cm 5cm 10cm
(a) 30 cm (c) 20 cm
(b) 25 cm (d) 40 cm
22. In an adiabatic process wherein pressure increases by 2 % , if CP = 3 ,then the volume decreases by about 3 CV 2 9 2 4 (a) % (b) % (c) 4% (d) % 4 3 9 23. A circular disc of radius R is removed from a bigger circular disc of radius 2R such that the centre of mass of new disc is aR from the centre of the bigger disc. The value of a is 1 1 1 1 (a) (b) (c) (d) 6 2 4 3 24. A wave equation which gives the displacement along the y-direction is given by y = 10–4 sin(60t + 2x) where x and y are in m and t in s. This represents a wave (a) travelling with a velocity of 60 m s–1 in the positive x-direction (b) of wavelength 2p m (c) of frequency (30/p) Hz (d) of amplitude 10–4 m travelling along the positive x-direction. 25. In the Young's double slit experiment, the interference pattern is found to have an intensity ratio between bright and dark fringes as 9. This implies that (a) the intensities at the screen due to two slits are 5 units and 4 units respectively (b) the intensities at the screen due to two slits are 4 units and 1 unit respectively (c) amplitude ratio is 3 (d) the amplitude ratio is 2. 26. A wire of density 9 × 103 kg m–3 is stretched between two clamps one m apart and is subjected to an extension of 4.9 × 10–4 m. What will be the lowest frequency of the transverse vibrations in the wire? (Y = 9 × 1010 N m–2) (a) 38 Hz (b) 36 Hz (c) 35 Hz (d) 32 Hz 27. A particle is executing linear simple harmonic motion. The fraction of the total energy that is potential, when 1
its displacement is of its amplitude is 2 1 1 1 1 (a) (b) (c) (d) 2 16 8 4 28. In Young's double slit experiment, the distance between two sources is 0.1 mm. The distance of the screen from the source is 20 cm. Wavelength
of light used is 5460 Å. The angular position of the first dark fringe is (a) 0.08° (b) 0.16° (c) 0.20° (d) 0.32° 29. A simple pendulum has a time period T1. The point of suspension is now moved vertically upward according to equation y = kt2 where k = 1 m s–2 and y is the vertical displacement. The T2 x time period now becomes T2. The ratio of 12 is . T2 5 What is the value of x? (g = 10 ms–2) (a) 6
(b) 3
(c) 1
(d) 4
30. In a Young's double slit experiment, red light of wavelength 6000 Å is used and the nth bright fringe is obtained at a point P on the screen. Keeping the same setting, the source is replaced by green light of 5000 Å and now (n + 1)th bright fringe is obtained at the point P. Calculate the value of n. (a) 4 (b) 6 (c) 3 (d) 5 31. The rms value of the electric field of the light coming from the sun is 720 N C–1. The rms value of magnetic field of the electromagnetic wave is (a) 220 × 10–8 T (b) 240 × 10–8 T (c) 200 × 10–8 T (d) 210 × 10–8 T 32. An electron is in an excited state in a hydrogen like atom. It has a total energy of –3.4 eV. The kinetic energy of the electron is E and its de Broglie wavelength is l. Then (a) E = 6.8 eV, l = 6.6 × 10–10 m (b) E = 3.4 eV, l = 6.6 × 10–10 m (c) E = 3.4 eV, l = 6.6 × 10–11 m (d) E = 6.8 eV, l = 6.6 × 10–11 m 33. A point source of electromagnetic radiation has an average power output of 1000 W. The maximum value of electric field at a distance 5.0 m from the source is (a) 34.6 V m–1 (c) 69.2 V m–1
(b) 49.0 V m–1 (d) 98.0 V m–1
34. Nine resistors each of resistance R are connected in the circuit as shown in figure. The effective resistance between A and B is
C
A
B
2 7 3 (a) R (b) R (c) R (d) R 9 6 5 35. A nucleus at rest disintegrates into two nuclear parts which have their velocities in the ratio 2 : 1. The ratio of their nuclear sizes will be Physics For you | April ‘17
33
(a) 21/3 : 1 (c) 31/2 : 1
(b) 1 : 31/2 (d) 1 : 21/3
R 36. Figure shows a circuit with known resistances R1 and R2. R2 R1 Neglect the internal resistance E E1 of the sources of current and resistance of the connecting wire. The magnitude of electro-motive force E1 such that the current through the resistance R is zero will be (a) ER1/R2 (b) ER2/R1 (c) E(R1 + R2)/R2 (d) ER1/(R1 + R2)
37. A surface irradiated with light of wavelength 480 nm gives out electrons with maximum velocity v m s–1 the cut off wavelength being 600 nm. The same surface would release electrons with maximum velocity 2v m s–1 if it is irradiated by light of wavelength (a) 325 nm (b) 360 nm (c) 384 nm (d) 300 nm 38. A zener diode when used as a voltage regulator is connected (i) in forward bias (ii) in reverse bias (iii) in parallel with load (iv) in series with load (a) (i) and (ii) are correct (b) (ii) and (iii) are correct (c) (i) only correct (d) (iv) only correct
39. The half life of a radioactive isotope X is 50 years. It decays to another element Y which is stable. The two elements X and Y were found to be in the ratio of 1 : 15 in a sample of a given rock. The age of the rock was estimated to be (a) 150 years (b) 200 years (c) 250 years (d) 100 years 40. Which of the following circuits provides full wave rectification of an ac input? (a) ac
(b) input
output
ac
(c) input
(d)
34
ac input
output
solutions
1. (b) : At the time of projection kinetic energy of the 1 stone, K = mu2 2 And m = 2 kg, K = 98 J 2K 2 × 98 ∵ u2 = = = 98 m 2 2 ∵ v2 = u2 – 2gh ⇒ h = u (∵ v = 0) ...(i) 2g 98 ∴ h= =5m 2 × 9. 8 1 1 Also, K = mu2 = m 2 gh (Using (i)) 2 2 1 1 K ′ = mu ′ = m × 2 gh ′ 2 2 K ′ h′ K ∴ = ∵K ′ = K h 2 K h′ h 5 = ⇒ h ′ = = m = 2. 5 m 2K h 2 2 n1 g 1 n2 g 2 + g1 − 1 g 2 − 1 2. (b) : g mixture = n1 n + 2 g1 − 1 g 2 − 1 7 5 Here, n1 = 1, g 1 = , n2 = 1, g 2 = 5 3 3 ∴ g mixture = = 1.5 2 T 3. (b) : Let a be common acceleration a of the system. m1 T m2 a The free body diagrams of two blocks is as shown in the figure. m2 Their equations of motion are T = m1 a ...(i) m2 g – T = m2 a ...(ii) From eqn. (i) and (ii), we get m2 g a= ...(iii) m1 + m2 1 Using, s = ut + at 2 2 1 m2 g ⇒d = 0×t + t2 2 (m1 + m2 ) ∴
(∵u = 0 and Using eqn. (iii))
output
Physics For you | April ‘17
∴
t=
2d(m1 + m2 ) m2 g
Physics For you | April ‘17
35
4. (d) ^
^
5. (a) : Here, u = i + 2 j and tan θ =
uy ux
=
11. (a) : At the surface of earth r = R, Gravitational potential energy,
2 1
Us = −
The required equation is gx 2 y = x tan θ − 2 2u cos2 θ 10 x 2 = x ×2− 2 2 1 2 12 + 22 × 5 ∴ y = 2x – 5x2
(
At the height, h = nR, r = R + h = R(1 + n)
)
−g
RT ⇒ P = Constant ⇒ P ∝ T 1− g P g ∵ P ∝ T C ∴ C= For monatomic gas, ( g − 1) 5 g= . 3 53 5 Therefore C = = 5 3 −1 2
( )
3 8. (a) : ∵mg sin θ = mg sin 37° = mg 5 3 3 4 f max = mmg cos θ = mg = mg 4 5 5 Block will move upwards when 6 kx0 = f max + mg sin θ = mg ...(i) 5 From conservation of mechanical energy 6 kx0 5 mg 3 1 2 = = m (Using (i)) Mgx0 = kx0 ⇒ M = 2g 2g 5 2 Cq 9. (c) : As AC = BC, therefore, from E D symmetry q q Potential at D = Potential at E B A i.e., VD = VE ∴ Work done in taking charge Q from D to E = Q (VE – VD) = Zero 10. (a) : From, VT = V0(1 + g∆T) ⇒
VT − V0 = g∆T V0
0.24 −5 −1 = g 40°C ⇒ g = 6 × 10 °C 100 Coefficient of linear expansion, g a = = 2 × 10 −5 °C −1 3 ∴
36
Physics For you | April ‘17
mgR GMm =− R(1 + n) (1 + n) Change in gravitational potential energy is mgR − (−mgR) ∆U = Uh – Us = − (1 + n) mgR n =− + mgR = mgR 1+ n 1 + n ∴
6. (d) 7. (d) : For adiabatic change, PVg = Constant g
GM ∵ g = 2 R
GMm = − mgR R
Uh = −
12. (c) : If θ is angle which each pith ball in equilibrium, makes with the vertical, then as it is clear from figure. F kq 2 / r 2 r /2 F mg = = tanθ = y mg mg or ∴ ∴
y=
mg r 3 2kq 2
r′ y′ = r y r r ′ = 1/3 2
θ y r/2
⇒ y ∝ r3 1/3
y / 2 = y
1/3
1 = 2
1/3
13. (b) : Here, m = 500 kg, u = 250 m s–1, a = 20 m s–2, g = 10 m s–2
dm =? dt
dm Thrust required, m(a + g) = u dt dm 500 (20 + 10) = 250 × dt 500 30 × dm = = 60 kg s −1 250 dt 14. (d) 15. (b) : Power, P = Fv = constant dv dv ∵ F = ma = m ∴ m v=P dt dt P or vdv = dt m Integrate both sides we get
∫ vdv =
P v2 P dt ⇒ = t + C1 m∫ 2 m
At t = 0, v = 0, so, C1 = 0
t ∴ θ − θ0 = °C 5 dQ 1 t °C ∴ = k(θ − θ0 ) = W °C −1 5 s dt 2
1/2
v2 P 2P ∴ = t or v = t m 2 m 1/2 1/2 ds 2Pt 2Pt ∴ = or ds = dt dt m m
Integrate both sides, we get 1/2
dQ 1 = W s −1 t . dt 10
1/2
2 2 P 3/ 2 2Pt ∫ ds = ∫ m dt ⇒ s = 3 m t + C2
∴ Q=∫
∴ u = ( 2 + 1) m s −1 and u ′ = 2u = 2( 2 + 1) m s
−1
19. (c) : Let θ be the temperature of the body. It is given that the temperature of the surroundings is θ0 = 20 °C dQ Rate of loss of heat, ...(i) = k(θ − θ0 ). dt For θ = 40 °C, dQ = 10 W = k(40 °C − 20 °C) dt 10 W 1 ...(ii) or k = = W °C −1 20 °C 2 When the temperature of the body is raised uniformly, its temperature θ, at time t is given by 80 °C − 20 °C 1°C θ = 20 °C + t = 20 °C + t, 300 s 5s
(Using eqn. (ii) and (iii)) 300 s
2 1 1 −1 −1 t W s t dt = W s 10 10 2 0 s
300 s
0s
At t = 0, s = 0, so C2 = 0 1/2 8P 3/2 ∴ s = t 3/ 2 ⇒ s ∝ t 9m 16. (b) : According to Stefan’s law, E = sT4 Power radiated, P = (4pr2)sT4 ∴ P ∝ r2 17. (c) 18. (c) : Let mass of boy be m. Then, mass of man is = 2m, 1 Kinetic energy of man = Kinetic energy of boy 2 1 1 1 u ′2 u′ ,u = ∴ (2m)u2 = × mu ′ 2 ⇒ u2 = 4 2 2 2 2 When man speeds up to 1 m s–1, As, Kinetic energy of man = Kinetic energy of boy 1 1 1 (2m) (u + 1)2 = mu ′ 2 = m(2u)2 2 2 2 1 (u + 1)2 = 2u2 ⇒ u = 2 −1
...(iii)
2 2 (300 s) − (0 s) 1 = 4500 J = W s −1 10 2
20. (c) 21. (a) : For the image formed by the first lens, u1 = –30 cm, f1 = 10 cm 1 1 1 − = ∴ ⇒ v1 = 15 cm v1 −30 10 The image formed by the first lens serves as the object for the second. This is at a distance of (15 – 5) cm = 10 cm to the right of the second lens. Though the image is real it serves as a virtual object for the second lens, which means that the rays appear to come from it for the second lens, 1 1 1 − = ∴ ⇒ v2 = ∞ v2 10 −10 The virtual image is formed at an infinite distance to the left of the second lens. This acts as an object of the third lens. 1 1 1 − = cm ⇒ v3 = 30 cm v3 ∞ 30 The final image is formed 30 cm to the right of the third lens. ⇒
22. (a) 23. (b) : In figure, O is the centre of a circular disc of radius 2R and mass M. M = p(2R)2s, where s is mass per unit area of disc. C1 is centre of disc of radius R, which is removed.
x C2 O
C1
M Mass of removed disc, M1 = p(R)2 s = 4 Mass of remaining disc, M 3M = M2 = M − M1 = M − 4 4 Let its centre of mass be at C2, where OC2 = x ∴ M1 × OC1 = M2 × OC2 Physics For you | April ‘17
37
M 3M ×R= x 4 4 R 1 x = = aR ∴ a = 3 3
or
24. (c) : Here, y = 10–4 sin(60t + 2x) The standard equation of wave travelling in negative direction of x-axis, y = A sin (wt + kx). Compare the two equations, we get A = 10–4, w = 60, k = 2 w 60 30 ∴ Frequency v = = = Hz 2p 2p p 30 or v = p 2p 2p ∴ Wavelength, l = = =pm k 2 30 Velocity, v = ul ⇒ v = × p or v = 30 m s–1 p 25. (b) 26. (c) : The lowest frequency of the transverse vibrations 1 T in the wire, u = 2L m where T is the tension, L is the length and m is the mass per unit of the wire. As m = rA TL YA∆L ∵ Young’s modulus, Y = or T = A∆L L Now, u=
1 Y ∆L T Y ∆L u= = 2L rL m Lr
9 × 1010 × 4.9 × 10 −4 1 ⇒ u = 35 Hz 2 ×1 9 × 103 × 1
27. (d) 28. (b) : For first dark fringe (n = 1) lD lD x = (2 n − 1) = 2 d 2 d x l Angular position, θ = = D 2d 5460 × 10 −10 = radian 2 × 10 −4 180 degree = 0.16° = 2730 × 10 −6 × p 29. (a) : Given, y = kt2 The velocity and acceleration are given by dy d2 y (where k = 1 m s–2) = 2kt and = 2k dt dt 2 38
Physics For you | April ‘17
Thus, the point of suspension of the pendulum is moving upwards with an acceleration a = 2 m s–2. This is the case where the pseudo acceleration is acting downwards. Hence, effective acceleration due to gravity g′ = g + a = 10 + 2 = 12 m s–2 l l ∴ T1 = 2p and T2 = 2p g g′ ∴
T12
T22
=
g ′ 12 6 = = g 10 5
x 6 = ⇒x=6 5 5 30. (d) : Let x be the distance of point P from the centre of the screen. When red light (l = 6000 Å) is used, nth bright fringe is obtained at point P. ∴
nDl nD × 6000 × 10 −10 = d d When green light (l′ = 5000 Å) is used, (n + 1)th bright fringe is obtained at the same point P. (n + 1) Dl ′ (n + 1) D × 5000 × 10 −10 ∴ x= = d d Equating the two values of x, we get −10 nD × 6000 × 10 −10 (n + 1) D × 5000 × 10 = d d or 6n = 5(n + 1) ⇒ n = 5 31. (b) : The rms value of magnetic field of electromagnetic wave is E 720 Brms = rms = = 240 × 10–8 T c 3 × 108 32. (b) : The potential energy = – 2 × Kinetic energy = – 2K. ∴ Total energy = – 2K + K = – K = – 3.4 eV ∴ x=
As Kinetic energy (K) p2 or p = 2mK . ∴ K= 2m
1 1 1 m (2v )2 = hc − 2 l ′ l0 1 1 − 1 l l0 1 or ∴ = = 1 1 4 4 − l ′ l0
de Broglie wavelength, h h l= = p 2mK On substituting the values, we get 6.63 × 10 −34 l= 2 × 9.1 × 10 −31 × 3.4 × 1.6 × 10 −19
l′ = 300 nm 38. (b) 39. (b) : Let at a time t, NX be the number of nuclei of X in a sample and NY be the number of nuclei of Y in that sample. NX 1 Given, or NY = 15 NX = NY 15
= 6.6 × 10–10 m 33. (b) : Intensity of the electromagnetic wave is P 1 I= = ε0 E02 c 2 2 4 pr ∴ E0 =
2P 4 pε0 r 2 c
Part of N X =
2 × 1000 × (9 × 109 )
=
(5)2 × (3 × 108 ) or
≈ 49 V m–1 C
34. (d) :
A
R/3
R/3
Then,
A
B
R/3
(2R / 3)(R / 3) 2 Effective resistance is Reff = = R 2R R 9 + 3 3 35. (d) : As the nucleus at rest disintegrates, linear momenta of the two parts must be equal and opposite, i.e., m v 1 m1v1 = m2v2 or 1 = 2 = m2 v1 2 As nuclear size, r ∝ A1/3 ∴
r1 A1 = r2 A2
1/3
m = 1 m2
1/3
1 = 2
1/3
1/3
= 1: 2
36. (c) 37. (d) : According to Einstein’s photoelectric equation hc hc 1 1 2 2 mvmax = hu − hu0 or mvmax = − l l0 2 2 where l is the wavelength of incident radiation and l0 is threshold wavelength. 1 1 1 ∴ mv 2 = hc − 2 l l0
NX 1 1 = = N X + NY 16 2
4
n
⇒
B
1 [N + N Y ] 16 X
If n is the number of half lives of isotope X.
2R/3 R/3
1 1 − 480 600 1 1 − l ′ 600
NX 1 1 = = 2 N X + NY 2
4
or
n=4
If t is the age of rock, and T is the half life period of X, then t = nT = 4 × 50 = 200 years. 40. (d)
EXAM DATES 2017
SRMJEEE JEE MAIN
1st April to 30th April (Online) 2nd April (Offline) 8th & 9th April (Online)
VITEEE
5th April to 16th April (Online)
NATA
16th April
WBJEE
23rd April
Kerala PET AMU (Engg.) Karnataka CET
24th April (Physics & Chemistry) 25th April (Mathematics) 30th April 2nd May (Biology & Mathematics) 3rd May (Physics & Chemistry)
NEET
7th May
MHT CET
11th May
COMEdK (Engg.)
