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ne had always assumed that the advance of science is logical and based on assumptions that are verified by experimentation. This was
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true during the days of classical physics. However, modern physics in the
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Managing Editor Editor
: :
is the Advance of science by Logic, intuition or Experimentation?
June 2015
Corporate Office :
rial
twentieth century has grown by leaps and bounds flouting the ancient logical methods. If one were to analyse the march of physics, einstein,
Mahabir Singh Anil Ahlawat (Be, MBA)
Bohr, Louis de Broglie, Dirac, Schrodinger, Max Born and many other did not wait for experimentation for the formulation of theories.
contents Physics Musing (Problem Set-23)
According to einstein’s photoelectricity, photon is both a particle and a wave. Diffraction and interference of light as well as that of electrons were
8
unified by Max Born. Max Born developed quantum mechanics based on
AIPMT Solved Paper 2015
10
the probability waves suggested by einstein. Dirac’s contribution of his
Exam Prep
23
Physics Musing (Solution Set-22)
29
Target PMTs Practice Questions 2015
31
Kerala PMT Solved Paper 2015
41
Brain Map
46
Ace Your Way CBSE XII Series 1
51
Core Concept
62
Intuition has proved to be more powerful than logic, assumptions and
WB JEE Solved Paper 2015
65
experimentation.
Thought Provoking Problems
76
Concept Based FAQs
80
You Ask We Answer
84
Crossword
85
idea of ‘Dirac vacuum’ consisting of all particles and antiparticles is near the idea of ‘Soonya’ of Indian Philosophy. However our philosophy says that it is “Poorna” or infinity. Infinity + infinity = infinity, according to modern mathematics. But infinity – infinity is undefined. According to our concept, if one takes Poorna from Poorna, Poorna remains. einstein’s concept of c, the velocity of light and his enunciation that the mass of a body increases with velocity are beyond classical physics and transcends normal logic. einstein, Bohr, Louis de Broglie and Max Born set right the controversy of whether light is a particle or wave, by their correction that particles and light are simultaneously matter and wave.
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Physics for you | june ‘15
7
P
PHYSICS
MUSING
hysics Musing was started in August 2013 issue of Physics For You with the suggestion of Shri Mahabir Singh. The aim of Physics Musing is to augment the chances of bright students preparing for JEE (Main and Advanced) / AIIMS / 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.
2.
A disc of radius R is rolling purely on a flat horizontal surface, with constant angular velocity. The angle between the velocity and acceleration vectors at point P is (a) zero (b) 45° (c) 135° (d) tan–1(1/2) A solid ball of radius r rolls inside a hemispherical shell of radius R without slipping. It is released from rest from point A as shown in figure. The angular velocity of centre of the ball in position B about the centre of the shell is (a) 2 (c)
3.
4.
g 5(R − r) 2g 5(R − r)
(b) (d)
2
a
5.
Two glass plates are touching at one end and separated by a thin wire at the other end. When a monochromatic parallel beam of wavelength 4200 Å incident normally on the glass plates is reflected, an interference pattern of 30 fringes is observed. If the wavelength of light used is taken 7000 Å instead of 4200 Å, the number of fringes observed will be (a) 50 (b) 40 (c) 30 (d) 18
7.
Consider a YDSE that has different slits width, as a result, amplitudes of waves from slits are A and 2A, respectively. If I0 be the maximum intensity of the interference pattern, then intensity of the pattern at a point where phase difference between waves is f, is (a) I0 cos2 f
5g 2(R − r)
(b) w =
a (c) w = 0 w0
6.
10 g 7(R − r)
An object is moving towards a converging lens on its axis. The image is also found to be moving towards the lens. Then, the object distance u must satisfy (a) 2f < u < 4f (b) f < u < 2f (c) u > 4f (d) u < f A uniform solid brass sphere of radius a0 and mass m is set spinning with angular speed w0 about a diameter at temperature T0. If its temperature be increased to T without disturbing the sphere, its new angular speed, assuming that its new radius is a, will be (a) w = w0
same diameter and attain the same terminal velocity, the ratio of viscosity of water to that of the liquid is (a) 2.0 (b) 0.5 (c) 4.0 (d) 0.25
T w T0 0
T − T0 w0 T0
(d) w =
A metal ball A (density 3.2 g cm–3) is dropped in water, while another metal ball B (density 6.0 g cm–3) is dropped in a liquid of density 1.6 g cm–3. If both the balls have the
(c)
I0 f sin 2 3 2 I0 [5 + 8 cos f] (d) 9
(b)
I0 [5 + 4 cos f] 9
subjective type
8.