14th May
BITSAT
16th May to 30th May (Online)
JEE Advanced
21st May
J & K CET
27th May to 28th May
AIIMS
28th May
JIPMER
4th June
Physics For you | April ‘17
39
PRACTICE PAPER
AIIMS 1. A ring and a disc of different masses are rotating with the same kinetic energy if we apply a retarding torque on the ring and ring stops after making n revolutions, then in how many revolutions will the disc stop under the same retarding torque? (a) n (b) 2n (c) 4n (d) n/2 2. A hollow cylinder has a charge q C within it. If is the electric flux in units of V m associated with the curved surface B, the flux linked with the plane surface A in units of V m will be
(a)
q 20
q (c) 0
(b)
3
1 q (d) 2 0
3. A particle is thrown with a speed u at an angle with the horizontal. When the particle makes an angle with the horizontal, its speed changes to v, which is equal to (a) u cosθ (b) u cosθ cos (c) u cosθ sec (d) u secθ cos 4. Two charges of magnitude 10 units and 20 units are separated by certain distance. Now both the charges are brought into contact and again separated to initial position. What will be the ratio of initial and final force? 4 9 3 8 (a) (b) (c) (d) 3 8 2 9 5. Two small satellites move in circular orbits around the earth, at distances r and r + r from the centre of the earth. Th eir time periods of rotation areT and T + T, (r r, T T). Th enT is equal to 40
PHYSICS FOR YOU | APRIL ‘17
3 r T 2 r 3 r (c) T 2 r (a)
Exam on 28th May
2 r T 3 r r (d) T r
(b)
6. A vessel with water is placed on a weighing pan and it reads 600 g wt. A spherical body of mass 50 g and density 0.8 g cm–3 is sunk into the water with a pin of negligible volume, as shown in figure, keeping it sunk. Th e weighing pan will show a reading which is equal to (a) 600 g wt (b) 620 g wt (c) 640 g wt (d) 580 g wt 7. Th e inductanceL of a solenoid depends upon its radius R as (a) L R (b) L 1/R (c) L R2 (d) L R3 8. Figure shows a sinusoidal wave on a string. If the frequency of the wave is 150 Hz and the mass per unit length of the string is 0.2 g m–1, the average power transmitted by the wave is
(a) 2.34 W (c) 4.80 W
(b) 3.84 W (d) 5.78 W
9. A glass prism of refractive index 1.5 is immersed in water, shown in figure. A light beam incident normally on the face AB is totally reflected to reach the face BC if
(a)
2 8 < sin q < 3 9
(b) sin q ≤
2 3
(c) cos q ≥
8 9
(d) sin q >
8 9
v
10. In Young’s double slit experiment, the fringe width is b. If the entire arrangement is placed in a liquid of refractive index m, the fringe width becomes (a) mb
(b)
b b (c) m−1 m+1
(d)
b m
11. Two circular coils can be arranged in any of the three situations shown in the figure. Their mutual inductance will be
(i)
(a) (b) (c) (d)
(ii)
(iii)
maximum in situation (i) maximum in situation (ii) maximum in situation (iii) the same in all situations.
12. A bob of mass M is suspended by B a massless string of length L. The horizontal velocity v at position A is just sufficient to make it reach L the point B. The angle θ at which the speed of the bob is half of that v A at A, satisfies p p p (a) q = (b) < q < 4 4 2 3p 3p p (c) (d)
P
Q Floor
(a) P and Q both move upwards with speeds 2v. (b) P and Q both move upwards. (c) P moves upwards and Q moves downwards with equal speeds. (d) Both P and Q are at rest. 15. A, B and C are three points B in a uniform electric field. C The electric potential is (a) maximum at C (b) same at all the three points A, B and C (c) maximum at A (d) maximum at B
A
E
16. Two springs of spring constants k1 and k2 are connected as shown in the figure. The effective spring constant ke is k1k2 (a) k1 + k2 (b) k1 + k2 (c) k1k2 (d) k1/k2 17. A point source of light is kept at a depth of h in water of refractive index 4/3. The radius of the circle at the surface of water through which light emits is 7 3 7 3 h (a) h (b) h (c) h (d) 3 3 7 7 18. Brewster’s angle in terms of refractive index (m) of the medium is (a) tan −1 m (b) sin–1m (c) sin −1 m
(d) tan–1m
19. If reaction on the block is R and coefficient of friction between the block and surface is µ, the work done against friction in moving a body by distance d is
(a) mRd/4 (b) 2µRd (c) µRd
(d) µRd/2
Physics For you | April ‘17
41
20. The ionization energy of Li++ is equal to (a) 9 hcR (b) 6 hcR (c) 2 hcR (d) hcR 21. At absolute 0 K temperature, the band of maximum energy in which electrons are present is called (a) valence band (b) forbidden band (c) conduction band (d) fermi band. 22. If a-current gain of a transistor is 0.98. The value of b-current gain of the transistor is (a) 4.9 (b) 0.49 (c) 5 (d) 49 23. A paramagnetic substance of susceptibility 3 × 10–4 is placed in a magnetic intensity of 4 × 10–4 A m–1. Then the intensity of magnetization in the units of A m–1 is (a) 1.33 × 108 (b) 0.75 × 10–8 –8 (c) 12 × 10 (d) 14 × 10–8 24. The root-mean-square speed of molecules in still air at room temperature is closest to (a) walking speed of 2 m s–1 (b) the speed of fast car of 30 m s–1 (c) the speed of a supersonic airplane of 500 m s–1 (d) escape speed from Earth of 1.1 × 104 m s–1. 25. A body performs S.H.M. Its kinetic energy K varies with time t as indicated by graph (a)
(c)
(b)
(d)
26. In an experiment of photoelectric effect the stopping potential was measured to be V1 and V2 with incident light of wavelength l and l/2 respectively. The relation between V1 and V2 may be (a) V2 < V1 (b) V1 < V2 < 2V1 (c) V2 = 2V1 (d) V2 > 2V1 27. Frequency of the series limit of Balmer series of hydrogen atom in terms of Rydberg constant R and speed of light c is 4 Rc (a) Rc (b) 4Rc (c) (d) Rc 4 28. Einstein’s mass-energy conversion relation E = mc2 is illustrated by (a) nuclear fission (b) b-decay (c) rocket propulsion (d) steam engine 42
Physics For you | April ‘17
29. The amount of original radioactive material is left after 3 half-lives is (a) 6.5% (b) 12.5% (c) 25.5% (d) 33.3% 30. The frequencies of two tuning forks A and B are respectively 1.5% more and 2.5% less than that of the tuning fork C. When A and B are sounded together, 12 beats are produced in 1 s. The frequency of the tuning fork C is (a) 200 Hz (b) 240 Hz (c) 360 Hz (d) 300 Hz 31. A sound wave y = A0sin(wt – kx) is reflected from a rigid wall with 64% of its amplitude. The equation of the reflected wave is 64 A0 sin ( wt + kx ) (a) y = 100 64 A0 sin ( wt + kx ) 100 64 A0 sin ( wt − kx ) (c) y = 100 64 A0 cos ( wt − kx ) (d) y = 100 (b) y = −
32. In the reaction identify X 14 17 1 7N + a → 8X + 1p (a) an oxygen nucleus with mass 17 (b) an oxygen nucleus with mass 16 (c) a nitrogen nucleus with mass 17 (d) a nitrogen nucleus with mass 16. 33. X-rays, gamma rays and microwaves travelling in vacuum have same (a) wavelength but different velocities (b) frequency but different velocities (c) velocity but different wavelengths (d) velocity and same frequency 34. A smooth rectangular platform ABCD is moving towards right with a uniform speed u. At what angle θ must a particle be projected from A with speed v so that it strikes the point D? C
D
A
u (a) sin −1 v −1 v (c) sin u
u
v B
−1 v (b) cos u −1 u (d) cos v
35. A thin wire of resistance 4 W is bent to form a circle. The resistance across any diameter is (a) 4 W (b) 2 W (c) 1 W (d) 8 W 36. In the circuit shown, if the resistance 5 W develops a heat of 42 J s–1, the heat developed in 2 W must be about (in J s–1) 2
6
9 5
(a) 25
(b) 20
(c) 30
(d) 35
37. A carbon film resistor has colour code green, black, violet and gold. The value of the resistor is (a) 500 ± 5% MW (b) 50 MW (c) 500 ± 10% MW (d) 500 MW 38. A body of mass 8 kg is hanging from another body of mass 12 kg. The T1 combination is being pulled by a string 12 kg a with an acceleration of 2.2 m s–2 shown T2 in figure. The tension T1 and T2 will be 8 kg respectively (a) 260 N and 102 N (b) 336 N and 194 N (c) 220 N and 48 N (d) 240 N and 96 N 39. The mobility of free electrons (charge e, mass m and relaxation time t) in a metal is proportional to m e e m t (a) (b) e t (c) (d) mt m et 40. A thin metal plate M is inserted between the plates of a parallel plate capacitor as M shown in figure. The new capacitance in terms of initial capacitance C is (a) 2C (b) C/2 (c) C (d) infinity Directions : In the following questions (41-60), a statement of assertion is followed by a statement of reason. Mark the correct choice as : (a) If both assertion and reason are true and reason is the correct explanation of assertion. (b) If both assertion and reason are true but reason is not the correct explanation of assertion. (c) If assertion is true but reason is false. (d) If both assertion and reason are false. 41. Assertion : The length of the day is slowly increasing. Reason : The dominant effect causing a slow down in the rotation of the earth is the gravitational pull of other planets in the solar system.
42. Assertion : Mass is not conserved, but mass and energy are conserved as a single entity called mass-energy. Reason : Mass and energy are inter-convertible in accordance with Einstein’s relation, E = mc2. 43. Assertion : Impulsive force is large and acts for a short time. Reason : Finite change in momentum should be produced by the force. 44. Assertion : In javelin throw, the athlete throws the projectile at an angle slightly more than 45°. Reason : The maximum range does not depend upon angle of projection. 45. Assertion : Efficiency of a Carnot engine decreases with decrease in temperature difference between the source and the sink. Reason : Efficiency of cannot engine, T T −T η =1− 2 = 1 2 T1 T1 46. Assertion : At the centre of earth, a body has centre of mass, but no centre of gravity. Reason : At the centre of earth, acceleration due to gravity = 0. 47. Assertion : At critical temperature, surface tension of a liquid becomes zero. Reason : At critical temperature, intermolecular forces are maximum for liquids. 48. Assertion : When a dielectric medium is filled between the plates of a condenser, its capacitance increases. Reason : The dielectric medium reduces the potential difference between the plates of the condenser. 49. Assertion : For a point on the axis of a circular coil carrying current, magnetic field is maximum at the centre of the coil. Reason : Magnetic field is directly proportional to the distance of a point from the centre of the circular coil. 50. Assertion : If amplitude of simple pendulum increases, its time period also increases. Reason : The simple pendulum covers more distance in each vibration when its amplitude is large. 51. Assertion : The force between the plates of parallel plate capacitor is proportional to charge on it. Reason : Force is equal to charge per unit area per ε. Physics For you | April ‘17
43
52. Assertion : In an electromagnetic wave, magnitude of magnetic field vector is much smaller than the magnitude of electric field vector. Reason : Energy of electromagnetic waves is shared equally by the electric and magnetic fields. 53. Assertion : No interference pattern is detected when two coherent sources are infinitely close to each other. Reason : The fringe width is inversely proportional to the distance between the two slits. 54. Assertion : Violet shift indicates that a star is approaching the earth. Reason : Violet shift indicates decrease in apparent wavelength of light. 55. Assertion : Light emitting diode (LED) emits spontaneous radiation. Reason : LED are forward-biased p-n junction. 56. Assertion : If the frequency of the light incident on a metal surface is doubled, the maximum kinetic energy of emitted photo-electrons also gets doubled. Reason : Kinetic energy of particle is proportional to frequency. 57. Assertion : In an adiabatic process, change in internal energy of a gas is equal to work done on or by the gas in the process. Reason : Temperature of gas remains constant in a adiabatic process. 58. Assertion : Terminal voltage of a cell is greater than e.m.f. of the cell, during charging of the cell. Reason : The emf of a cell is always greater than its terminal voltage. 59. Assertion : If objective and eye lenses of a microscope are interchanged then it can work as telescope. Reason : The objective of telescope has small focal length. 60. Assertion : All the radioactive elements are ultimately converted to lead. Reason : All the elements above lead are unstable. solutions
1. (a) : Work done = change in kinetic energy = initial kinetic energy i.e., τ θ = K = constant. As τ is same in the two cases, θ must be same, i.e., number of revolutions must be same. 2. (d) : Let fA, fB and fC are the electric flux linked with A, B and C. 44
Physics For you | April ‘17
According to Gauss's theorem, q f A + f B + fC = e0 Since fA = fC, q q \ 2fA + fB = − fB or 2f A = e0 e0 q −f or, 2f A = e0 (Given fB = f) 1 q \ fA = − f 2 e0 3. (c) : In projectile motion, the horizontal component of velocity remains constant throughout the motion, so u cosθ = v cosφ or v = u cosθ /cosφ = u cosθsecφ 4. (d) : According to Coulomb's law 1 (10)(20) Initial force, Fi = ...(i) 4 pe0 r2 1 (15)(15) Final force, Ff = ...(ii) 4 pe0 r2 Divide eqn. (i) by (ii), we get Fi (10)(20) 8 = = Ff (15)(15) 9 5. (a) : By Kepler's third law, T2 ∝ r3 or T2 = k r3 ...(i) Differentiating it, we have, 2 T ∆T = 3 k r2 ∆r ...(ii) Divide eqn. (ii) by (i), we get 2T ∆T 3 k r 2 ∆r 3 ∆r or ∆T = T = 2 3 2 r T kr 6. (c) : The upward thrust i.e., (buoyancy force) acts on the body and an equal and opposite force acts on the water so the weight will be the sum of the two = 600 + 40 = 640 g wt. 7. (c) : The inductance of a solenoid, L = m0n2 Al where A is the area of cross-section of the solenoid, l its length and n is the number of turns per unit length. As A = pR2; where R is the radius of the solenoid. \ L = m0n2 pR2 l ⇒ L ∝ R2 8. (b) : Mass per unit length of the string, m = 0.2 g m–1 = 0.2 × 10–3 kg m–1 Frequency of the wave, u = 150 Hz From the figure, A = 0.06 m and 5 l = 20 cm 2 40 \ l = cm = 8 cm 5 = 8 × 10–2 m
Velocity of the wave, v = ul = (150 Hz) (8 × 10–2 m) = 12 m s–1 The average power transmitted by the wave is P = 2p2u2A2mv Substituting the given values, we get P = 2 × (3.14)2 × (150)2 × (0.06)2 × (0.2 × 10–3) × 12 = 3.84 W 3 a 4 a 9. (d) : Here, m g = 1.5 = , m w = 2 3 a mw 4/3 8 As sinC = = = a m g 3/2 9 For total internal reflection, sin q > sin c 8 \ sin q > 9 10. (d) : When the entire arrangement of Young’s double slit experiment is placed in a liquid of refractive index m, then the fringe width will become l ′D lD b lD b′ = = = Q b = d md m d 11. (a) : Flux linkage is maximum in the arrangement shown in diagram (i). Therefore mutual inductance will be maximum in case (i). 12. (d) : By conservation of energy, 2
1 2 1 v mv = m + mgh 2 2 2 1 1 5 gL m × 5 gL = m + mgL(1 − cos q) or 2 2 4 7 5 5 = + 1 − cos q or cos q = − 8 2 8 3p Hence
13. (c) : Here, N P = 1 , VP = 230 V, I S = 2 A NS
25
For an ideal transformer N 25 N P IS or I P = I S × S = 2 A × = 50 A = NP 1 NS IP 14. (b) : When P is pushed downwards with speed v, then Q moves upwards with same speed. Just after collision of P with floor, the block P starts moving upward. As string becomes loose so the block Q continue to move upwards. 15. (d) : In the direction of electric field, electric potential decreases. \ VB > VC > VA 16. (a) : The springs are connected in parallel. \ Effective spring constant ke = k1 + k2
17. (a) : sin C =
1 ⇒ m
R 2
R +h
2
=
3 4
2
Squaring, 16R = 9R2 + 9h2 3 7 R 2 = 9h 2 ⇒ R = h 7 18. (d) : When light is reflected at a particular angle of incidence such that the angle of incidence is tan–1m, the reflected light is completely polarised and the refracted light is also polarised and both will be making 90° to each other. The angle of incidence in this condition is called Brewster’s angle. 19. (c) :
To move the block without acceleration, the applied force F = f = µR ∴ Work done against friction = work done by applied force W = F × d = µRd 20. (a) : Q Ionization energy of hydrogen like atom = Z2hcR \ Ionization energy of Li++ = 9hcR (as Z = 3 for Li++) 21. (a) 22. (d) : Q
b=
a 1− a
\
b=
0.98 = 49 1 − 0.98
23. (c) : I = cH = (4 × 10–4) × (3 × 10–4) = 12 × 10–8 A m–1 3RT 3 × (25 / 3) × 300 ; 508 m s–1 = − 3 M 29 × 10 25. (a) : Kinetic energy varies with time but is never negative. 26. (d) : According to Einstein’s photoelectric equation hc f0 hc f0 − − ⇒ Vs = ⇒ V1 = le e le e f hc 2hc f0 hc f f − 0 = V2 = − =2 − 0 + 0 e l le e le e e e 2 24. (c) : v =
V2 = 2V1 + f0/e
\ V2 > 2V1 Physics For you | April ‘17
45
27. (d) : Frequency of a spectral line of the Balmer series of hydrogen atom is 1 1 u = Rc 2 − 2 ; n = 2, 3, ............ 2 n For series limit, n = ∞ 1 1 1 Rc \ u = Rc 2 − 2 = Rc = 4 4 2 ∞ 28. (a)
30.
31.
32. 33.
34.
35.
46
n
3
1 N 1 1 = = = 2 8 N 0 2 % amount of original material left 1 = × 100% = 12.5% 8 (d) : Let the frequencies of tuning forks A, B and C are uA, uB and uC respectively. According to given problem 1.5 uA = uC + uC ...(i) 100 2.5 and u B = uC − uC ...(ii) 100 Also, uA – uB = 12 Hz 1.5 2.5 \ uC + uC − uC − uC = 12 100 100 (Using eqn. (i) and (ii)) 12 × 100 4uC = 300 Hz = 12 ⇒ uC = 4 100 (b) : In case of a sound wave, the reflection at a rigid boundary will take place with a phase reversal of p but the reflection at an open boundary takes place without any phase change. \ The reflected wave is 64 64 y= A0 sin ( wt + kx + p) = − A0 sin ( wt + kx ) 100 100 (a) (c) : In vacuum, X-rays, gamma rays and microwaves travelling with same velocity, i.e., with the velocity of light c (= 3 × 108 m s–1) but different wavelengths or frequencies. (d) : The particle will strike the point D if velocity of particle with respect to platform is along AD or component of its relative velocity along AB is zero. It will be so if v cosq = u or q = cos–1(u/v) (c) : The resistance of the total wire is 4 W. Here, ACB (2 W) and ADB (2 W) are in parallel. 1 1 1 2 ⇒ = + = =1⇒ R =1W R 2 2 2
29. (b) :
36. (c) : Let R = 5 W In AB, (6 + 9) W = 3R
Physics For you | April ‘17
37.
38.
39. 40.
41.
42. 44.
45. 46. 48.
Given I12 R = 42 J s −1 9×5 ⇒ I2 = 42 16 42 × 16 \ I2 = 9×5 42 × 16 × 2 \ Heat developed in 2 W = ≈ 30 J s −1 45 (a) : The number assigned to green, black and violet are 5, 0 and 7. For gold, tolerance is 5%. \ Resistance value of the given resistor is R = 50 × 107 W ± 5% = 500 ± 5% MW (d) : From the figure the equation of motion of 8 kg block, T2 – 8 g = 8a T1 T2 = 8 a + 8 g = 8 (a + g) 12 kg = 8 × (2.2 + 9.8) = 96 N a The equation of motion of 12 kg block is T2 12 × a = T1 – 12 g – T2 8 kg T1 = 12 (a + g) + T2 = 12 (2.2 + 9.8) + 96 = 240 N 8g v et (a) : m = d = E m (d) : When the two plates of capacitor are joined by a metal plate, both the plates acquire the same potential. Therefore, potential difference, V = 0. Q Q \ C = = = infinity V 0 (d) : The length of the day depends on the speed of rotation of the earth about its axis. As it remains constant, the length of the day is not changing. Further, the earth is not slowing down due to the gravitational pull of other planets in the solar system. (a) 43. (a) (d) : When a body is projected from a place above the earth’s surface, the angle of projection must be slightly less than 45° for the maximum horizontal angle. T T −T (a) : As η = 1 − 2 = 1 2 , T1 T1 therefore, η will decrease if (T1 – T2) decreases. (a) 47. (c) (a) : The dielectric molecules are polarised, producing an opposite electric field. Thus the effective electric field and hence the potential difference between the plates is reduced and
49.