Two identical sonometer wires have a fundamental frequency of 500 Hz when kept under the same tension. What fractional increase in the tension of one wire would cause an occurrence of 5 beats per second, when both wires vibrate together? 9. Find the temperature at which the fundamental frequency of an organ pipe is independent of small variation in temperature in terms of the coefficient of linear expansion (a) of the material of the tube. 10. A uniform rope of length 12 m and mass 6 kg hangs vertically from a rigid support. A block of mass 2 kg is attached to the free end of the rope. A transverse pulse of wavelength 0.06 m is produced at the lower end of the rope. What is the wavelength of the pulse when it reaches the top of the rope?
By Akhil Tewari, Author Foundation of Physics for jee Main & Advanced, Senior Professor Physics, RAO IIT ACADeMY, Mumbai.
8
physics for you | june ‘15
Here, the references of few are given :
Exact Questions
Exam Q. No.
MTG Book
Q. No.
P. No.
Exam Q. No.
MTG Book
Q. No.
P. No.
1
AIPMT Guide
28
103
24
AIPMT Guide
39
651
3
AIPMT Guide
33
650
27
AIPMT Guide
26
294
6
AIPMT Guide
45
720
29
NCERT Fingertips
38
119
12
AIPMT Guide
28
579
36
AIPMT Guide
81
217
13
AIPMT Guide
56
449
37
AIPMT Guide
56
141
18
AIPMT Guide
21
200
43
AIPMT Guide
120
20
22
AIPMT Guide
1
314
44
AIPMT Guide
153
456
Q. No.
P. No.
Similar Questions Exam Q. No.
MTG Book
Q. No.
P. No.
Exam Q. No.
2
NCERT Fingertips
17
291
20
MTG Book Physics For You May'15
39
34
5
AIPMT Guide
118
615
32
NCERT Fingertips
54
94
7
AIPMT Guide
88
345
34
AIPMT Guide
73
256
8
21
609
39
NCERT Fingertips
38
208
9
AIPMT Guide AIPMT Guide
34
535
41
AIPMT Guide
85
345
15
Physics For You Jan'15
21
15
45
AIPMT Guide
18
446
and more such questions …… 1.
Three blocks A, B and C, of masses 4 kg, 2 kg and 1 kg respectively, are in contact on a frictionless surface, as shown. If a force of 14 N is applied on the 4 kg block, then the contact force between A and B is A
(a) 8 N (c) 2 N 2.
RAl
(d)
p
(a)
1/3
13 (b) 53
RAl
5 R 3 Al
Which of the following figures represent the variation of particle momentum and the associated de-Broglie wavelength? p
(b) 18 N (d) 6 N
Physics for you | june ‘15
3.
C
If radius of the 27 13 Al nucleus is taken to be RAl, then the radius of 125 53Te nucleus is nearly 3 (a) RAl 5
10
B
1/3
53 (c) 13
(b) p
(c)
p
(d)
4.
5.
6.
The two ends of a metal rod are maintained at temperatures 100°C and 110°C. The rate of heat flow in the rod is found to be 4.0 J/s. If the ends are maintained at temperatures 200°C and 210°C, the rate of heat flow will be (a) 8.0 J/s (b) 4.0 J/s (c) 44.0 J/s (d) 16.8 J/s For a parallel beam of monochromatic light of wavelength ‘l’, diffraction is produced by a single slit whose width ‘a’ is of the order of the wavelength of the light. If ‘D’ is the distance of the screen from the slit, the width of the central maxima will be Da 2Da (a) (b) l l Dl 2Dl (c) (d) a a Which logic gate is represented by the following combination of logic gates?
(a) AND (c) OR 7.
A particle is executing SHM along a straight line. Its velocities at distances x1 and x2 from the mean position are V1 and V2, respectively. Its time period is (a) 2 p
(c) 2 p 8.
9.