50.
51.
52.
53.
54.
55.
consequently the capacitance is increased (QC = Q/V). (c) : The magnitude of the magnetic field produced due to flow of current through a circular loop at a point on its axis is given by m 2pIa2 ⇒ B= 0 4 p (a2 + x 2 )3/2 This is maximum at centre of loop, where x = 0 m 2pI m0 I ⇒ B= 0 = 4p a 2a l (d) : Time period of simple pendulum T = 2p , g is independent of the amplitude of vibration, when amplitude is small (Which is usually assumed). The motion is simple harmonic motion when the restoring force is directly proportional to the displacement. In simple pendulum, F = – mgsinθ. F is proportional to θ or angular displacement only when sinθ ; θ or θ is very small. If θ is very large or amplitude is very large, then motion of pendulum no more remains S.H.M. (d) : The potential energy stored between the plates of capacitor, 2 e0 A 1 Q2 = 1 Q x Q C = U= x 2 e0 A 2 C −∂U ∂ 1 Q2x 1 Q2 Force F = = − = − ∂x ∂x 2 e0 A 2 e0 A Negative sign shows force is attractive. (b) : At every instant the ratio of the magnitudes of the electric field to the magnetic field of an electromagnetic wave is given by E/B = c. From this equation, the magnitude of electric vector is much greater than the magnitude of magnetic vector. Also electromagnetic waves carry energy which is equally shared by electric and magnetic field. (a) : When d is negligibly small, fringe width b which is proportional to 1/d may become too large. Even a single fringe may occupy the whole screen. Hence the pattern cannot be detected. (b) : As lv < lr so, violet shift means apparent wavelength of light from a star decreases. Obviously, apparent frequency increases. This would happen when the star is approaching the earth. (b) : In semiconductor there may be energy bands due to donor impurities (ED) near the conduction band or acceptor impurities (EA) near the valence band. When electron falls from higher to lower
energy level containing holes, the energy is released in the form of radiation. The energy of radiation emitted by LED is equal or less than the band gap of the semiconductor used. The radiation released lies in range of visible light whose colour depends on the semiconductor used.
56. (d) : Let Kmax and K′max be the maximum kinetic energies corresponding to frequencies u and 2u of the incident light. Let φ0 be the work function of the metal. ∴ hu = Kmax + φ0 and 2hu = K′max + φ0 K′ + f On dividing, 2 = max 0 K max + f0 or, 2Kmax + 2φ0 = K′max + φ0 or, K′max = 2Kmax + φ0 It means the maximum kinetic energy of electron will become more than double if the frequency of the incident radiation is doubled. 57. (c) : In an adiabatic process, no exchange of heat is permissible i.e., dQ = 0 As, dQ = dU + dW = 0 ∴ dU = – dW 58. (c) : During charging of the cell, current flows inside the cell from positive to negative, supporting the electrostatic field inside. Therefore V > e. 59. (d) : We cannot interchange the objective and eye lens of a microscope to make a telescope. The reason is that the focal length of lenses in microscope are very small, of the order of mm or a few cm and the difference (fo – fe) is very small, while the telescope objective have a very large focal length as compared to eye lens of microscope. 60. (b) : All those elements which are heavier than lead are radioactive. This is because in the nuclei of heavy atoms, besides the nuclear attractive forces, repulsive forces between the protons are also effective and these forces reduce the stability of the nucleus. Hence, the nuclei of heavier elements are being converted into lighter and lighter elements by emission of radioactive radiation. When they are converted into lead, the emission is stopped because the nucleus of lead is stable (or lead is most stable elements in radioactive series). However there are series changing to stable bismuth Physics For you | April ‘17
51
Carnot's Cycle Isothermal expansion:
THERMODYNAMICS
W1 = mRT1 loge
CLASS XI Zeroth Law Two systems in thermal equilibrium with a third system separately are in thermal equilibrium with each other.
First Law Heat supplied to a gas may (i) raise its internal energy (ii) enable it to expand and thereby do external work. dQ = dU + dW
DQ mDT 1 DQ Molar specific heat capacity, C = m DT DQ = msDT = ms(Tf – Ti) or s =
Laws of Thermodynamics
here, (Q1 < Q2)
THERMODYNAMICS Heat Engine The efficiency h of the engine is ;
State Variables and Equation of State The relation between the state variables (P, V, T) of the system is called equation of state. For m moles of an ideal gas, equation of state is PV = mRT and for 1 mole of an ideal gas it is PV = RT. Isothermal Process Isothermal process : A thermodynamic process in which the temperature remains constant. · Equation of isothermal process, PV = constant. · Work done during isothermal process, Vf Pi W = mRT ln ; W = mRT ln Vi Pf · The slope of isothermal curve on a P-V diagram at any point on the curve is given by dP P =– dV V
Refrigerator The coefficient of performance of a refrigerator W is ; Hot Q2 Q2 Q1 Q2 Cold Reservoir Reservoir a= = W Q1– Q2 T1 T2
Thermodynamic Processes
h=
W Q Hot Q 1 = 1 – 2 Reservoir Q1 Q1 T 1
W Q2
Cold Reservoir T2
here (Q1 > Q2)
Adiabatic Process Adiabatic Process : A thermodynamic process in which no heat flows between the system and the surroundings. · Equation of adiabatic process, PVg = constant, where g = CP/CV. · Work done during adiabatic process, (PiVi – PfVf) mR(Ti – Tf) W= ;W= (g – 1) g–1 · The slope of adiabatic curve on a P-V diagram at any point on the curve is given by dP P = –g dV V
Isochoric Process Isochoric (isometric) process : A thermodynamic process in which volume remains constant. · Equation of isochoric process : P = constant. T · No work is done by the gas in an isochoric process. · The slope of the isochoric curve on a P-V diagram is infinite.
P A (P1, V1, T1) Isoth expa ermal nsion T1, Q1
B (P2, V2, T1) c ati iab sion Ad pan ex
Specific Heat Capacity
Second Law There is no heat engine can have efficiency h equal to 1 or no refrigerator can have coefficient of performance is equal to infinity.
Adiabatic expansion: R(T1 – T2) W2 = m g–1 Isothermal compression: V3 W3 = mRT2 loge V4 Adiabatic compression: R(T1 – T2) W4 = m g –1 c bati Adia ression p Com
Thermal Equilibrium The macroscopic variables such as pressure, temperature, volume, mass, composition, etc., which characterize a system, do not change with time.
V2 V1
D (P4, V4, T2)
Isothe Compr rmal ession T2, Q2
O
E
F
C (P3, V3, T2)
G
H
V
Ne t w o r k d o n e d u r i n g t h e complete cycle, W = W1 + W2 + (–W3) + (–W4) = W1 – W3 = Area ABCD (As W2 = W4) Work done W = Efficiency, h = Heat input Q1 Q W h=1– 2 = Q1 Q1 Relation between Coefficient of Performance and Efficiency of Refrigerator 1–h a= h Isobaric Process Isobaric process : A thermodynamic process in which pressure remains constant. · Equation of isobaric process : V = constant. T · Work done during isobaric process, W = P(Vf – Vi) = mR(Tf – Ti). · The slope of the isobaric curve on a P-V diagram is zero.
DUAL NATURE OF RADIATION AND MATTER
Transmission electron microscope
Electron source Condenser lens Objective lens Specimen
CLASS XII
Diffraction lens
+
ct
fe
Current indicator
c tri
lec
oe
t ho
n
io
at
ic pl
P of
Ef
Photoelectric Effect The phenomenon of emission of electrons from a metal surface when an electromagnetic wave of suitable frequency is incident on it is called photoelectric effect.
Photoelectric Equation ·
DUAL NATURE OF RADIATION AND MATTER
Particle Nature of Radiation
Ap
·
·
· · ·
de Broglie Wavelength h l= p · For electron having K.E. (K) is h Ap l= , here p = 2mK pl ic 2mK at io n · For a charged particle accelerated of by potential V is de Br h og l= , here p = 2qmV lie 2qmV W av
Davisson-Germer Experiment
Photoelectric Effect Electrons ejected Light from the shining surface Showed on clean sodium particle properties metal in a of light vacuum
es
de Broglie Hypothesis 50°
·
Due to symmetry in nature, the particle in motion also possesess wave-like properties. And these waves are called matter waves.
Nickel crystal
Sodium metal
E = Kmax + f0, where f0 = work function E = energy of incident light Kmax = maximum KE of e– 1 2 2 + hu Þ hu = 1 mvmax 0 Þ 2 mvmax = h(u – u0) 2
At constant frequency u and potential V (Photo-current) ip µ I (intensity) At constant frequency and intensity, the minimum negative potential at which the photocurrent becomes zero is called stopping potential (V0). At stopping potential V0, Kmax of e– = eV0 For a given frequency of the incident radiation, the V0 is independent of I. The V0 varies linearly with u.
Wave Nature of Matter
Nature's Love with symmetry arises the matter-wave duality
Davisson and Germer Experiment ·
q d
Experimental Study and Conclusion of Photoelectric Effect ·
Electron microscope is a device designed to study very minute objects · Based on principle of de Broglie wave and the fast moving electrons can be focussed by E or B field in a same way as beam of light is focussed by glass lenses.
An electrical device which converts light energy into electrical energy, is called as photocell or photoelectric cell. · It works on the principle of photoelectric emission of electrons
54 V beam
Voltage source
Fluorescent screen
·
·
Metal surface Positive terminal
Electron Microscope
Photoelectric Cell
Evacuated chamber
Lattice spacing
q
Study of wave nature of electron. · At a suitable potential V, the fine beam of electrons from electron gun is allowed to strike on the nickel crystal. The electrons are scattered in all directions and following assumptions were made : Ø Intensity of scattered electrons depends over scattering angle f. Ø A kink occurs in curve at f = 50° for 54 eV beam. Ø The intensity is maximum at accelerating voltage 54 V. After this voltage, intensity starts decreasing.
Bragg condition for constructive interference nl = 2dsin q
f Scattering angle f Accelerated electrons
u = constant
ip
I3 > I2 > I1 u > u > u ip 3 2 1 I3 I2 I1
–V0
V
u3
u2
u1
–V03 –V02–V01
I Saturation current
Photocell Radiant energy
V
Ø Here, q = 1 (180° – f) Þ q = 65° at f = 50° 2 From Bragg’s law (particle nature), l = 2d sin q Þ l = 1.65 Å. Also, from wave nature at V = 54 volt, l =
12.27 = 1.65 Å 54
AMU Engg.
1. If gravitational potential (V) of a solid sphere of radius a is plotted as a function of distance from its centre (r) which is the correct curve? (a)
(b)
(c)
(d)
2. Identify the wrong statement. (a) The electrical potential energy of a system of two protons shall increase if the separation between the two is decreased. (b) The electrical potential energy of a protonelectron system will increase if the separation between the two is decreased. (c) The electrical potential energy of a protonelectron system will increase if the separation between the two is increased. (d) The electrical potential energy of a system of two electrons shall increase if the separation between the two is decreased. 3. A physical quantity Q is related to four observables x 2 /5 z 3 x, y, z and t by the relation Q = y t The percentage errors of measurement in x, y, z, and t are 2.5%, 0.5%, 2% and 1% respectively. The percentage error in Q will be (a) 5% (b) 4.5% (c) 8% (d) 7.75% 4. An aircraft executes a horizontal loop at a speed of 720 km h–1 with its wings banked at 15°. The radius of the loop is (a) 10 km (b) 15 km (c) 12 km (d) 23 km 54
Physics For you | April ‘17
Exam Date 30th April 2017
5. An ideal cell is connected to a capacitor through a voltmeter. The reading V of the voltmeter is plotted against time (t). Which of the following best represents the resulting curve? (a)
(c)
(b)
(d)
V
V t
t
6. At a metro station, a girl walks up a stationary escalator in time t1. If she remains stationary on the escalator, then the escalator take her up in time t2. The time taken by her to walk up on the moving escalator will be t1t2 (t + t ) (a) 1 2 (b) (t2 − t1 ) 2 t1t2 (c) (d) t1 – t2 (t1 + t2 )
7. Two inclined frictionless tracks, one gradual and the other steep meet at A from where two stones are allowed to slide down from rest, one on each track as shown in figure. Which of the following statements is correct? I
A
II h
B
C
(a) Both the stones reach the bottom at the same time but not with the same speed. (b) Both the stones reach the bottom with the same speed and stone I reaches the bottom earlier than stone II. (c) Both the stones reach the bottom with the same speed and stone II reaches the bottom earlier than stone I. (d) Both the stones reach the bottom at different times and with different speeds.
8. Find the time constant (in ms) for the given R - C circuits in the given order respectively. R1 R2
V C1
C2
C2 C1
V R1
y
A
P
V R2
R1
C1
R2
C2
R1 = 1 W, R2 = 2 W, C1 = 4 mF, C2 = 2 mF. (a) 18, 4, 8/9 (b) 18, 8/9, 4 (c) 4, 18, 8/9 (d) 4, 8/9, 18
9. The rms value of the electric field of the light coming from the sun is 720 N C–1. The average total energy density of the electromagnetic wave is (a) 3.3 × 10–3 J m–3 (b) 4.58 × 10–6 J m–3 –9 –3 (c) 6.37 × 10 J m (d) 81.35 × 10–12 J m–3 10. For a particle a graph of x x versus t is shown in figure. A B E C Choose correct alternatives. (a) The particle was not at t D rest initially. (b) The speed at E exceeds that at D. (c) At C, the velocity and the acceleration vanish. (d) Average velocity for the motion between A and D is positive. 11. What is the angle of incidence for an equilateral prism of refractive index 3 so that the ray is parallel to the base inside the prism? (a) 30° (b) 45° (c) 60° (d) 90° 12. Figure shows the variation of energy with the orbit radius r of a satellite in a circular motion. Mark the correct statement. (a) C shows the total energy, B the kinetic energy and A shows the potential energy of the satellite. (b) A shows the kinetic energy, B shows the total energy and C shows the potential energy of the satellite. (c) A and B show the kinetic and potential energies respectively and C shows the total energy of the satellite. (d) C and A show the kinetic and potential energies respectively and B shows the total energy of the satellite. 13. A particle slides down a frictionless parabolic (y = x2) track (A – B – C) starting from rest at point A. Point B is at the vertex of parabola and point C is at a height less than that of point A. After C, the particle moves freely in air as a projectile. If the particle reaches highest point at P, then
v0
–x2
(a) (b) (c) (d)
–x1
C
B –x0 (x = 0)
x
KE at P = KE at B Height at P = Height at A Total energy at P = Total energy at A Time of travel from A to B = Time of travel from B to P.
y 14. The magnetic field in a region is given by B = B0 x k . d a d A square loop of side d is x placed with its edges along the x and y axis. The loop is moved with a constant velocity v = v0 i. The emf induced in the loop is B v d2 (a) B0v0d (b) 0 0 2a 3 B v d2 Bv d (c) 0 0 (d) 0 0 a a2 15. The length of a potentiometer wire is l. A cell of emf e is balanced at a length l/3 from the positive end of the wire. If the length of the wire is increased by l/2. The distance at which the same cell give a balance point is l 2l l 4l (a) (b) (c) (d) 6 3 2 3 16. Particles of masses 2M, m and M are respectively at points A, B and C with AB = 1/2(BC). m is much-much smaller than M and at time t = 0, they are all at rest. At subsequent times before any B C A collision takes place m (a) m will remain at rest. 2M M (b) m will move towards M. (c) m will move towards 2M. (d) m will have oscillatory motion.
17. Two plane mirrors are A inclined at angle q as shown in figure. If a ray P parallel to OB strikes the other mirror at P and O finally emerges parallel to OA after two reflections then q is equal to (a) 90° (b) 60° (c) 45° (d) 30° Physics For you | April ‘17
B
55
19. Figure shows two coherent sources S1 and S2 emitting wavelength l. The separation S1S2 = l.5 l p and S1 is ahead in phase by relative to S2. Then 2 the maxima occur in direction q given by sin–1 of 1 S1 (i) 0 (ii) 2 1 5 S2 (iii) − (iv) − 6 6 Correct options is/are (a) (i), (iii) and (iv) (b) (i), (ii) and (iii) (c) (ii) only (d) All the above 20. Three photo diodes D1, D2 and D3 are made of semiconductors having band gaps of 2.5 eV, 2 eV and 3 eV, respectively. Which ones will be able to detect light of wavelength 6000 Å? (a) D3 (b) D2 (c) D1 (d) None of these 21. A cricket ball of mass 150 g has an initial ^ ^ velocity u = (3 i + 4 j) m s −1 and a final velocity ^ ^ v = − (3 i + 4 j ) m s −1 after being hit. The change in momentum (in kg m s–1) is ^ ^ (a) zero (b) −(0.45 i + 0.6 j) ^ ^ ^ ^ (c) −(0.9 i + 1.2 j) (d) −5(i + j) 22. In a YDSE experiment if a slab whose refractive index can be varied is placed in front of one of the slits, then the variation of resultant intensity at mid-point of screen with ‘m’ will be best represented by (m ≥ 1) [Assume slits of equal width and there is no absorption by the slab] (a)
I0
(b) =1
(c)
56
I0
(d) =1
I0
Physics For you | April ‘17
=1
=1
I0
23. A train moves towards a stationary observer with speed 34 m s–1. The train sounds a whistle and its frequency registered by the observer is u1. If the train’s speed is reduced to 17 m s–1, the frequency registered is u2. If the speed of sound is 340 m s–1, then the ratio u1 / u2 is (a) 18/19 (b) 1/2 (c) 2 (d) 19/18 24. The graph between two temperature scales A and B is shown in figure. Between upper fixed point and lower fixed point there are 150 equal division on scale A and 100 on scale B. The relationship for conversion between the two scales is given by TA − 180 TB = (a) 100 150 TA − 30 TB = (b) 150 100 TB − 180 TA = (c) 150 100 TB − 40 TA B = (d) 100 180 25. RMS speed of a monatomic gas is increased by 2 times. If the process is done adiabatically then the ratio of initial volume to final volume will be (a) 4 (b) 42/3 (c) 23/2 (d) 8 A
18. The SHM of a particle is given by p x(t ) = 5 cos 2 pt + . Calculate the displacement 4 and the magnitude of acceleration of the particle at t = 1.5 s. (a) –3.0 m, 100 m s–2 (b) +2.54 m, 200 m s–2 (c) –3.54 m, 140 m s–2 (d) +3.55 m, 120 m s–2
26. The dimensions of length are expressed as Gx cy hz, where G, c and h are the universal gravitational constant, speed of light and Planck’s constant respectively, then (a) x = (1/2), y = (1/2), z = (5/2) (b) x = (1/2), y = (5/2), z = (1/2) (c) x = (1/2), y = (–3/2), z = (1/2) (d) x = (5/2), y = (1/2), z = (3/2) 27. Three identical rods, each of length l, are joined to form a rigid equilateral triangle. Its radius of gyration about an axis passing through a corner and perpendicular to the plane of the triangle is l l l 3 (a) (b) (c) (d) l 2 3 2 2 28. A concave lens forms the image of an object such that the distance between the object and image is 10 cm and the magnification produced is 1/4. The focal length of the lens will be (a) 8.6 cm (b) 6.2 cm (c) 10 cm (d) 4.4 cm
29. The magnetic field at the centre of an equilateral triangular loop of side 2L and carrying a current I is 9m 0 I 3 3m0 I (a) (b) 4 pL 4 pL 2 3m0 I 3m0 I (d) pL 4 pL 30. To get an output 1 from the circuit shown in the figure, the input must be (c)
A B Y C
(a) A = 0, B = 1, C = 0 (c) A = 1, B = 0, C = 1
(b) A = 1, B = 0, C = 0 (d) A = 1, B = 1, C = 0
31. A current carrying circular loop of radius R is placed in the x-y plane with centre at the origin. Half of the loop with x > 0 is now bent so that it now lies in the y-z plane. (a) The magnitude of magnetic moment diminishes. (b) The magnetic moment does not change. (c) The magnitude of B at (0, 0, z), z >>R increases. (d) The magnitude of B at (0, 0, z), z >> R is unchanged. 32. The wires A and B shown in the A figure are made of the same material, and have radii rA and rB respectively. m The block between them has a mass m. When the force F is mg/3, one of the B wires breaks. F (a) A will break before B if rA = rB (b) A will break before B if rA > 2rB. (c) Either A or B may break if rA ≠ 2rB. (d) The lengths of A and B must be known to predict which wire will break. 33. A concave mirror is placed on horizontal table with its axis directed vertically upward. Let O be the pole of the mirror, C its center of curvature and F is the focus. A point object is placed at C. It has a real image, also located at C. If the mirror is now filled with water, the image will be (a) real and will remain at C. (b) virtual and located at a point between C and ∞. (c) real and located at a point between C and O. (d) real and located at a point between C and F.
34. An electron (mass m) with an initial velocity ^ ^ v = v0 i (v0 > 0) is in an electric field E = − E0 i (E0 = constant > 0). It’s de Broglie wavelength at time t is given by (l0 = initial de Broglie wavelength of electron) l0 eE t (a) (b) l0 1 + 0 eE0t mv0 1 + mv (c) l0
0
(d) l0t
35. A column of air and tuning fork produce 4 beats per second when sounded together. The tuning fork gives the lower note. The temperature of air is 15°C. When the temperature falls to 10°C, the two produce 3 beats per second. The frequency of the fork is (a) 110 Hz (b) 160 Hz (c) 190 Hz (d) 90 Hz 36. Mass m1 moves on a slope making an angle q with the horizontal and is attached to mass m2 by a string passing over a frictionless pulley as shown in figure. The co-efficient of friction between m1 and the sloping surface is m.
m1
m2
Which of the following statement is true? (a) If m2 > m1sinθ , the body will move up the plane. (b) If m2 > m1 (sinθ + mcosθ), the body will move up the plane. (c) If m2 < m1 (sinθ + mcosθ), the body will move up the plane. (d) If m2 > m1 (sinθ – mcosθ), the body will move down the plane. 37. A body with mass 5 kg is acted upon by a force ^ ^ F = (−3 i + 4 j) N. If its initial velocity at t = 0 is ^ ^ u = (6 i − 12 j) m s–1, the time at which it will just have a velocity along the y-axis is (a) Not possible (b) 10 s (c) 2 s (d) 15 s 38. A thin film of refractive index 1.5 and thickness 4 × 10–5 cm is illuminated by light normal to the surface. What wavelength within the visible spectrum will intensified in the reflected beam? (a) 4800 Å (b) 5800 Å (c) 6000 Å (d) 6800 Å Physics For you | April ‘17
57
39. Two hydrogen atoms in the ground state collide inelastically. The maximum amount by which their combined kinetic energy is reduced is (a) 10.2 eV (b) 20.4 eV (c) 13.6 eV (d) 27.2 eV 40. The deuteron is bound by nuclear forces just as H-atom is made up of p and e bound by electrostatic forces. If we consider the force between neutron and proton in deuteron as given in the form of a coulomb potential but with an effective charge e′ 1 e′2 V= 4 pe0 r estimate the value of (e′/e) given that the binding energy of a deuteron is 2.2 MeV (a) 3.64 (b) 4.05 (c) 3.05 (d) 1 41. A body of mass 0.5 kg travels in a straight line with velocity v = kx3/2 where k = 5 m–1/2 s–1. The work done by the net force during its displacement from x = 0 to x = 2 m is (a) 1.5 J (b) 50 J (c) 10 J (d) 100 J 42. A car is moving in a circular horizontal track of radius 10 m with a constant speed of 10 m s–1. A plumb bob is suspended from the roof of the car by a light rigid rod of length 1 m. The angle made by the rod with the vertical is (a) zero (b) 30° (c) 45° (d) 60° 43. Two coherent point sources S1 and S2 vibrating in phase emit light of wavelength l. The separation between the sources is 2l. The smallest distance from S2 on the line passing through S2 and perpendicular to S1S2, where a minimum of intensity occurs is l 15 l 7l 3l (a) (b) (c) (d) 2 12 4 4 44. Consider two containers A and B containing identical gases at the same pressure, volume and temperature. The gas in container A is compressed to half of its original volume isothermally while the gas in container B is compressed to half of its original value adiabatically. The ratio of final pressure of gas in B to that of gas in A is g −1 1 g–1 (a) 2 (b) 2 1 (c) 1 − g
2
1 (d) g + 1
Physics For you | April ‘17
4(2 + b) 3(3 + b)
(b)
4(3 + b) 3(3 + b) (d) 3(2 + b) 4(2 + b) 46. Three sinusoidal waves of the same frequency travel along a string in the positive x-direction. Their amplitudes are y, y/2 and y/3 and their phase constants are 0, p/2 and p respectively. The amplitude of the resultant wave is (a) 0.63y (b) 0.72y (c) 0.83y (d) 0.52y (c)
47. The plates of a parallel plate capacitor are charged upto 100 V. A 2 mm thick insulator sheet is inserted between the plates. Then to maintain the same potential difference the distance between the capacitor plates is increased by 1.6 mm. The dielectric constant of the insulator is (a) 6 (b) 8 (c) 5 (d) 4 48. A large drop of oil, whose density is less than that of water, floats up through a column of water. Assume that the oil and the water do not mix. The coefficient of viscosity of the oil is h0 and that of water is hw. The velocity of the drop will depend on (a) both h0 and hw (b) h0 only (c) hw only (d) neither h0 nor hw 49. A battery of emf 8 V with internal resistance 0.5 W is being charged by a 120 V dc supply using a series resistance of 15.5 W. The terminal voltage of the battery is (a) 20.5 V (b) 15.5 V (c) 11.5 V (d) 2.5 V 50. I-V characteristics of four devices are shown in figure. (i)
(ii)
I
I V
V
(iii) I
(iv)
2
45. The density of a non-uniform rod of length 1 m is given by r(x) = a(1 + bx2) where a and b are constants and 0 ≤ x ≤ 1. The centre of mass of the rod will be at 58
3(2 + b) 4(3 + b)
(a)
V
I V
Identify devices that can be used for modulation (a) (i) and (iii) (b) Only (iii) (c) (ii) and some regions of (iv) (d) All the devices can be used
solutions
1. (c) : V =
GM 3
(3a2 − r 2 ) inside and −
GM outside r
2a the sphere. 2. (b) : The electrical potential energy of a protonelectron system will decrease as the separation Q Q between the two is decreased as U = 1 2 4pe0r Q1 = charge on proton, Q2 = charge on electron = –Q1 3. (c) :
∆Q × 100 = Q
∆z ∆y 1 ∆t 2 ∆x 5 x + 3 z + y + 2 t × 100
2 = × 2.5% + (3 × 2%) + (0.5%) + (0.5%) 5 = 7 + 1 = 8%. 4. (b) : Speed of aircraft, 5 v = 720 km h–1 = 720 × = 200 m s–1 18 angle of banking, q = 15°, radius of the loop, r = ? v2 v2 As, tanq = ⇒ r= rg g tan q \ r=
(200)2 40000 = = 15 km 10 × tan 15° 10 × 0.2679
5. (b) : This is basically an RC circuit, charging from a capacitor. The resistance R of the voltmeter is the resistance in the circuit. The voltage across R = current × R = reading of the voltmeter V. Thus, the nature of the V-t curve is the same as the nature of the I-t curve. 6. (c) : Let L be the length of escalator. L Velocity of girl w.r.t. stationary escalator, v ge = t1 L Velocity of escalator, ve = t2 \ Required velocity of girl w.r.t. ground would be
\
1 1 v g = v ge + ve = L + t1 t2 The desired time is tt L L t= = = 12 vg 1 1 t +t L + 1 2 t t 1
2
7. (c) : In figure, AB and AC are two smooth planes inclined to the horizontal at ∠q1 and ∠q2 respectively.
As height of both the planes is the same, therefore, both the stones will reach the bottom with same speed. According to law of conservation of mechanical energy, PE at the top = KE at the bottom 1 1 \ mgh = mv12 … (i), mgh = mv22 … (ii) 2 2 From eqns. (i) and (ii), we get v1 = v2 As is clear from figure, acceleration of the two stones are a1 = gsinq1 and a2 = gsinq2 respectively. As q2 > q1 \ a2 > a1 v From v = u + at = 0 + at or t = a 1 As t ∝ , and a2 > a1 \ t2 < t1 a Hence, stone II will take lesser time and reach the bottom earlier than stone I. 8. (b) : Time constant, t = CR For first case t1= (C1 + C2)(R1 + R2) = (4 + 2)(1 + 2)ms = 18 ms CC RR 8 For second case t 2 = 1 2 1 2 = ms C1 + C2 R1 + R2 9 For third case
t 3 = (C1
R1R2 6 × 2 = = 4 ms 3 R1 + R2
+ C2 )
9. (b) 10. (c) : (a) At time t = 0, particle is at A, and graph around A is straight line. So particle is at rest initially. (b) Slope of the graph represent velocity. Slope of graph at point D is more than that of at point E, hence speed at D is more than that at E. (c) Point C represents a point of inflexion, hence the velocity and the acceleration vanish. (d) Average velocity for the motion between A and D is negative because net displacement (0 – xA) is negative. Hence, it is incorrect. 11. (c) GMm 12. (c) : Kinetic energy = 2r GMm Potential energy = − r Physics For you | April ‘17
59
GMm and the total energy, E = − 2r 1 Kinetic energy is always positive and KE ∝ r 1 Potential energy is negative and | PE |∝ r Similarly total energy is also negative and |E| < |PE| 13. (c) : (a) Let m = mass of particle v = speed of the particle at point B 1 1 KE at B = mv 2 , KE at P = mv02 cos2 q 2 2 Here, v > v0 \ (KE at B) > (KE at P) (b) The motion of the particle is a constrained motion along ABC but after point C to point P it moves freely under gravity. Hence, Height at P ≠ Height at A. (c) There is no external force on the system hence its total energy is conserved. Hence, total energy at P = total energy at A. (d) Time of travel from A to B ≠ time of travel from B to P. Reason is given in (b) 14. (d) : Total flux linked with the coil at time t = 0 is d
d
B x B d3 φi = ∫ dφ = ∫ 0 ⋅ (d )(dx ) = 0 2a a 0 0
and flux linked with the coil at time t = 1 s is φf =
v0 + d
∫
dφ =
v0
v0 + d
∫
v0
B0 x ⋅ (d )(dx ) a
y
z d
d B0d 2 x [d + 2v0d] dx 2a Now, induced emf = change in flux in unit time B v d2 \ e = φ f − φi = 0 0 a 15. (b) 16. (c) 17. (b) : ∠OPQ =∠OQP = q
=
x
A
P
O
Q
B
or 3q = 180° \ q = 60° 18. (c) : The given equation of SHM motion is p x(t ) = 5 cos 2 pt + 4 Compare the given equation with standard equation of SHM x(t ) = A cos (wt + φ ) 60
Physics For you | April ‘17
We get, w = 2p s–1 At t = 1.5 s
p (i) Displacement, x(t ) = 5 cos 2 p × 1.5 + 4 p = − 5 cos = – 3.54 m 4 (ii) Acceleration, a = – w2 × displacement = – (2p s–1)2 × (–3.54 m) ≈ 140 m s–2 19. (a) 20. (b) 21. (c) 22. (c) 23. (d) : According to Doppler’s effect, u1 = u
v (v − v S )
or
u1 = u
or
u2 = u
, (Observer is stationary) 340
=
u × 10
(340 − 34) 9 v , (observer is stationary) Again, u2 = u (v − v S′ )
\
340
=
u × 20
340 − 17 19 u 19 u × 10 19 = × or 1 = . u2 9 u × 20 u2 18 u1
24. (b) : From the graph For the scale A, Lower fixed point = 30°C Upper fixed point = 180°C, For the scale B, Lower fixed point = 0°C, Upper fixed point = 100°C \ The relationship between the two scales A and B is given by TA − 30 TB − 0 TB = = 150 100 100 25. (d) : vrms ∝ T vrms is increased by 2 times, i.e., its temperature is increased by 4 times. In adiabatic process, TV g−1 = constant 1
\
5 Vi T f g −1 (For monatomic gas g = ) = V T 3 f i
\
Vi = (4)3/2 = 8 Vf
26. (c) : Let L = Gx cy hz [M0L1T0] = [M–1L3T–2]x [LT–1]y [ML2T–1]z or \ \
[M0L1T0] = M–x + z L3x + y + 2z T–2x – y – z – x + z = 0, 3x + y + 2z = 1, – 2x – y – z = 0, 1 −3 1 x = , y = , z = 2 2 2
27. (c) : I = I AB + I AC + I BC ml 2 ml 2 = + + (I D + mh2 ) 3 3
Clearly, focal length of the new optical system is less than the original. So, the object is effectively at a distance greater than twice the focal length. So, the real image will be formed between F and 2F. 34. (a) : Here, E = − E0 i ; Initial velocity, v = v0 i Force acting on electron due to electric field F = (− e)(− E i) = eE i
2 ml 2 = ml 2 + + m(l sin 60)2 3 12 2 1 3 3 = ml 2 + + = ml 2 3 12 4 2 \
0
3 2 l ml = (3m)k 2 or k = 2 2
28. (d) 29. (a) : Magnetic field at O is m I B = 3 0 (sin 60° + sin 60°) 4 pr
r
I
9m I m = 3 0 (I ) 3 ( 3 ) = 0 4p L 4 pL
30°
L 3
I
O
60° 60° r
L
I
30. (c) 31. (a) : For a circular loop of radius R carrying current I placed in x-y plane, the magnetic moment M = I × pR2. It acts perpendicular to the loop i.e., along z-direction. When half of the current loop is bent in y-z plane, then magnetic moment due to half current loop in x-y plane, M1 = I(pR2/2) acting along z-direction. Magnetic moment due to half current loop in y-z plane, M2 = I(pR2/2) along x-direction. Effective magnetic moment due to entire bent current loop, M ′ =
M12
+ M22
= (I pR / 2) + (I pR / 2) = 2
2
2
2
I pR2 2
35. (a) : Let frequency of the tuning fork = u Frequency of air column at 15°C, u15 = u + 4 Frequency of air column at 10°C, u10 = u + 3 As v = ul \ u15 = (u + 4)l and u10 = (u + 3)l u Hence 15 = u + 4 ...(i) u10 u + 3 Also
1 1 1 1 = + + F f water lens f concave mirror f water lens or
1 1 1 =2× + F f water lens f concave mirror
1 1 2 = 2(m − 1) + F R R 1 2m R or = ⇒ F= F R 2m
v15 273 + 15 288 = = v10 273 + 10 283
From eqns. (i) and (ii), we have 1/2
u + 4 288 = u + 3 283
i.e., Magnetic moment diminishes.
32. (a) 33. (d) : When the mirror is filled with water, then the equivalent focal length F is given by
0
Acceleration produced in the electron, F eE a = = 0 i m m Now, velocity of electron after time t, eE0t eE t vt = v + a t = v0 + 0 i or | vt | = v0 + m m h h h = = Now, lt = eE0t mvt eE0t m v0 + mv0 1 + m mv0 l0 \ lt = h eE0t l0 = mv + 1 mv 0 0
or or 36. (b) :
...(ii)
1/2
5 = 1 + 283
1 5 1+ × 2 283
(u + 3) + 1 5 1 5 =1+ ⇒ = 566 u+3 (u + 3) 566 u + 3 = 113.2 ⇒ u= 110 Hz
C
or
O
Normal by inclined plane on m1 (at equilibrium) N1 = m1gcosq, Friction force, f = mN1 ⇒f = mm1gcosq. Physics For you | April ‘17
61
37. 38.
39.
40. 42.
Since, friction opposes the relative motion, so when block is moving up f will be down along the plane and when it is moving downward f will be up along the plane. If mass m1 is moving up along the inclined plane then, m2g > (m1gsinq + mm1gcosq) ⇒ m2 > m1(sinq + mcosq) (b) (a) : Condition for observing bright fringe is 2 md 12 × 10−7 1 2 md = m + l \ l = = 1 1 2 m+ m + 2 2 The integer m that gives the wavelength in the visible region (4000 Å - 7000 Å) is m = 2. In that case, 12 × 10−7 l= = 4.8 × 10−7 = 4800 Å 1 2+ 2 (a) : In inelastic collision, Reduction in KE = KE before collision – KE after collision Total KE of both Hydrogen atoms before collision = 2 × 13.6 = 27.2 eV For maximum reduction in KE, one H atom goes over to first excited state (n1 = 2) and other remains in ground state(n2 = 1) then their combined KE after collision is 13.6 13.6 = 2 + 2 = 3.4 + 13.6 = 17 eV (2) (1) Hence, reduction in KE = 27.2 – 17 = 10.2 eV. (a) 41. (b) (c) : The radial acceleration is ar 2 2 −2 v 2 10 m s = = = 10 m s −2 10 m r
Tcos
T Tsin
If T = tension in the rod then ar Tcosq = mg, Tsinq = mar, mg a \ tan q = r = 1 or q = 45° g 43. (a) 44. (a) : The gas in container A is compressed isothermally, \ P1V1 = P2V2 PV V or P2 = 1 1 = P1 1 = 2P1 ( V2 = V1 / 2) V2 V1 / 2 Again the gas in container B is compressed adiabatically, \ P1V1g = P2′ (V2′ )g 62
Physics For you | April ‘17
P2' = P1
Hence
V1g
(V2')g
g
V = P1 1 = 2 g P1 V1 / 2
P2' 2 g P1 = = 2 g −1 P2 2 P1
45. (a) : Mass of a small element of length dx of the rod at a distance x from the one end of the rod is dm = rdx = a(1 + bx2)dx The centre of mass of the rod is 1
1
2 ∫ x dm ∫ xa(1 + bx ) dx 3(2 + b) XCM = 01 = 01 = 4(3 + b) 2 ∫ dm ∫ a(1 + bx ) dx 0
0
46. (c) q C V and q are same. Therefore, capacity should not change i.e., Ci = Ct e A e0 A t or 0 = ⇒ 1. 6 − t + = 0 t d k d + 1. 6 − t + k 2 Here t = 2 mm, \ 2 − 1.6 = ⇒ k=5 k 48. (c) 49. (c) : When the battery of emf 8 V is charged from a dc supply of 120 V, the effective emf in the circuit is e = 120 V – 8 V = 112 V Total resistance of the circuit = 15.5 W + 0.5 W = 16 W \ Current in the circuit during charging 112 V e I= = = 7A R + r 16 W \ Terminal voltage of the battery, V = Emf of the battery + Voltage drop across battery = 8 + Ir = 8 V + (7 A) (0.5 W) = 11.5 V 50. (c) : Here, (i) represents ideal modulation related to triode valve for linear plate modulation. (ii) represents square law diode modulation. (iii) represents constant current source. (iv) represents part of it as square law modulation. Thus, for modulation, (ii) and some regions of (iv) can be used. 47. (c) : As, V =
{
}
solution set-44
1. If N0 is the initial level of radiation and N is the N permissible level, then N = 0 50 We know that, N = N0e–lt ...(i) 0.693 0.693 −1 = = 0.0693 day where l = t1/2 10 From equation (i), we have N0 = N 0e −0.0693t or t = 56.45 days 50 2. The angle of incidence on the inclined face of the prism is 30°. 30° r A i If r is the angle of α 30° refraction, then by Snell’s law 2α sin r 120° F = 2 o sin 30 1 or sin r = α 2 \ r = 45° Angle a = r – i = 45° – 30° = 15° The required angle = 2a = 2 × 15° = 30° 3. The first image (I1) is formed at the focus of the lens. This image will work as the object for the plane mirror and second image (I2) is formed behind the mirror. Now, this image will work as object for the lens. For the third case, 1 1 1 − = I1 I2 v u f 20 cm 1 1 1 − =+ v −30 −10 10 cm 1 −1 1 4 ⇒ v = –7.5 cm ⇒ = − =− v 10 30 30 The final image distance = 2.5 cm in front of the mirror
α
1
4.