(b) NOR (d) NAND
V12 + V22 x12
+ x22
x12 V12
+ x22 + V22
(b) 2 p
(d) 2 p
m0n2e r m0ne (c) 2 pr (a)
(b)
(d) Zero
10. A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to v(x) = bx–2n, where b and n are constants and x is the position of the particle. The acceleration of the particle as a function of x, is given by (a) –2b2 x–2n + 1 (b) –2nb2 e–4n + 1 2 –2n – 1 (c) –2nb x (d) –2nb2 x–4n – 1 11. The electric field in a certain region is acting radially outward and is given by E = Ar. A charge contained in a sphere of radius ‘a’ centred at the origin of the field, will be given by (a) 4pe0Aa3 (b) e0Aa3 2 (c) 4pe0Aa (d) Ae0a2 12. A radiation of energy ‘E’ falls normally on a perfectly reflecting surface. The momentum transferred to the surface is (C = Velocity of light) 2E E (a) 2 (b) 2 C C 2E E (c) (d) C C 13. A, B and C are voltmeters of resistance R, 1.5R and 3R respectively as shown in the figure. When some potential difference is applied between X and Y, the voltmeter readings are VA, VB and VC respectively. Then
V12 − V22 x12
− x22
x22 V12
− x12 − V22
Two identical thin plano-convex glass lenses (refractive index 1.5) each having radius of curvature of 20 cm are placed with their convex surfaces in contact at the centre. The intervening space is filled with oil of refractive index 1.7. The focal length of the combination is (a) –50 cm (b) 50 cm (c) –20 cm (d) –25 cm An electron moving in a circular orbit of radius r makes n rotations per second. The magnetic field produced at the centre has magnitude
m0ne 2r
B X
A
(a) VA = VB ≠ VC (c) VA = VB = VC
C
Y
(b) VA ≠ VB ≠ VC (d) VA ≠ VB = VC
14. A rod of weight W is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at a distance d from each other. The centre of mass of the rod is at distance x from A. The normal reaction on A is (a)
W (d − x ) x
(b)
W (d − x ) d
(c)
Wx d
(d)
Wd x
Physics for you | june ‘15
11
15. A wire carrying current I has the shape as shown in adjoining figure. Linear parts of the wire are very long and parallel to X-axis while semicircular portion of radius R is lying in Y-Z plane. Magnetic field at point O is Z
I R I
O
Y I
X m I ^ ^ (a) B = − 0 p i + 2 k 4p R
m I ^ ^ (b) B = 0 p i − 2 k 4p R m I ^ ^ (c) B = 0 p i + 2 k 4p R m I ^ ^ (d) B = − 0 p i − 2 k 4p R 16. A wind with speed 40 m/s blows parallel to the roof of a house. The area of the roof is 250 m2. Assuming that the pressure inside the house is atmospheric pressure, the force exerted by the wind on the roof and the direction of the force will be (rair = 1.2 kg/m3) (a) 2.4 × 105 N, upwards (b) 2.4 × 105 N, downwards (c) 4.8 × 105 N, downwards (d) 4.8 × 105 N, upwards 17. In a double slit experiment, the two slits are 1 mm apart and the screen is placed 1 m away. A monochromatic light of wavelength 500 nm is used. What will be the width of each slit for obtaining ten maxima of double slit within the central maxima of single slit pattern? (a) 0.5 mm (b) 0.02 mm (c) 0.2 mm (d) 0.1 mm 18. A mass m moves in a circle on a smooth horizontal plane with velocity v0 at a radius R0. The mass is attached to a string which passes through a smooth hole in the plane as shown. v0 R0
12
Physics for you | june ‘15
m
The tension in the string is increased gradually R and finally m moves in a circle of radius 0 . The 2 final value of the kinetic energy is
1 2 mv 2 0 1 2 2 (c) mv0 (d) mv0 4 19. Kepler’s third law states that square of period of revolution (T) of a planet around the sun, is proportional to third power of average distance r between sun and planet i.e. T2 = Kr3 here K is constant. If the masses of sun and planet are M and m respectively then as per Newton’s law of gravitation force of attraction between them is GMm , here G is gravitational constant. F= r2 The relation between G and K is described as 1 (a) K = G (b) K = G (c) GK = 4p2 (d) GMK = 4p2 2 (a) 2mv0
(b)