Differentiating w.r.t. x, −2u u2 u2 + 2uf − \ dx = + d u = du u + f (u + f )2 (u + f )2 f 2 − (u + f )2 + f 2 dx = d u = − 1 du 2 u + f (u + f ) du = a <<|u| = |3f| 2 f 3 dx = − 1 a = − a 4 −3 f + f 3 a dv = du + dx = a − a = 4 4 a 2 dv = \ Required distance = 2 5. Here, energy loss during the collision is used to excite the atom or ion. According to quantum mechanics, loss in energy for H atom: (0, 10.2 eV, 12.09 eV, ...., 13.06 eV) For He+ ion: (0, 40.8 eV, 48.36 eV, ..., 54.4 eV) To excite the hydrogen atom and He+ ion to the first excited state, minimum energy required = 40.8 eV + 10.2 eV = 51 eV Now, according to Newtonion mechanics, minimum loss is 0 (for elastic collision). Maximum loss will be there in perfectly inelastic collision. Now, let mass of H atom be m, then mass of He+ ion = 4m, Initial kinetic energy = K Let v0 be the initial speed of H atom and vf the final common speed. According to momentum v conservation: mv0 = 4mv f + mv f ⇒ v f = 0 5 v02 1 1 2 K 1 Final kinetic energy = (5m ) = mv = 2 25 5 2 0 5 Now, for minimum value of K so that the electron excite to the first excited state of H atom and He+ ion. K 4K \ Change in kinetic energy = K − = 5 5 51 × 5 4K = 51 eV or K = eV or K = 63.75 eV 5 4 6. Let the lengths of the three A rods be l1, l2, and l3. From cosine law, we have
...(i) (putting in (i))
−u2 1 1 1 1 1 1 u+ f − = ⇒ = + = ;x= u+x u f u+x u f uf u+ f
or
α2
1 1 1 − = v u f v – u = x; v = u + x
l2 + l2 − l2 cos q = 1 2 3 2l1l2
θ
2l1l2 cos q = l12 + l22 − l32
B
Differentiating eqn. (i), we get 2l1cosqdl2 + 2l2cosqdl1 – 2l1l2sinqdq = 2l1dl1 + 2l2dl2 – 2l3dl3 Physics For you | April ‘17
α1
C
...(i)
...(ii) 63
The changes in length of respective rods due to temperature rise are dl1 = l1a1DT, dl2 = l2a1DT, dl3 = l3a2DT ...(iii) For an equilateral triangle l1 = l2 = l3 = l and q = 60°. So from eqns. (ii) and (iii) we get 2l2 cos60°a1DT + 2l2 cos60°a1DT – 2l2 sin60°dq = 2l2a1DT + 2l2a1DT – 2l2a2DT 3 a a dq = a1DT + a1DT - a2DT or 1 DT + 1 DT − 2 2 2 3 dq = 2a1DT − a 2 DT 2 2(a2 − a1)DT 3 or (a2 − a1)DT = dq or dq = 3 2 7. We consider the equilibrium behavior of a disc of fluid thickness dy, experiencing a pressure difference arising from the fluid. (P + dP)A + rgAdy = PA, dP= –rgdy, dP = −rg ...(i) dy rRT From ideal gas equation, we have P = ...(ii) M We eliminate r from eqns. (i) and (ii), dP Mg P dP Mg dy Mg dy =− , =− =− dy R T P R T R T0 (1 − cy ) or a1DT −
Integrating the above equation, we obtain Mg lnP = ln (1 − cy ) + C RcT0 Initially y = 0, P = P0 Mg \ ln P − ln P0 = ln (1 − cy ) RT0c Mg P n ln = ln (1 − cy ) ; where n = RT0c P0
P = P0(1 – cy)n = P0(1 – cy)Mg/cRT0 8. Let A be the area of cross-section of the tube. Consider a differential x dx volume Adx of the capillary at a distance x from left end. The temperature at this cross-section is x=0 x=L TL − T0 x T = T0 + L Volume of the capillary V = AL, is constant. Applying ideal gas equation to this differential volume, we have T −T P(Adx) = dnRT = dnR T0 + L 0 x L 64
Physics For you | April ‘17
Rearranging the expression, we have L n R dx =∫ dn ∫0 0 T −T PA T0 + L 0 x L L
L nR T −T ln T0 + L 0 x = L 0 PA TL − T0 T PAL 1 TL PV n= ln = ln L TL − T0 R T0 (TL − T0 ) R T0 9. Let the temperature at B be T. Total thermal 3L resistance from F to B or E to B is 2KA respectively. 2KA Heat per second flowing to B from F = (T − T ) 3L 3 2KA Heat per second flowing to B from E = (T2 − T ) 3L KA Heat travelling from B to A per second = (T − T1 ) L In steady state, from conservation of energy, net heat arriving at B must be equal to net heat flowing out from B; thus we have 2KA(T3 − T ) 2KA(T2 − T ) KA(T − T1 ) + = 3L 3L L which on solving for T we get 2T + 2T2 + 3T1 T= 3 7 10. Let heat transferred per second is Q, then dq dT a dT Q= = −K 2πrl or, Q = 2πl dt dr r dr
R2
∫R1 r dr =
2πla T2 dT Q ∫T1
4πla(T2 − T1) ....(i) (R22 − R12 ) 2πla = (T2 − T1) ; Q = 2 Q (R22 − R12 )
(
)
R 2 − R12 Q 2πla T dT or T − T = 1 ∫R1 Q ∫T1 4πla ....(ii) (R is the distance from the axis of the cylinder) R
r dr =
From eqns. (i) and (ii), T = T1 +
(R2 − R12 )(T2 − T1) R22 − R12
Class XI
T
his specially designed column enables students to self analyse their extent of understanding of complete syllabus. Give yourself four marks for correct answer and deduct one mark for wrong answer. Self check table given at the end will help you to check your readiness.
Total Marks : 120
Time Taken : 60 min
NEET / AiiMs / PMTs Only One Option Correct Type
1. An open pipe is in resonance in its 2nd harmonic with a tuning fork of frequency u1. Now it is closed at one end. If the frequency of tuning fork is increased slowly from u1 then again a resonance is obtained with a frequency u2. If in this case the pipe vibrates in nth harmonic, then 3 5 (a) n = 5, u2 = u1 (b) n = 5, u2 = u1 4 4 3 9 (c) n = 3, u2 = u1 (d) n = 3, u2 = u1 4 4 2. The radii of the two columns of U-tube are r1 and r2. When a liquid of density r (angle of contact is 0°) is filled in it, the difference of liquid level in two arms is h. The surface tension of liquid is rghr1r2 rgh(r2 − r1 ) (a) (b) 2(r2 − r1) 2r1r2 2(r2 − r1 ) rgh (c) (d) rghr1r2 2(r2 − r1 ) 3. An engine has an efficiency of 1/6. When the temperature of sink is reduced by 62 °K, its efficiency is doubled. Temperature of the source is (a) 37°C (b) 62°C (c) 124°C (d) 99°C 4. Two capillaries of radii r1 and r2 and lengths l1 and l2 respectively are in series. A liquid of viscosity h is flowing through the combination under a pressure difference P. What is the rate of volume flow of liquid ? pP l1 l2 + (a) 8h r14 r24
−1
(b)
8pP l1 l2 + h r14 r24
−1
−1
pP r14 r24 8pP l1 l2 + (c) (d) + 8h l1 l2 h r14 r24 5. n drops of a liquid, each with surface energy E, join to form a single drop. (a) Some energy will be absorbed in the process. (b) The energy released will be E(n – n2/3). (c) The energy released will be nE(22/3 – 1). (d) None of these 6. A bag of mass M is suspended by a rope. A bullet of mass m is fired at it with speed v and gets embedded in it. The loss of energy of the system is (a)
Mmv 2 2( M + m)
(b)
Mv 2 2( M + m)
m2 v 2 (d) 1 ( M + m) v 2 2( M + m) 2 7. A point P lies on the axis of a ring of mass M and radius r at a distance r from its centre C. A small particle starts from P and reaches C under gravitational attraction. Its speed at C will be : (c)
(a)
2GM r
(b) zero
(
)
2GM 2GM 1 2 −1 1 − (d) r r 2 8. A machine gun fires 360 bullets per minute, with a velocity of 600 m s–1. If the power of the gun is 5.4 kW, mass of each bullet is (a) 5 kg (b) 0.5 kg (c) 5 g (d) 0.5 g (c)
Physics For you | April ‘17
65
9. For the same cross-sectional area and for a given load, the ratio of depressions for the beam of a circular cross-section and square cross-section is (a) 1 : p (b) p : 1 (c) p : 3 (d) 3 : p 10. A particle starts from rest, whose acceleration versus displacement graph is shown in the figure. The velocity of the particle at t = 15 s is
14. Assertion : When a solid sphere and a solid cylinder are allowed to roll down an inclined plane, the sphere will reach the ground first even if the mass and radius of the two bodies are different. Reason : The acceleration of body rolling down the inclined plane is directly proportional to the radius of the rolling body. 15. Assertion : Banking of road reduces the wear and tear of tyres of vehicles taking turn on curved roads. Reason : Component of normal reaction of ground provides necessary centripetal force to the tyres passing over a banked curved road which is otherwise to be provided by frictional force. JEE MAiN / JEE AdvANcEd / PETs Only One Option Correct Type
–1
–1
(a) 5 m s (b) 5 3 m s 5 (c) m s–1 (d) zero 2 11. v, T, r and l denote speed, surface tension, mass density and wavelength, respectively. In an experiment, v depends on T, r and l respectively. The value of v is proportional to l T T T (a) (b) (c) (d) rT l rl rl 12. A disc of mass m and radius R has a concentric hole of radius r. Its moment of inertia about an axis through its centre and perpendicular to its plane is 1 1 (a) m(R – r)2 (b) m(R2 – r2) 2 2 1 1 (c) m(R + r)2 (d) m(R2 + r2) 2 2 Assertion & Reason Type
Directions : In the following questions, a statement of assertion is followed by a statement of reason. Mark the correct choice as : (a) If both assertion and reason are true and reason is the correct explanation of assertion. (b) If both assertion and reason are true but reason is not the correct explanation of assertion. (c) If assertion is true but reason is false. (d) If both assertion and reason are false. 13. Assertion : The first derivative of a vector of constant magnitude (either zero or a non-zero) is perpendicular to the vector itself. Reason : Scalar product of two vectors obeys commutative law. 66
Physics For you | April ‘17
16. Two uniform strings A and B made of steel are made to vibrate under the same tension. If the first overtone of A is equal to the second overtone of B and if the radius of A is twice that of B, the ratio of the lengths of the strings is (a) 2 : 1 (b) 3 : 2 (c) 3 : 4 (d) 1 : 3 17. Three discs A, B and C having radii 2, 4 and 6 cm respectively are coated with carbon black. Wavelengths for maximum intensity for the three discs are 300, 400 and 500 nm respectively. If QA, QB and QC are powers emitted by A, B and C respectively, then (a) QA will be maximum (b) QB will be maximum (c) QC will be maximum (d) QA = QB = QC 18. A loop of 6.28 cm long thread is put gently on a soap film in a wire loop. The film is pricked with a needle inside the soap film enclosed by the thread. If the surface tension of soap solution is 0.030 N m–1, then the tension in the thread is (a) 1 × 10–4 N (b) 2 × 10–4 N –4 (c) 3 × 10 N (d) 4 × 10–4 N 19. Certain amount of an ideal gas is contained in a closed vessel. The vessel is moving with a constant velocity v. The molecular mass of gas is M. The rise in temperature of the gas when the vessel is suddenly stopped is (g = CP/CV) (a)
Mv 2 (g − 1) 2R(g + 1)
(b)
(c)
Mv 2 2R(g + 1)
(d)
Mv 2 (g − 1) 2R
Mv 2 2R(g − 1)
More than One Options Correct Type
Integer Answer Type
20. A double star is a system of two stars of masses m and 2m, rotating about their centre of mass only under their mutual gravitational attraction. If r is the separation between these two stars then their time period of rotation about their centre of mass will be proportional to
24. A man standing on a trolley pushes another identical trolley (both trolleys are at rest on a rough surface), so that they are set in motion and stop after some time. If the ratio of mass of first trolley with man to mass of second trolley is 3, then find the ratio of the stopping distances of the second trolley to that of the first trolley. (Assume coefficient of friction to be the same for both the trolleys)
3
(a) r 2
1
(b) r
(c) m 2
−
1
(d) m 2
21. A student uses a simple pendulum of exactly 1 m length to determine g, the acceleration due to gravity. He uses a stop watch with the least count of 1 sec for this and records 40 seconds for 20 oscillations. For this observation, which of the following statement(s) is (are) true? (a) Error DT in measuring T, the time period, is 0.05 seconds. (b) Error DT in measuring T, the time period, is 1 second. (c) Percentage error in the determination of g is 5%. (d) Percentage error in the determination of g is 2.5%. U 22. The potential energy curve for interaction between two molecules is shown r A B C O in figure. Which of the following statements are true ? D (a) The molecules have maximum attraction for r = OA. (b) The molecules have maximum kinetic energy for r = OB. (c) The intermolecular force is zero for r = OB. (d) For the gaseous state, the depth BD of the potential energy curve is much smaller than KBT.
23. Two blocks A and B of masses 5 kg and 2 kg, respectively, connected by a spring of force constant = 100 N m–1 are placed on an inclined plane of inclination 30° as shown in figure. If the system is released from rest (a) There will be no compression or elongation in the spring if all surfaces are smooth. (b) there will be elongation in the spring if A is rough and B is smooth. (c) Maximum elongation in the spring 35 cm if all surfaces are smooth. (d) There will be elongation in the spring if A is smooth and B is rough.
25. Two particles A and B are projected from point O with equal speeds. They both hit the point P of an inclined plane of inclination 15°. Particle A is projected at an angle 30° with inclined plane. If the ratio of time of flight of particles A and B is 1 : Find the value of n.
n.
26. A cylindrical vessel of height 500 mm has an orifice (small hole) at its bottom. The orifice is initially closed and water is filled in it upto height H. Now the top is completely sealed with a cap and the orifice at the bottom is opened. Some water comes out from the orifice and the water level in the vessel becomes steady with height of water column being 200 mm. Find the fall in height (in mm) of water level due to opening of the orifice. [Take atmospheric pressure = 1.0 × 105 N m–2, density of water = 1000 kg m–3 and g = 10 m s–2. Neglect any effect of surface tension.] Comprehension Type Figure shows a capillary tube of radius r dipped into water. The atmospheric pressure is P0 and the capillary rise of water is h. S is the surface tension for waterglass. A h
Physics For you | April ‘17
67
27. Initially, h = 10 cm. If the capillary tube is now inclined at 45°, the length of water rising in the tube will be (a) 10 cm (b) 10 2 cm (c)
10 2
cm
(d) None of these
(a) (b) (c) (d) 30.
28. Which of the following graphs may represent the relation between the capillary rise h and the radius r of the capillary? h
A R P S Q
r
r h
(c)
(B) Buoyant force on the cube, (Q) 1 Arv2 r = density of liquid 2 h
(d) r
29. A loudspeaker diaphragm 0.2 m in diameter is vibrating at 1 kHz with an amplitude of 0.01 × 10–3 m. Assume that the air molecules in the vicinity have the same amplitude of vibration. Density of air is 1.29 kg m–3. Match the column I with the column II. Take velocity of sound = 340 m s–1. Column-I Column-II (A) Pressure amplitude (P) 2.7 × 10–2 immediately in front of the diaphragm (in N m–2) (B) Sound intensity in front of the diaphragm (in W m–2)
(Q) 2.15 × 10–5
(C) The acoustic power radiated (in W)
(R) 27.55
(D) Intensity at 10 m from the (S) 0.865 loud speaker (in W m–2)
A
h
r
Matrix Match Type
D Q S P P
A
h
(b) h
C P Q Q S
Column I Column II (A) Hydrostatic force on the (P) Arv2 side wall of the cubical vessel, r = density of liquid
h
(a)
B S R R R
(C) Impact (reaction force on (R) rghA the vessel by the liquid coming out of the vessel, r = density of liquid A
v
(D) Aerodynamic force acting on the flat roof surface of (S) area A, r = density of air A
(a) (b) (c) (d)
A R S S P
B Q Q R Q
1 rghA 2
v
C P P P R
D S R Q S
Keys are published in this issue. Search now! J
Check your score! If your score is > 90%
ExcEllEnt work !
You are well prepared to take the challenge of final exam.
No. of questions attempted
……
90-75%
Good work !
You can score good in the final exam.
No. of questions correct
……
74-60%
satisFactory !
You need to score more next time.
Marks scored in percentage
……
< 60%
68
Physics For you | April ‘17
not satisFactory! Revise thoroughly and strengthen your concepts.
7
Time Allowed : 3 hours
Maximum Marks : 70 GENERAL INSTRUCTIONS
(i) All questions are compulsory. There are 26 questions in all. (ii) This question paper has five sections: Section A, Section B, Section C, Section D and Section E. (iii) Section A contains five questions of one mark each, Section B contains five questions of two marks each, Section C contains twelve questions of three marks each, Section D contains one value based question of four marks and Section E contains three questions of five marks each. (iv) There is no overall choice. However, an internal choice has been provided in one question of two marks, one question of three marks and all the three questions of five marks weightage. You have to attempt only one of the choices in such questions.
section-A
1. Does the charge given to a metallic sphere depend on whether it is hollow or solid. Give reason for your answer. 2. A long straight current carrying wire passes normally through the centre of circular loop. If the current through the wire increases, will there be an induced emf in the loop? Justify. 3. At a place, the horizontal component of earth's magnetic field is B and angle of dip is 60°. What is the value of horizontal component of the earth's magnetic field at equator? 4. Name the junction diode whose I-V characteristics are drawn below :
I
V
5. How is the speed of em-waves in vaccum determined by the electric and magnetic fields? section-B
6. How does Ampere-Maxwell law explain the flow of current through a capacitor when it is being charged by a battery? Write the expression for the displacement current in terms of the rate of change of electric flux. 7. Define the distance of closest approach. An a-particle of kinetic energy 'K ' is bombarded on a
thin gold foil. The distance of the closest approach is 'r'. What will be the distance of closest approach for an a-particle of double the kinetic energy? OR Write two important limitations of Rutherford nuclear model of the atom. 8. Find out the wavelength of the electron orbiting in the ground state of hydrogen atom. 9. Define the magnifying power of a compound microscope when the final image is formed at infinity. Why must both the objective and the eyepiece of a compound microscope has short focal lengths? Explain. 10. Which basic mode of communication is used in satellite communication? What type of wave propagation is used in this mode? Write, giving reason, the frequency range used in this mode of propagation. section-c
11. (i) Find the value of the phase difference between the current and the voltage in the series LCR circuit shown below. Which one leads in phase: current or voltage? (ii) Without making any other change, find the value of the additional capacitor C1, to be connected in parallel with the capacitor C, in order to make the power factor of the circuit unity. physics for you | APRIL ‘17
69
(ii) Depict the orientation of the dipole in (a) stable, (b) unstable equilibrium in a uniform electric field. 19. (i) A radioactive nucleus 'A' undergoes a series of decays as given below: 12. Write the two processes that take place in the formation of a p-n junction. Explain with the help of a diagram, the formation of depletion region and barrier potential in a p-n junction. 13. (i) Obtain the expression for the cyclotron frequency. (ii) A deuteron and a proton are accelerated by the cyclotron. Can both be accelerated with the same oscillator frequency? Give reason to justify your answer. 14. (i) How does one explain the emission of electrons from a photosensitive surface with the help of Einstein's photoelectric equation? (ii) The work function of the following metals is given : Na = 2.75 eV, K = 2.3 eV, Mo = 4.17 eV and Ni = 5.15 eV. Which of these metals will not cause photoelectric emission for radiation of wavelength 3300 Å from a laser source placed 1 m away from these metals? What happens if the laser source is brought nearer and placed 50 cm away? V 15. A r e s i s t a n c e o f R draws current from R0 a potentiometer. The A B potentiometer wire AB, has a total resistance of R 0. A volt age V i s R supplied to the potentiometer. Derive an expression for the voltage across R when the sliding contact is in the middle of the potentiometer wire.
16. Define the term 'amplitude modulation'. Explain any two factors which justify the need for modulating a low frequency base-band signal. 17. (i) Find the equivalent capacitance between A and B in the combination given below. Each capacitor is of 2 mF capacitance. A
C1
P
C2
C3 R
S
C4 T
C5
B
(ii) If a dc source of 7 V is connected across AB, how much charge is drawn from the source and what is the energy stored in the network? 18. (i) Derive the expression for electric field at a point on the equatorial line of an electric dipole. 70
physics for you | APRIL ‘17
A
A1
A2
A3
A4
The mass number and atomic number of A2 are 176 and 71 respectively. Determine the mass and atomic numbers of A4 and A. (ii) Write the basic nuclear processes underlying b+ and b– decays. A
20. (i) A ray of light incident on face AB of an equilateral glass prism, shows minimum deviation of 30°. Calculate the speed of light B C through the prism. (ii) Find the angle of incidence at face AB so that the emergent ray grazes along the face AC.
21. For a CE-transistor amplifier, the audio signal voltage across the collector resistance of 2 kW is 2 V. Given the current amplification factor of the transistor is 100, find the input signal voltage and base current, if the base resistance is 1 kW. 22. Describe the working principle of a moving coil galvanometer. Why is it necessary to use (i) a radial magnetic field and (ii) a cylindrical soft iron core in a galvanometer? Write the expression for current sensitivity of the galvanometer. Can a galvanometer as such be used for measuring the current? Explain. OR (a) Define the term 'self-inductance' and write its S.I. unit. (b) Obtain the expression for the mutual inductance of two long co-axial solenoids S1 and S2 wound one over the other, each of length L and radii r1 and r2 and n1 and n2 number of turns per unit length, when a current I is set up in the outer solenoid S2. section-D
23. Mrs. Rashmi Singh broke her reading glasses. When she went to the shopkeeper to order new specs, he suggested that she should get spectacles with plastic lenses instead of glass lenses. On getting the new spectacles, she found that the new ones were thicker than the earlier ones. She asked this question to the shopkeeper but he could not offer satisfactory explanation for this. At home, Mrs. Singh raised the same question to her daughter Anuja who explained why plastic lenses were thicker.
(a) Write two qualities displayed each by Anuja and her mother. (b) How do you explain this fact using lens maker's formula? section-e
24. (a) Draw a labelled diagram of AC generator. Derive the expression for the instantaneous value of the emf induced in the coil. (b) A circular coil of cross-sectional area 200 cm2 and 20 turns is rotated about the vertical diameter with angular speed of 50 rad s –1 in a uniform magnetic field of magnitude 3.0 × 10–2 T. Calculate the maximum value of the current in the coil. OR (a) Draw a lab el le d diag ram of a step-up transformer. Obtain the ratio of secondary to primary voltage in terms of number of turns and currents in the two coils. (b) A power transmission line feeds input power at 2200 V to a step-down transformer with its primary windings having 3000 turns. Find the number of turns in the secondary to get the power output at 220 V. 25. (a) Distinguish between unpolarised light and linearly polarised light. How does one get linearly polarised light with the help of a polaroid? (b) A narrow beam of unpolarised light of intensity I0 is incident on a polaroid P1. The light transmitted by it is then incident on a second polaroid P2 with its pass axis making angle 60° relative to the pass axis of P1. Find the intensity of the light transmitted by P2.
OR (a) Explain two features to distinguish between the interference pattern in Young's double slit experiment with the diffraction pattern obtained due to a single slit. (b) A monochromatic light of wavelength 500 nm is incident normally on a single slit of width 0.2 mm to produce a diffraction pattern. Find the angular width of the central maximum obtained on the screen. Estimate the number of fringes obtained in Young's double slit experiment with fringe width 0.5 mm, which can be accommodated within the region of total angular speed of the central maximum due to single slit. 26. (i) Derive an expression for drift velocity of electrons in a conductor. Hence deduce Ohm's law. (ii) A wire whose cross-sectional area is increasing
linearly from its one end to the other, is connected across a battery of V volts. Which of the following quantities remain constant in the wire? (a) drift speed (b) current density (c) electric current (d) electric field Justify your answer. OR (i) State the two Kirchhoff 's laws. Explain briefly how these rules are justified. (ii) The current is drawn B from a cell of emf E A C and internal resistance r connected to the E, r network of resistors each of resistance r as shown in the figure. Obtain the expression for (a) the current drawn from the cell and (b) the power consumed in the network. solutions
1. No, the charge given to a metallic sphere does not depend on whether it is hollow or solid because all the charges will move to the outer surface of the sphere. Charge will be distributed uniformly over the surface of the sphere. 2. The magnetic lines of force due to current are parallel to the plane of the loop. So angle between magnetic field and area vector is 90°. Hence, the flux linked with the loop is zero. Hence, there will be no induced emf in the loop. 3. Given: BH = B, d = 60°; BH = BT cos d B = BT ⇒ BT = 2B 2 At equator d = 0 \ BH = BT cos d ⇒ BH = 2B eq
eq
4. The junction diode is solar cell.
5. The speed of em-waves in vacuum determined by E the electric (E0) and magnetic fields (B0) is, c = 0 B0 6. According to Ampere-Maxwell law, The total current is the sum of displacement current and the conduction current, i.e.; df i = ie + id = ic + e0 E dt When a capacitor charged through a battery then inside the capacitor plates there is no conduction current, i.e.; ic = 0 and there is only displacement current, so that id = i df The displacement current is, id = e0 E dt physics for you | APRIL ‘17
71
7. The distance from the nucleus, where all kinetic energy of a-particles is completely converted into potential energy is known as the distance of closest approach. 1 2 Ze 2 1 ⋅ r= or r ∝ 4 pe0 K K If kinetic energy will be double, then the distance of closest approach will become half. OR The two important limitations of Rutherford nuclear model of the atom are : (i) This model cannot explain about the stability of matter. (ii) It cannot explain the characteristic line spectra of atoms of different elements. 8. The wavelength of the electron orbiting in the ground state of hydrogen atom is hc 6.62 × 10 −34 × 3 × 108 l= = E −13.6 eV (Q Ground state energy of hydrogen atom = –13.6 eV) =
6.62 × 10 −34 × 3 × 108 −13.6 × 1.6 × 10 −19
−1.986 × 10 −25
= 9.126 × 10–8 m = 912.6 Å 21.76 × 10 −19 9. The magnification of compound microscope when L D the final image is formed at infinity, M = f0 f e =
Both the objective and the eyepiece of a compound microscope has short focal length so as to produce L D large magnifying power as, M = 1+ f0 fe
10. Point to point is the basic mode of communication used in satellite communication. Line-of-sight (LOS) communication is used in satellite communication. Space wave propagation is used in this mode. The frequency range used in space wave mode is between 54 MHz to 4.26 GHz because at such a high frequency both the ground wave and sky wave propagation fail. 11. (i) Given : V = V0 sin(1000t + f), R = 400 W, L = 100 mH, C = 2 mF The standard equation is given as V = V0 sin (wt + f) \ w = 1000 XL = wL \ = 1000 × 100 × 10–3 = 102 = 100 W 1 1 = XC = = 500 W wC 1000 × 2 × 10 −6 72
physics for you | APRIL ‘17
Phase difference between the current and the voltage in the series LCR circuit is given as, X − XL f = tan −1 C R 500 − 100 \ f = tan −1 = tan −1 1 400 f = 45° Since XC > XL is greater, therefore current leads in phase. (ii) To make power factor of the circuit unity, XC = XL C = 2 µF R = 400 Ω L = 100 mH 1 = 100 w(C + C1 ) C1 1 ⇒ = 100 1000(C + C1 ) ~ V = V0 sin (1000 t + φ) 1 or C + C1 = 105 or C1 = 10–5 – C = 10–5 – 0.2 × 10–5 = 0.8 × 10–5 ⇒ C1 = 8 mF 12. Two process that take place in the formation of a p-n junction are diffusion and drift.
When p-n junction is formed, then at the junction free electrons from n-type diffuse over to p-type, thereby filling in the holes in p-type. Due to this a layer of positive charge is built on n-side and a layer of negative charge is built on p-side of the p-n junction. This layer sufficiently grows up within a very short time of the junction being formed, preventing any further movement of charge carriers (i.e., electrons and holes) across the junction. Thus a potential difference V0 of the order of 0.1 to 0.3 V is set up across the p-n junction called potential barrier or junction barrier. The thin region around the junction containing immobile positive and negative charges is known as depletion layer. 13. (i) Expression for the cyclotron frequency: A charged particle produced at point P by a source is accelerated towards dee D1 due to applied electric field, but moves along semi-circular path of radius, mv r= in D1 due to force of magnetic field on it. qB
When it reaches the gap between the two dees, polarities of the dees is changed by oscillator and now the charged particle is accelerated towards D2, where it follows semi-circular path of increased radius with increased velocity.
High frequency oscillator
~
N
D1
D2
W
B S
Energetic proton beam Target
\ Mo and Ni will not cause photoelectric emission. If the laser source is brought nearer and placed 50 cm away, then photoelectric emission will not effect, since it depends upon the work function and threshold frequency. 15. While the slide is in the middle of the potentiometer only half of its resistance (R0/2) will be between the points A and C. Hence, the total resistance between A and C, R1, will be given by the following expression, V 1 1 1 R0 = + B A R1 R (R0 / 2) C
R R0 R R0 + 2R The total resistance between A and C will be sum of resistance between A and C and C and B, i.e., R1 + R0/2 \ The current flowing through the potentiometer will be 2V V = I= R1 + R0 / 2 2R1 + R0
R1 =
This process repeats itself again and again and charged particle spends the same time inside a dee irrespective of its velocity or the radius of circular path, as t=
pr p mv pm = = v v qB qB
2pm So time period of its motion is T = 2t = and qB hence frequency of its motion is v 1 qB = = T 2mp This is known as ‘cyclotron frequency’. (ii) No, they do not accelerate with the same frequency as they have different mass. 14. (i) The Einstein's photoelectric equation is given as Kmax = hu – f0 Since Kmax must be non-negative implies that photoelectric emission is possible only if hu > f0 f or uf > u0 where u0 = 0 , h This shows that the greater the work function f0, higher the threshold frequency u0 needed to emit photoelectrons. Thus, there exists a threshold f frequency u0 = 0 for the metal surface, below h which no photoelectric emission is possible. (ii) Condition for photo electric emission, hu > f0 hc or > f0 l for l = 3300 Å hc 1.989 × 10−25 6.03 × 10−19 = = = 3.77 eV l 3300 × 10−10 1.6 × 10−19
The voltage V1 taken from the potentiometer will be the product of current I and resistance R1, 2V V1 = IR1 = × R1 2R1 + R0 Substituting for R1, we have a R ×R 2V × 0 V1 = R0 + 2R R ×R 2 0 + R0 R0 + 2R V1 =
2VR 2VR or V1 = 2R + R0 + 2R R0 + 4 R
16. (i) Amplitude Modulation : In this modulation the amplitude of a carrier wave changes in accordance with the amplitude of message signal. Only the amplitude of the carrier wave is changed but the frequency of the modulated wave remains the same. (ii) Two factors which justify the need for modulating a low frequency base-band signal are l2 (a) Power radiated by antenna is P ∝ 2 there is l a need of higher frequency conversion for effective power transmission by the antenna. physics for you | APRIL ‘17
73
(b) For effective transmission, the size of the l antenna should be at least of the size , where l 4 is wavelength of signal to be sent. So we can obtain transmission with reasonable antenna lengths if transmission frequency is high. 17. (i) In the circuit C2, C3 and C4 are in parallel \ Cparallel = C2 + C3 + C4 = 2 + 2 + 2 = 6 mF C2
A
C1
C3 C4
\
C5
B=A
C1
Cparallel
C5
B
Equivalent capacitance between A and B is 1 1 1 1 = + + Cequivalent C1 Cparallel C5 =
1 1 1 3 +1+ 3 7 + + = = 6 2 6 2 6
6 = 0.86 mF 7 (ii) Q = CequivalentV = 0.86 × 7 = 6 mC. 1 1 Energy, E = QV = × 6 × 7 = 21 J 2 2 18. (i) Electric field on the equatorial line of an electric → dipole : → EPB sinθ EPB Electric field at any point → on the perpendicular → EPB cosθ θ P bisector of an electric E → EPA cosθ θ dipole at distance r from → → EPA its centre is EPA sinθ Enet = Ex → p θ = EPA cos q + EPB cos q A θ B –q +q (Vertical component cancel 2a each other) or Enet = 2EPA cos q (EPA = EPB) \
Cequivalent =
q a 1 . Enet = 2 . 2 2 2 4 pe0 (r + a ) (r + a2 )1/2 q . 2a 1 Enet = 2 4 pe0 (r + a2 )3/2 p 1 or Enet = 2 4 pe0 (r + a2 )3/2 directed antiparallel to dipole moment p . For short dipole, when r >>> a, then electric field at point P is 1 p E= 4 pe0 r 3 In vectorial form, the electric field intensity at point
74
physics for you | APRIL ‘17
P on the perpendicular bisector of short electric 1 −p . r dipole is then given by E = 4 pe0 r 3 (ii) Work done in rotating the dipole through an angle q in uniform electric field, U = –pE cosq. When q = 0°, then Umin = –pE So, potential energy of an electric dipole is minimum, when it is placed with its dipole moment p parallel to the direction of electric field E and so it is called its most stable equilibrium position. When q = 180°, then Umax = + pE So, potential energy of an electric dipole is maximum, when it is placed with its dipole moment p anti parallel to the direction of electric field E and so it is called its most unstable equilibrium position. 19. (i) For b– decay, – 180A 176A 72 70 1
176A 71 2
γ
→
b+
For decay, + 180 A 176A 74 72 1
172A 69 3
176A 71 2
172A 69 3
172A 69 4
γ
→ 172 69A4 \ The mass number and atomic number of A4 is 172 and 69 and the mass number and atomic number of A is 180 and 72 in b– decay and 180 and 74 is b+ decay. (ii) The emission of electron in b– decay is accompanied by the emission of an antineutrino ( u ) in b+ decay a neutrino is generated. 20. (i) The refractive index of the material of prism, A + dm sin 2 m= A sin 2 Given : A = 60°, dm = 30° sin 45° 1 = . 2 ⇒ m = 2 sin 30° 2 c 3 × 108 c Q m= ⇒v= = = 2.12 × 108 m s–1 m 1.414 v 1 1 A (ii) sin iC = = m 2 δm iC = r = 45° i r1 r A = r1 + r ⇒ r1 = 15°
m=
sin i
sin r1
= 2
B
C
sin i =
2 sin 15° =
( 3 − 1) 2 2
× 2
3 −1 2 3 −1 i = sin–1 2
sin i =
21. Given, RC = 2 kW, VCC = 2 V, b = 100 IB = ?, RB = 1 kW The current amplification factor, I V 2 b = C , IC = CC = , IC = 1 mA R IB 2 × 103 C
1 mA I V bR IB = C = \ IB = 10 mA Q o = c b 100 Vi Ri 2 2000 = 100 × or Vi = 0.01 V Vi 1000
22. Working principle of moving coil galvanometer : It works on the principle that a current-carrying coil placed in a magnetic field experiences a torque, the magnitude of which depends on the strength of current. (i) Radial magnetic field : To maintain the plane of the coil always remains parallel to the field B and to have maximum torque. (ii) A cylindrical soft iron core : This has high permeability and it intensifies the magnetic field and hence increases the sensitivity of the galvanometer. a NBA Current sensitivity, IS = = I k Yes, a galvanometer can be used for measuring the current. By measuring the deflection produced in the galvanometer coil one can obtain the current in the galvanometer. OR (a) Self-inductance : Emf is induced in a single isolated coil due to change of flux through the coil by means of varying the current through the same coil. This is called self-inductance. This S.I. unit of self-inductance is Henry (H). (b) L r2
r1 n1 turns n2 turns
S1
S2
When a current I is set up through S2, it in turn sets up a magnetic flux through S1. Let us denote it by f1. The corresponding flux linkage with solenoid S1 is N1 f1 = M12I ...(i) M12 is called the mutual inductance of solenoid S1 with respect to solenoid S2. The resulting flux linkage with coil S1 is, N1f1 = (n1L)(pr12)(m0n2I) = m0n1n2pr12lI ...(ii) where n1L is the total number of turns in solenoid S1. Thus, from eqn. (i) and eqn. (ii) M12 = m0n1n2pr12L ...(iii) We now consider the reverse case. A current I1 is passed through the solenoid S1 and the flux linkage with coil S2 is, N2f2 = M21I1 ...(iv) M21 is called the mutual inductance of solenoid S2 with respect to solenoid S1. The flux due to the current I1 in S1 can be assumed to be confined solely inside S1 since the solenoids are very long. Thus, flux linkage with solenoid S2 is N2f2 = (n2L)(pr12)(m0n1I1) where n2L is the total number of turns of S2, from eqn. (iii) M21 = m0n1n2pr12L ...(iv) Using eqn. and (iii) and eqn. (iv), we get M12 = M21 = M 23. (a) Anuja is a good explainer and a very knowledgeable student, and her mother Mrs. Rashmi Singh is curious and observant. (b) Lens maker's formula 1 1 1 = (m − 1) − f R1 R2 m = refractive index of lens material Refractive index of glass (mg) > Refractive index of plastic (mp) \ (mg – 1) > (mp – 1) To keep focal length same for glass and plastic lens. (Radius of curvature of glass) > (Radius of curvature of plastic) ⇒ Thickness of plastic lens > thickness of glass lens. 24. (a) Refer to point 4.8 (2) page no. 275 (MTG Excel in Physics). (b) Here, A = 200 cm2 = 2 × 10–2 m2 N = 20, w = 50 rad s–1, B = 3 × 10–2 T, Imax = ? Maximum emf induced in the coil, physics for you | APRIL ‘17
75
e0 = NBAw = 20 × 3 × 10–2 × 2 × 10–2 × 50 = 0.6 V If R is the resistance of the coil the maximum value of the current is, 0.6 V e Imax = 0 = R R OR (a) Refer to point 4.8 (1) page no. 274 (MTG Excel in Physics) (b) Np = 3000, Vp = 2200 V, Vs = 220 V, Ns = ? N pVs Vs N s = As or, Ns = Vp n p Vp
3000 × 220 = 300 2200 25. (a) Refer to point 6.15 (2, 3, 7) page no. 453 (Excel in Physics) (b) Intensity of unpolarised light = I0 I Intensity of light transmitted by P1 = 0 2 Angle between pass axis of P1 and that of P2, q = 60° So, intensity of light transmitted by P2 I I = 0 cos2 60° = 0 8 2 \
Ns =
OR (a) Interference pattern In interference pattern obtained by Young's double slit experiment. (i) All the bright and dark fringes are of same width. (ii) A l l t h e b r i g h t fringes are of same intensity.