20. Three identical spherical shells, each of mass m and radius r are placed as shown in figure. Consider an axis XX′ which is touching to two shells and passing through diameter of third shell. Moment of inertia of the system consisting of these three spherical shells about XX′ axis is 16 2 mr (a) (b) 4mr2 5 11 2 mr (c) (d) 3mr2 5 21. A ship A is moving Westwards with a speed of 10 km h–1 and a ship B 100 km South of A, is moving Northwards with a speed of 10 km h–1. The time after which the distance between them becomes shortest, is (a) 5 2 h (b) 10 2 h (c) 0 h (d) 5 h 22. The ratio of the specific heats
Cp
Cv degrees of freedom (n) is given by 2 (a) 1 + n
n (b) 1 + 2
1 (c) 1 + n
n (d) 1 + 3
= γ in terms of
23. If in a p–n junction, a square input signal of 10 V is applied, as shown,
(a) 90 J (c) 100 J
(b) 1 J (d) 99 J
28. One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in the figure, then the output across RL will be (a)
(c)
–5 V
(b)
A
5
5V
P(in kPa) 2
B 4
(d)
V(in m3)
24. A certain metallic surface is illuminated with monochromatic light of wavelength, l. The stopping potential for photo-electric current for this light is 3V0. If the same surface is illuminated with light of wavelength 2l, the stopping potential is V0. The threshold wavelength for this surface for photo-electric effect is l p (a) (b) 6 4 (c) 6l (d) 4l 25. A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A dielectric slab of dielectric constant K, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect? 1 2
1 −1 . K
(a) The change in energy stored is CV 2
(b) The charge on the capacitor is not conserved. (c) The potential difference between the plates decreases K times. (d) The energy stored in the capacitor decreases K times. 26. A particle of mass m is driven by a machine that delivers a constant power k watts. If the particle starts from rest the force on the particle at time t is 1 mk t −1/2 (a) 2mk t −1/2 (b) 2 mk −1/2 t (c) (d) mk t −1/2 2 1 27. A Carnot engine, having an efficiency of η = as 10 heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is
6
The change in internal energy of the gas during the transition is (a) 20 J (b) –12 kJ (c) 20 kJ (d) –20 kJ 29. A block of mass 10 kg, moving in x direction with a constant speed of 10 m s–1, is subjected to a retarding force F = 0.1x J/m during its travel from x = 20 m to 30 m. Its final KE will be (a) 275 J (b) 250 J (c) 475 J (d) 450 J 30. Consider 3rd orbit of He+ (Helium), using nonrelativistic approach, the speed of electron in this orbit will be [given K = 9 × 109 constant, Z = 2 and h (Plack’s Constant) = 6.6 × 10–34 J s] (a) 0.73 × 106 m/s (b) 3.0 × 108 m/s 6 (c) 2.92 × 10 m/s (d) 1.46 × 106 m/s 31. A resistance ‘R’ draws power ‘P’ when connected to an AC source. If an inductance is now placed in series with the resistance, such that the impedance of the circuit becomes ‘Z’, the power drawn will be R (a) P (b) P Z R (c) P Z
2
(d) P
R Z
32. A block A of mass m1 rests on a horizontal table. A light string connected to it passes over a frictionless pully at the edge of table and from its other end another block B of mass m2 is suspended. The coefficient of kinetic friction between the block and the table is mk. When the block A is sliding on the table, the tension in the string is m m (1 − mk ) g m m (1 + mk ) g (a) 1 2 (b) 1 2 (m1 + m2 ) (m1 + m2 ) (c)
(m2 + mk m1 ) g (m1 + m2 )
(d)
(m2 − mk m1 ) g (m1 + m2 )
Physics for you | june ‘15
13
33. The refracting angle of a prism is A, and refractive index of the material of the prism is cot (A/2). The angle of minimum deviation is (a) 90° – A (b) 180° + 2A (c) 180° – 3A (d) 180° – 2A 34. On observing light from three different stars P, Q and R, it was found that intensity of violet colour is maximum in the spectrum of P, the intensity of green colour is maximum in the spectrum of R and the intensity of red colour is maximum in the spectrum of Q. If TP, TQ and TR are the respective absolute temperatures of P, Q and R, then it can be concluded from the above observations that (a) TP < TR < TQ (b) TP < TQ < TR (c) TP > TQ > TR (d) TP > TR > TQ 35. A conducting square frame of side ‘a’ and a long straight wire carrying current I are located in the same plane as shown in the figure. The frame moves to the right with a constant velocity ‘V’. The emf induced in the frame will be proportional to
(a) (c)