Diffraction pattern In diffraction pattern obtained due to a single slit. (i) C ent ra l br ig ht fringe is twice the width of any other secondary br ig ht or dark fringe. (ii) Intensity of central bright fringe is maximum and it decreases with increase in the order of secondary bright fringes.
(b) l = 500 nm = 5 × 10–7 m, a = 0.2 mm = 2 × 10–4 m Angular width of central maxima, 2l 2 × 5 × 10−7 q0 = = 5 × 10–3 rad = a 2 × 10−4 For Young's double slit experiment, b = 0.5 mm 76
physics for you | APRIL ‘17
= 5 × 10–4 m
q D Required number of fringes, N = 0 b −3 5 × 10 × 1 Assume D = 1 m, N = = 10 5 × 10−4 26. (i) Refer to point 2.1(3, 4, 5) page no. (92) (MTG Excel in Physics) (ii) (c) Electric current : The rate of flow of charge I through (a non-uniform conductor) a conductor is same, hence current remains constant. As area of cross-section of the conductor is varying so current density through wire and drift velocity of electron will not be same. OR (i) Refer to point 2.5 (1, 2) page no. 99 (MTG Excel in Physics) (ii) Circuit can be redrawn as
For net resistance between point A and B. Here, r, 2r, 2r and r are in parallel. 1 1 1 1 1 So, = + + + RAB r r 2r 2r 1 3 r = or, RAB = RAB r 3 Net resistance of the circuit, r 4r R = r + RAB = r + = 3 3 (i) Current drawn from the cell E E 3E I= = = R (4r / 3) 4r (ii) Power consumed in network, P = I2RAB \
2
3E r 3E 2 P= = 4r 3 16r
Solution Senders of Physics Musing SET-44
1. Suhas Sheikh, Mumbai 2. Arun Aggarwal, Kanpur 3. Abhinav Singh, Patna
ON
The emf induced across the ends of a conductor due to its motion in a magnetic field is called motional emf.
Let us consider a straight conductor MN moving with constant velocity vector v towards right as shown in the figure. The magnetic field B is uniform and directed into the page and as we move the conductor right with constant velocity vector v then free electrons present in the conductor also move with same velocity vector v in addition to their random velocities. As moving charge in magnetic field experiences a force, so these free electrons also experience a magnetic force. Force due to random velocities averages out to zero. But due to constant velocity vector v towards right electrons experience a magnetic force in downward direction. This magnetic force causes the free electrons in the conductor to move from M to N. Hence free electrons will accumulate at the lower end and there will be a deficiency of free electrons and hence a surplus of positive charge at the upper end. As electrons moves in downward direction, end M acquires a positive potential as compared to N. The potential difference which is appearing across the ends of the straight conductor is called induced emf. These charges at the ends will produce an electric field in downward direction which will exert an upward force on electron. The strength of this electric field is increased until the electric force produced by this field equal in magnitude to the magnetic force. At this point the downward flow of electrons stops and the equilibrium is attained.
v N t1 = 0
M
M
M e
v
FE
FB t2 = t
N
F E e E FB
v N
Steady state
MoTional eMf
Er. Sandip Prasad*
Calculation of Potential Difference across the Conductor:
In steady state or equilibrium position, net force on the electron is zero. \ FE = Fm qE = q (v × B) E = (v × B) M \ VM − VN = ∫ E ⋅ dl M
N
VM − VN = ∫ (v × B).dl N
Calculation of Potential Difference across the ends of a Straight conductor:
Case-1 : Straight rod translating perpendicular to a uniform magnetic field and its velocity vector is perpendicular to the length of the rod Consider a straight rod of length l moving in a plane perpendicular to the plane of the magnetic field of strength B with a velocity v . As shown in the figure magnetic field is directed into plane of the paper while the rod translates in the plane of the paper. The velocity vector of the rod is perpendicular to the length of the rod. M y
l
v x N
M VM − VN = ∫ (v × B) ⋅ dl N
^ ^ ^ Now, v = v(i ), B = B(− k), dl = dl ( j) \ VM − VN =
M
^
^
^
∫ [{v(i ) × B(− k)} ⋅ dl( j)}
N
*Author is Director of Sandip Physics Classes and motivational speaker physics for you | april ‘17
77
=
M
^
^
∫ [vB(j ) ⋅ dl ( j) =
N
\ VM − VN = vBl
M
∫
v Bdl = vB
N
M
∫
dl
N
Case-2 : Straight rod translating perpendicular to a uniform magnetic field and its velocity vector is along the length of the rod Consider a straight rod of length l moving in a plane perpendicular to the plane of the magnetic field of strength B with a velocity v . As shown in the figure magnetic field is directed into plane of the paper while the rod translates in the plane of the paper. The velocity vector of the rod is along the length of the rod. y
v
N
l
N M
^
^
M
^
v
N
M
l
^
^
y
v
M
l
^ ^ B = B(i ) ; dl = dl (i )
^ v = v (i ) ;
M As, VM − VN = ∫ (v × B) ⋅ dl N
M
\ VM − VN = ∫ [{v (i ) × B(i )} ⋅ dl (i )] ^
^
N
\ VM – VN = 0 78
physics for you | april ‘17
^
^
Case-4 : Straight rod translating perpendicular to the uniform magnetic field and its velocity vector makes an angle θ with the length of the rod
Case-3 : Straight rod translating in the direction of the uniform magnetic field A) Velocity vector is along the length of the rod : Consider a straight rod of length l moving in the region of uniform magnetic field of strength B with a velocity v . As shown in the figure magnetic field and velocity vector along the length of the rod.
N
^
N
M
v
\ VM – VN = 0
^
\ VM – VN = 0
N
B
x
^ ^ v = v( j) ; B = B (i^) ; dl = dl (i ) M As, VM − VN = ∫ (v × B) ⋅ dl
\ VM − VN = ∫ [{v(i ) × B(− k )} ⋅ dl(i )] = ∫ vB( j) ⋅ dl (i ) N
y
B
\ VM − VN = ∫ [{v ( j ) × B(i )} ⋅ dl(i )]
^ dl = dl(i )
^ B = B(− k) ; M As, VM − VN = ∫ (v × B) ⋅ dl ^ v = v (i ) ;
N M
x
M
B) Velocity vector is perpendicular to the length of the rod : Consider a straight rod of length l moving in the region of uniform magnetic field of strength B with a velocity v . As shown in the figure velocity vector is perpendicular to the length of the rod and magnetic field.
x
l
y
x
N
Consider a straight rod of length l moving in a plane perpendicular to the plane of the magnetic field of strength B with a velocity v . As shown in the figure magnetic field is directed into the paper while the rod translates in the plane of the paper. However the rod itself is not oriented perpendicular to the velocity vector and instead makes an angle θ as shown in the figure. ^ ^ v = [v sin θ(i ) + v cos θ( j)] ; B = B(− k^) ^ dl = dl( j) M \ VM − VN = ∫ (v × B) ⋅ dl M
N
^
^
^
^
= ∫ [{v sin θ(i ) + v cos θ( j ) × B(− k)} ⋅ dl( j)] N
= vB sinθ
M
∫ dl
N
= vBlsinθ ⇒ VM – VN = Bv(lsinθ) = Magnetic field × velocity of the rod × projection of the length of the rod on a plane perpendicular to the velocity vector or, VM – VN = Bl (vsinθ) = Magnetic field × length of the rod × component of velocity vector perpendicular to the length of the rod on a plane perpendicular to the velocity Important points : In case of straight conducting rod moving in a uniform magnetic field with constant velocity and if any one of the two physical quantities v , B and l are parallel then no induced emf is produced. In case of straight conducting rod moving in a uniform magnetic field with constant velocity and if any one of the physical quantities v , B and l are not perpendicular to other two physical quantities , then we need to find the component of that quantity, so that at least one component must be parallel to other two physical quantities. Case 5 : Induced emf across an arbitrary shaped rod translating in a uniform magnetic field Consider an arbitrarily M M shaped rod translating in dlsin a plane perpendicular to a dl dlcos magnetic field of strength B v FB with a velocity v . Assume the magnetic field is directed into the plane of the paper N N and the rod is moving to the right with velocity vector v perpendicular to the magnetic field. If the charge on an electron is e, then the magnetic force on the free electrons is FB = evB and it is directed downwards as shown in the figure. In order to calculate the induced emf in the conductor, consider a small element of the rod of length dl. Let this element be oriented at an angle θ to the velocity vector v . Emf induced across the element, dε = vBdl sinθ. Net induced emf across the ends of the rod is ε = ∫ dε = ∫ vBdl sin θ
= ∫ vBdy = vB ∫ dy
ε = vBy Here, y = Projection of wire along a line which is perpendicular to velocity vector = Effective length \ ε = Velocity × magnetic field × projection of the entire conductor onto a plane perpendicular to the velocity. Thus an arbitrarily shaped conductor can be replaced by a straight conductor whose length is equal to the projected length of the conductor on to a plane perpendicular to the direction of motion. EXAMPLE-1 : Find the emf across the point P and Q which are diametrically opposite points of a semicircular closed loop moving in a magnetic field as shown in the figure. Also draw the electrical equivalence B of each branch. Q Soln.: Join the end points P and P a Q and replace the semi-circular loop by a straight rod of length 2a. We now have a straight rod translating in a magnetic field therefore induced emf between P and Q will be ε = 2Bav. Positive terminal will be in the direction of (v × B). So P is positive terminal. 2Bav
P
2Bav
Q
Case-6 : Calculation of induced emf across the ends of a rod, if the rod is rotating in uniform magnetic field: Axis of rotation L e t u s c on s i d e r a conduc t ing ro d of length l rotating about an axis passing A through one of its O x dx ends with constant angular velocity ω in B B a uniform magnetic field B as shown in the figure. Consider a small element of length dx at a distance x from one of its ends. Emf across the ends of length element dx is dε = vBdx \ Emf across the rod ε = ∫ dε
= ∫ vBdx = B ∫ vdx = B ∫ xωdx physics for you | april ‘17
79
l
= Bω ∫ xdx = Bω ∫ xdx 1 \ ε = Bωl 2 2
0
A spoked wheel Axis of rotation of spoke length l is rotated about rim its axis with an spokes angular velocity ω in a plane normal to uniform B B magnetic field B as shown. The emf induced across the ends of each spoke is 1 ε = Bl 2ω, with axle (centre) at higher potential. 2 Since all the spoke are parallel between axle and rim, the emf induced between axle and rim is 1 ε = Bl 2ω 2 It is independent of number of spokes. EXAMPLE-2 : A copper rod of length 2 m is rotated with a speed of 10 rev s–1, in a uniform magnetic field of 1 T about a pivot at one end. The magnetic field is perpendicular to the plane of rotation. Find the emf induced across its ends. 1 1 Soln.: ε = Bl 2ω = B(2πυ)l 2 = πBυl 2 2 2 ε = 3.14 × 1 × 10 × 2 × 2 = 125.6 V EXAMPLE-3 : A wheel with rim 10 metallic spokes, each 0.5 m long is rotated with spokes a speed of 120 rev min –1 in a plane normal to the earth’s magnetic field at the place. If the magnitude of the field is 0.40 G, what is the induced emf between the axle and the rim of the wheel? Soln.: Here each spoke of wheel act as a source of an induced emf (cell) and emf ’s of all spokes are parallel. υ = 120 rev min–1 = 2 rev s–1, B = 0.40 G = 0.4 × 10–4 T, l = 0.5 m 1 1 Induced emf, ε = Bωl 2 = B(2πυ)l 2 = Bπυl 2 2 2 22 ε = 0.4 × 10−4 × × 0.5 × 0.5 × 2 7 –5 ε = 6.29 × 10 V 80
physics for you | april ‘17
(Induced emf in a wheel is independent of number of spokes.) EXAMPLE-4 : Find the emf induced in the rod in the following cases. v
(a)
v (b) B
B B (c)
A v
(d) C
L
B v
B
Soln.: (a) emf = 0 (b) emf = 0 (c) emf = 0 (d) Figure shows a closed coil ABCA moving in a uniform magnetic field B with a velocity v. The flux passing though the coil is constant and therefore the induced emf is zero. Now consider rod AB, which is part of the coil. Emf induced in the rod = vBL A Suppose the emf induced in part ACB C is ε, as shown in the figure. vBL Since the emf in the coil is zero, emf (in ACB) + emf (in BA) = 0 or –ε + vBL = 0 or ε = vBL B Thus emf induced in any path joining A and B is same, provided the magnetic field is uniform. Also the equivalent emf between A and B is vBL (here the two emf 's are in parallel). EXAMPLE-5 : A rod of length l is kept parallel to a long wire carrying constant current I. It is moving away I from the wire with a velocity v. Find the emf induced in the wire when its distance from the long wire is x. µ Ilv Soln.: ε = Blv = 0 2πx EXAMPLE-6 : A rectangular loop is moving parallel to a long wire carrying current I with a velocity v. Find the emf induced I in the loop, if its nearest end is at a distance a from the wire. Soln.: emf = 0
B
l
v
x
v l a
b
EXAMPLE-7 : A square wire of B length l, mass m and resistance R slides without friction down B the parallel conducting wires of negligible resistance as shown in figure. The rails are connected to each other at the bottom by a resistanceless rail parallel to the wire so that the wire and rails form a closed rectangular loop. The plane of the rails makes an angle θ with horizontal and a uniform vertical field of magnetic induction B exists throughout the region. Show that the wire acquires a steady state velocity of mgR sin θ magnitude v = 2 2 2 . B l cos θ Soln.: Force down the plane = mgsinθ At any instant if the velocity is v the induced emf = lBcosθ × v lB cosθv Current in the loop = R Force on the conductor in the horizontal direction Blv cosθ = ×B×l R B2l 2 cos θ Component parallel to the incline = (v cos θ) R mgR sin θ B2l 2 cos2 θ v = mg sin θ \ v = 2 2 2 If v is constant R B l cos θ EXAMPLE-8 : Magnetic field in a region is given by B ^ B = 0 y k where L = constant. A conducting rod of L
length L lies along y-axis between origin and point (0, L, 0). Rod moves with velocity v = v0 i^ then find emf induced across ends of the rod. Soln.: dε = vBdy B y = v0 0 dy L L
v B L ε = ∫ dε = 0 0 ∫ y dy L 0 0 ε=
y (0, L, 0) ^ B B = 0 y(k) L
dy
v0(î)
y
v0 B0L2 v0 B0L = 2L 2
x
(0, 0, 0)
EXAMPLE-9 : A rod of length l is rotating with angular speed ω about l a one of its end while at i a d i s t an c e a f rom an infinite long wire carrying current i. Find emf induced at the instant shown in figure. µ0i dr Soln.: dε = vB dr = ωr × 2π(r + a) ε = ∫ dε =
ωµ0i l rdr ∫ 2π 0 r + a
µ iω l + a = 0 l − a ln a 2π
i
a O
l r
dr
physics for you | april ‘17
81
Class XII
T
his specially designed column enables students to self analyse their extent of understanding of complete syllabus. Give yourself four marks for correct answer and deduct one mark for wrong answer. Self check table given at the end will help you to check your readiness.
Total Marks : 120
Time Taken : 60 min NEET / AiiMs / PMTs
Only One Option Correct Type
1. The output current of a 70% amplitude modulated generator is 1.6 A. To what value will the current rise if the generator is additionally modulated by another audio wave of modulation index 0.6 ? (a) 1.61 A (b) 1.71 A (c) 1.81 A (d) 1.91 A 2. An observer is 4 m from an isotropic point light source whose power is 500 W. The rms value of electric field due to the source at the position of observer is (a) 12.6 V m–1 (b) 15.6 V m–1 –1 (c) 25.6 V m (d) 30.6 V m–1 3. A circular beam of light of diameter of 2 cm falls on a plane surface of glass. The angle of incidence is 60° and refractive index of glass is 1.5. The diameter of the refracted beam is (a) 3.26 cm (b) 2.52 cm (c) 3.0 cm (d) 4.0 cm 4. A charged ball B hangs from a silk + thread S, which makes an angle q P + + S with a large charged conducting + + sheet P as shown in the figure. The + electric field due to the sheet is B + proportional to + (a) cosq (b) cotq (c) sinq (d) tanq 5. The refracting angle of a prism is A and refractive A index of the material of prism is cot . The angle 2 of minimum deviation will be (a) 180° – 3A (b) 180° + 3A (c) 90° – 3A (d) 180° – 2A 82
Physics For you | April ‘17
6. A rectangular loop of sides 25 cm and 10 cm carrying a current of 15 A is placed with its longer side parallel to a long straight conductor 2.0 cm apart carrying a current of 25 A. What is the net force on the loop? (a) 7.8 × 10–4 N (b) 6.6 × 10–4 N (c) 8.2 × 10–4 N (d) 5.4 × 10–4 N 7. If K1 and K2 are maximum kinetic energies of photoelectrons emitted when lights of wavelengths l1 and l2 respectively incident on a metallic surface and l1 = 3l2, then K K (a) K1 > 2 (b) K1 < 2 3 3 (c) K1 = 2K2
(d) K2 = 2K1
8. A copper rod of length 20 cm and cross-sectional area 2 mm2 is joined with a similar aluminium rod as shown in figure below. Find the resistance of the combination between the ends. Resistivity of copper = 1.7 × 10–8 W m and resistivity of aluminium = 2.6 × 10–8 W m. (a) 2.424 m W (b) 1.028 m W (c) 3.012 m W (d) 4.009 m W 9. An element with atomic number Z = 11 emits Ka-X-ray of wavelength l. The atomic number of element which emits Ka-X-ray of wavelength 4l is (a) 6 (b) 4 (c) 11 (d) 44 10. At a given instant there are 25% undecayed radio-active nuclei in a sample. After 10 s the number of undecayed nuclei reduces to 12.5%. Calculate the mean-life of the nuclei (a) 38 s (b) 20 s (c) 14.43 s (d) 13 s
11. Two positive charges of magnitude q are placed at the ends of one side of a square of length 2a. Two negative charges of the same magnitude are kept at the other corners. Starting from rest, if a charge Q moves from the middle of first side to the centre of square, its kinetic energy at the centre of square is (a)
1 2qQ 4 πε0 a
1 1 − 5
(b) zero (c) (d)
1 2qQ 4 πε0 a
1 1 + 5
1 2qQ 4 πε0 a
2 1 − 5
12. Truth table for the given circuit is
(a)
(b)
(c)
(d)
Assertion & Reason Type
Directions : In the following questions, a statement of assertion is followed by a statement of reason. Mark the correct choice as : (a) If both assertion and reason are true and reason is the correct explanation of assertion. (b) If both assertion and reason are true but reason is not the correct explanation of assertion. (c) If assertion is true but reason is false. (d) If both assertion and reason are false.
13. Assertion : A photocell is called an electric eye. Reason : Photocell responds in the form of electric current, when light is made to fall on it. This is similar to eye observing in the presence of light.
14. Assertion : The conduction current and displacement current together has the property of continuity. Reason : The conduction current is same as displacement current when source is only dc and not ac. 15. Assertion : The fringe obtained at the centre of the screen is known as zeroth order fringe, or the central fringe. Reason : Path difference between the waves from S1 and S2, reaching the central fringe (or zero order fringe) is zero. JEE MAiN / JEE AdvANcEd / PETs Only One Option Correct Type
16. Imagine an atom made of a proton and a hypothetical particle of double the mass of the electron but having the same charge as the electron. Apply the Bohr atom model and consider all possible transitions of this hypothetical particle to the first excited level. The longest wavelength photon that will be emitted has wavelength [in terms of the Rydberg constant R for the hydrogen atom] equal to 18 4 9 36 (a) (b) (c) (d) 5R 5R 5R 5R 17. In the figure, CP represents a wave front and AO and BP, the corresponding two rays. Find the condition on q for constructive interference at P between the ray BP and the reflected ray OP. 3l Q O R (a) cos q = 2d l C (b) cos q = d 4d A l (c) sec q − cos q = d P B 4l (d) sec q − cos q = d 18. A non-conducting ring having charge q uniformly distributed over its circumference is placed on a rough horizontal surface. A vertical time varying magnetic field B = 4t2 is switched on at time t = 0. Mass of the ring is m and radius is R. The ring starts rotating after 2 seconds. The coefficient of friction between the ring and the table is 2qmR 8qR qR 4qmR (a) (b) (c) (d) g mg 2 mg g 19. A rigid circular loop of radius r and mass m lies in the x-y plane on a flat table and has a current Physics For you | April ‘17
83
i flowing in it. At this particular place, the earth's magnetic field is B = Bxi + Bz k . The value of i so that one edge of the loop lifts from the table is (a) (c)
mg πr Bx2 + Bz2 mg πrBx
(b) (d)
mg πrBz
(a) a = b (b) v = 2v0
×
2mv0 sin a (c) PQ = Bq
v
P × Q ×
(d) particle remains in the
mg
field for time t =
πr Bx Bz
More than One Options Correct Type 20. A radioactive sample has initial concentration N0 of nuclei. Then, (a) the number of undecayed nuclei present in the sample decays exponentially with time (b) the activity (R) of the sample at any instant is directly proportional to the number of undecayed nuclei present in the sample at that time (c) the number of decayed nuclei grows exponentially with time (d) the number of decayed nuclei grows linearly with time.
×
v0 ×
×B
×
×
×
×
×
×
2m ( π − a)
Bq 23. A transistor is used in the common emitter mode as an amplifier. Then (a) the base-emitter junction is forward-biased. (b) the base-emitter junction is reverse-biased. (c) the input signal is connected in series with the voltage applied to bias the base-emitter junction. (d) the input signal is connected in series with the voltage applied to bias the base-collector junction Integer Answer Type
24. A Bohr hydrogen atom undergoes a transition n = 5 to n = 4 and emits a photon of frequency u. Frequency of circular motion of electron in n = 4 orbit is u4. The ratio u/u4 is found to be 18/5m. State the value of m. 25. A narrow monochromatic beam of light of intensity I is incident on a glass plate as shown in figure. Another identical glass plate is kept close to the first one and parallel to it. Each glass plate reflects 25% of the light incident on it and transmits the remaining.
21. Figure shows a balanced Wheatstone bridge.
I max the I min interference pattern formed by two beams obtained after one reflection at each plate. Find the ratio
(a) If P is slightly increased, the current galvanometer flows from C to A. (b) If P is slightly increased, the current galvanometer flows from A to C. (c) If Q is slightly increased, the current galvanometer flows from C to A. (d) If Q is slightly increased, the current galvanometer flows from A to C.
in the in the in the in the
22. A particle of charge –q and mass m enters a uniform magnetic field B (perpendicular to paper inwards) at P with a velocity v0 at an angle a and leaves the field at Q with velocity v at an angle b as shown in figure. Then 84
Physics For you | April ‘17
26. In the arrangement shown in figure, y = 1.0 mm, d = 0.24 mm and D = 1.2 m. The work function of the material of the emitter is 2.2 eV. Find the stopping potential V needed to stop the photocurrent (in × 10–1 V).
Comprehension Type
Matrix Match Type
In the given circuit the capacitor (C) may be charged through resistance R by battery V by closing switch S1. Also when S1 is opened and S2 is closed the capacitor is connected in series with inductor (L).
27. At the start, the capacitor was uncharged. When switch S1 is closed and S2 is kept open, the time constant of this circuit is t. Which of the following is correct? (a) After time interval t, charge on the capacitor is CV . 2 (b) After time interval 2t, charge on the capacitor is CV (1 – e–2) (c) The work done by the voltage source will be half of the heat dissipated when the capacitor is fully charged (d) After time interval 2t, charge on the capacitor is CV(1 – e–1).
29. A circular current carrying loop is placed in x-y plane as shown in figure. A uniform magnetic field B = B0 k is present in the region. Match the following. Column I (A) Magnetic moment (P) of the loop (B) Torque on the loop (Q) (C) Potential energy of (R) the loop (D) Equilibrium of the loop (S) (T) A B C D (a) R P T S (b) R P S T (c) T Q R P (d) Q P S R
t π (b) Q = Q0 cos − 2 LC
(d) Q = −
1
d 2Q dt
Maximum Along positive z-axis Stable None
30. Two spherical shells are as shown in figure. Suppose r is the distance of a point from their common centre. Then,
28. Given that the total charge stored in the LC circuit is Q0, for t ≥ 0, the charge on the capacitor is π t (a) Q = Q0 c o s + 2 LC
(c) Q = − LC
Column II Zero
2
d 2Q
LC dt 2
Column I (A) Electric field for r < R1 (B) Electric potential for r < R1 (C) Electric potential for R1 < r < R2 (D) Electric field for R1 < r < R2 A B C (a) P R Q (b) R P S (c) S R P (d) S P Q
Column II is constant for q2 and vary for q1 (Q) is zero for q2 and vary for q1 (R) is constant (P)
(S)
is zero D S Q Q R
Keys are published in this issue. Search now! J
Check your score! If your score is > 90%
ExcEllENT work !
You are well prepared to take the challenge of final exam.
No. of questions attempted
……
90-75%
Good work !
You can score good in the final exam.
No. of questions correct
……
74-60%
sATisFAcTory !
You need to score more next time.
Marks scored in percentage
……
< 60%
NoT sATisFAcTory! Revise thoroughly and strengthen your concepts.
Physics For you | April ‘17
85
Contd. from Page no. 30
on Exam ay st 21 M
2r 3 r 5 = π l ρv02 r − + dr R2 R 4 Hence, kinetic energy of the fluid, R l πR2ρv02 2r 3 r 5 K = π l ρv02 ∫ r − + dr = 6 R2 R 4 0
6. (a) : CP = CO = Radius of circle = R ∠CPO = ∠POC = 60° = ∠OCP OCP is an equilateral triangle. OP = R = Extended length of spring. Natural length of spring = 3R/4 3R R ∴ Extension, x = R − = 4 4 mg Let spring force = F = kx where Q k = R mg mg R or F = = 4 R 4 The free body diagram of the C N ring is shown in the figure. 60° Here F = kx = mg/4 , P F N = Normal Reaction. 60° O ax As soon as the ring is released, mg the net force along y-axis is zero. \ Fy = 0 \ N + F cos 60° = mg cos 60° or N = mg cos 60° – F cos 60° mg mg 1 3mg − or N = 2 or N = 2 4 8 y
7. (a, b, d) 8. (b, c, d) : Let dV be the volume flowing per second through the cylindrical shell of thickness dr then, r2 r3 dV = (2πr dr )v0 1 − = 2π v0 r − dr R2 R2 and the total volume, R r3 R2 π 2 V = 2πv0 ∫ r − = R v0 dr = 2πv0 4 2 R2 0
r
R
r
v
Here frictional force is the shearing force on the tube, dv exerted by the fluid, which equals − ηS , where S dr is the surface area of tube. r 2 dv r = −2v0 Given, v = v0 1 − So 2 dr R2 R and at r = R,
Then, viscous force is given by, dv dF = − η (2πRl ) dr r = R 2v = − 2 π R η l − 0 = 4 π η lv0 R Taking a cylindrical shell of thickness dr and radius r dv Viscous force, F = − η (2π r l ) , dr Let, DP be the pressure difference, then net force on dv the element = ∆ P π r 2 + 2 π η l r dr But, since the flow is steady, Fnet = 0 dv 2 π l ηr 2v r 0 2 4 η lv R 0 dr = or ∆P = = 2 2 2 πr πr R 9. (a) : Work done by electric field during motion of point charge +q from B to A (x is distance of closest approach) −2 π l η r
a
q.(VA − VB ) = q. ∫ x
dr
Let, dK be the kinetic energy, within the above cylindrical shell. Then 1 1 dK = (dm)v 2 = (2 π rlρ)v 2dr 2 2 2
r 1 = (2π l ρ) r v02 1 − dr 2 R2 2
86
2v dv =− 0 dr R
PHYSICS FOR YOU | APRIL ‘17
∴ W =q
λ dx 2πε0 x
λ a ln 2πε0 x
By work-energy theorem qλ a ln ⇒ K0 = 2πε0 x ⇒ x = ae −2 πε0 K 0 / λq
+
a
+
+ + +
x
A
B
10. (a, b, c) : x = a cos pt, y = b sin pt 2
2
x y ∴ cos2 pt + sin2 pt = + a b 2 2 y x or + = 1 It is an equation of ellipse. 2 a b2 x = a cos pt dx ∴ = − ap sin pt ⇒ v x = − ap sin pt dt
Q
2
d x 2
= −ap2 cos pt ⇒ ax = −ap2 cos pt
dt Q y = b sin pt dy ∴ = bp cos pt ⇒ v y = bp cos pt dt
d2 y
= −bp2 sin pt ⇒ a y = −bp2 sin pt . dt 2 When t = p/2p, ax = 0 and vy = 0; ay = – bp2, vx = – ap Thus velocity is along negative x-axis. Acceleration is along negative y-axis. At t = t, r (t ) = xi + yj = a cos pt i + b sin pt j
= − p2 [ xi + y j] = − p2 r (t )
Hence acceleration is always directed towards origin. At t = 0, the particle is at (a, 0) At t = p/(2p), the particle is at (0, b) y
b 0
or plate resistance, rp =
dV 1 = VA −1 −3 dI 0.125 × 10
or rp = 8 × 103 VA–1 = 8 × 103 W Again, I = (0.125 V – 7.5) mA
...(i)
I = (0.125 × 300 – 7.5) or
I = (37.5 –7.5) mA or I = 30 mA
In second case, Vg = –3 V, V = 300 V, I = 5 mA ∆I ∴ gm = at V constant ∆Vg (30 − 5) × 10−3
= 12.5 × 10–3 AV–1 ...(ii) m = rP × gm = 8 × 103 × 12.5 × 10–3 = 100 ...(iii) or g m =
[−1 − (−3)]
13. (a, b, c) : The optical path difference between the beams arriving at P, Dx = (l2 – l1) + d sinq The condition for maximum intensity is Dx = nl, here n = 0, ± 1, ± 2, .... 1 1 Thus, sin θ = [∆x − (l2 − l1 )] = [nλ − (l2 − l1 )] d d 1
10 × 10−6
[n × 500 × 10−9 − 20 × 10−6 ]
n = 2 − 1 40
= − p [a cos pt i + b sin pt j] 2
2p
dI = 0.125 × 10−3 AV −1 dV
=
a(t ) = ax i + a y j
t=
∴
a x
Therefore distance covered is one-fourth of elliptical path. It is not equal to a. 11. (a, b, c) 12. (b, c) : At Vg = –1 V, V = 300 V Here I is anode current and V is plate voltage. I = (0.125 V – 7.5) × 10–3 A
Hence, θ = sin −1 2 n − 1 40 sinq| ≤ 1 n ∴ − 1 ≤ 2 − 1 ≤ 1 40
or –20 ≤ (n – 40) ≤ 20 or 20 ≤ n ≤ 60 Hence, number of maxima = 60 – 20 = 40 At C, phase difference,
2π 2π φ = (l2 − l1 ) = (20 × 10−6 ) = 80 p 500 × 10−9 λ Hence, maximum intensity will appear at C. To obtain minimum intensity at C, λ (µ − 1) t = ...(i) 2 here m is refractive index and t is thickness of slab PHYSICS FOR YOU | APRIL ‘17
87
or t =
λ 500 × 10−9 = = 500 nm 2 (µ − 1) 2 × 0.5
...(i)
Substituting t and l is equation (i) We get 3 m = = 1.5 2 14. (a, b, c) : Here A and B represents the centre of mass of cone and sphere respectively y1 = OA = R2, y2 = OB = 4R2 + R1 If m1 = m2 = m (say) and R1 = R2 = R mR + m5R B yCM = = 3R R1 2m If their volume mass density are equal r1 = r2 = r (say) and R1 = R2 = R (say)
16. (b) : C2,eff = C2 + Cadd = 365 + 36 = 401 pF = C(say) Necessary inductance is 1 1 L= = 2 2 2 6 4 π f C 4 π (0.54 × 10 Hz)2 (401 × 10−12 F) = 2.2 × 10–4 H = 0.22 mH 17. (c) : For refraction in surface A, [at M] µ2 µ1 µ2 − µ1 − = v1 u2 R 3 −1 1 3 1 2 = − = 2v1 (−2R) (−0.25) −0.5 3 1 1 = − = −4 2v1 −0.5 0.5
A O
R2
Axis
4 3 πR ρ = m 3 m2 = Mass of solid cone 1 = πR2 4 Rρ = m1 = m 3
(From centre of mass of solid cone) my + my2 yCM = 1 − OA = 3R − R = 2R m1 + m2 If r1 = 2r2 4 4 m1 = πR3 2ρ2 = 2m, m2 = πR3ρ = m (say) 3 3
1.60 = 2.96 . 0.54
Q
Solving C2 eff = C1 eff(2.96)2, PHYSICS FOR YOU | APRIL ‘17
B A
OOO 2 1
X Y
So image is formed at O1, a distance v1 = 0.375 m from X and act as a virtual object for the second surface. 18. (d) : For refraction in the second surface B, [Pole v] 3 µ1 µ2 µ1 − µ2 1 − 2 − = = (−0.5) v2 v1 R2 ⇒
−1 1 3 − = v2 2 × (−0.25 − 0.375) 2(−0.5)
1 1 3 = − v2 1 (0.625 × 2)
Adding a capacitor in parallel will result in an effective capacitance given by C1 eff = C1 + Cadd with a similar expression for C2. We want to choose Cadd so that
C2 eff f1 = = 2.96 f2 C1 eff
M
3 ⇒ v1 = − m = – 0.375 m. 8
mR + 2m5R 11R = 3m 3
15. (d) : The desired ratio is
88
C − 8.76 C1 (365 pF) − 8.76 (10 pF) Cadd = 2 = 36 pF = 7.76 7.76
4R2
m1 = Mass of sphere =
yCM =
C2 + Cadd = (C1 + Cadd)8.76,
f =
. 2 π LC 1
v2 = –0.714 Shift in point of observation OO2 = YO – YO2 = 3R – v2 = 0.750 – 0.714 = 0.03
MPP CLASS XI 1. (b) 6. (a) 11. (b) 16. (d) 21. (a,c) 26. (6)
2. (a) 7. (c) 12. (d) 17. (b) 22. (a,c,d) 27. (b)
ANSWER KEY
3. (d) 8. (c) 13. (b) 18. (c) 23. (a,d) 28. (c)
4. (a) 9. (c) 14. (c) 19. (b) 24. (9) 29. (c)
5. (c) 10. (a) 15. (a) 20. (a,d) 25. (2) 30. (a)
Readers can send their responses at
[email protected] or post us with complete address by 25th of every month to win exciting prizes. Winners' name with their valuable feedback will be published in next issue.
ACROSS
1. 2. 4. 5. 7.
12. 15. 17. 21. 22. 24. 25. 26. 27. 28.
The shift towards lower frequencies of the spectra of galaxies moving away from us. (3, 5) A curve expressing the relation between the melting point of ice and the applied pressure. (3, 4) The S.I. unit of pressure. (6) A unit of nuclear cross-section (4) The inertial force associated with the change in the tangential component of a particle’s velocity. (8, 5) The science that aims at specifying and reproducing colours as a result of measurement. (11) A steep wave that moves up narrowing channels produced either by regular tidal events, or as a result of tsunami. (4) A material with a strong chemical affinity for other materials. (6) The photographic process for producing threedimensional images. (10) A measure of the rate of decay of a periodic quantity. (4, 4) The maximum demand on a power generating source. (4, 4) The bright central spot in the system of diffraction rings formed by an optical system with light from a point source. (4, 4) The phenomenon of melting under pressure and freezing again when the pressure is reduced. (10) A single vibration of current, light or other wave. (5) It is also known as far-sightedness. (9)
DOWN
1. 3. 5. 6. 7. 8. 9.
An S.I. unit of angle. (6) An instrument for studying thin films on solid surfaces. (12) An isolating circuit used to minimize reaction between two circuits. (6) The study of forces and the resulting motion of objects through the air. (12) The flavour of fourth quark. (5) An electronic circuit used to implement a variety of simple two-state devices. (13) A model of the universe with the Earth as its centre. (10, 5)
10. The removal of any unwanted ac components from a circuit. (10) 11. It is a device fitted on a satellite which receives the signal and retransmits it after amplification. (11) 12. The study of the production and effects of very low temperatures. (10) 13. A material, such as potassium chloride, that darkens under electron bombardment and recovers on heating. (9) 14. Former name for Sonar. (5) 16. An instrument for amplifying and directing sound. (9) 18. These are rod-shaped cells of the retina that are sensitive to the intensity of light. (4) 19. A measure of the degree of wetness of the atmosphere. (8) 20. An electronic circuit that produces an output pulse when a specified number of input pulses have been received. (6) 23. Transient air glow events observed near 90 km, nearly simultaneously with a strong cloud-to-ground lightning stroke (5)
Physics For you | APRIL ‘17
89
90
Physics For you | APRIL ‘17