1 2
(2 x + a) 1 x2
(b) (d)
1 (2 x − a)(2 x + a) 1 (2 x − a)2
36. Two spherical bodies of mass M and 5M and radii R and 2R are released in free space with initial separation between their centres equal to 12R. If they attract each other due to gravitational force only, then the distance covered by the smaller body before collision is (a) 7.5R (b) 1.5R (c) 2.5R (d) 4.5R 37. Two similar springs P and Q have spring constants KP and KQ, such that KP > KQ. They are stretched first by the same amount (case a), then by the same force (case b). The work done by the springs WP and WQ are related as, in case (a) and case (b) respectively (a) WP > WQ ; WQ > WP (b) WP < WQ ; WQ < WP (c) WP = WQ ; WP > WQ (d) WP = WQ ; WP = WQ 14
Physics for you | june ‘15
38. Two particles of masses m1, m2 move with initial velocities u1and u2. On collision, one of the particles get excited to higher level, after absorbing energy e. If final velocities of particles be v1 and v2 then we must have 1 1 1 1 (a) m1u12 + m2u22 − e = m1v12 + m2v22 2 2 2 2 1 2 2 1 2 2 1 2 2 1 2 2 (b) m u1 + m u2 + e = m v1 + m2 v2 2 1 2 2 2 1 2 (c) m12u1 + m22u2 − e = m12v1 + m22v2 1 1 1 1 (d) m1u12 + m2u22 = m1v12 + m2v22 − e 2 2 2 2 39. The approximate depth of an ocean is 2700 m. The compressibility of water is 45.4 × 10–11 Pa–1 and density of water is 103 kg/m3. What fractional compression of water will be obtained at the bottom of the ocean? (a) 1.2 × 10–2 (b) 1.4 × 10–2 –2 (c) 0.8 × 10 (d) 1.0 × 10–2 40. Figure below shows two paths that may be taken by a gas to go from a state A to a state C.
In process AB, 400 J of heat is added to the system and in process BC, 100 J of heat is added to the system. The heat absorbed by the system in the process AC will be (a) 460 J (b) 300 J (c) 380 J (d) 500 J 41. The fundamental frequency of a closed organ pipe of length 20 cm is equal to the second overtone of an organ pipe open at both the ends. The length of organ pipe open at both the ends is (a) 120 cm (b) 140 cm (c) 80 cm (d) 100 cm 42. When two displacements represented by y1 = a sin(wt) and y2 = b cos(wt) are superimposed the motion is a 2 + b2 (a + b) (b) simple harmonic with amplitude 2 (c) not a simple harmonic a (d) simple harmonic with amplitude b (a) simple harmonic with amplitude
43. If energy (E), velocity (V) and time (T) are chosen as the fundamental quantities, the dimensional formula of surface tension will be (a) [EV–2T–2] (b) [E–2V–1T–3] –2 –1 (c) [EV T ] (d) [EV–1T–2]
h 3. (d) : de-Broglie wavelength, l = p 1 or l ∝ , lp = constant p This represents a rectangular hyperbola.
44. Across a metallic conductor of non-uniform cross section a constant potential difference is applied. The quantity which remains constant along the conductor is (a) drift velocity (b) electric field (c) current density (d) current
4. (b) : Rate of heat flow through a rod is given by dQ dT = − KA dt dx Let length of the rod be L. 110 − 100 10 Case I : dT = DT = = dx Dx L L dQ1 10 …(i) \ = − KA dt L dQ1 Also, = 4 J s −1 …(ii) dt dT DT 210 − 200 10 Case II : = = = dx Dx L L dQ2 10 …(iii) \ = − KA dt L So, from equations (i), (ii) and (iii) dQ2 dQ1 = = 4 J s −1 dt dt
45. A potentiometer wire has length 4 m and resistance 8 W. The resistance that must be connected in series with the wire and an accumulator of e.m.f. 2 V, so as to get a potential gradient 1 mV per cm on the wire is (a) 44 W (b) 48 W (c) 32 W (d) 40 W solutions 1. (d) : Here, MA = 4 kg, MB = 2 kg, MC = 1 kg, F = 14 N Net mass, M = MA + MB + MC = 4 + 2 + 1 = 7 kg Let a be the acceleration of the system. Using Newton’s second law of motion, F = Ma 14 = 7a \ a = 2 m s–2 Let F′ be the force applied on block A by block B i.e. the contact force between A and B. Free body diagram for block A
5. (c): Given situation is shown in the figure. For central maxima, l sin q = a Also, q is very-very small so y sin q ≈ tan q = D lD y l \ = , y= a D a Width of central maxima = 2 y =
Again using Newton’s second law of motion, F – F ′ = 4a 14 – F′ = 4 × 2 14 – 8 = F′ \ F′ = 6 N A1/3
2. (d) : Radius of the nucleus R = R0 \
1/3
RAl AAl = RTe ATe
Here, AAl = 27, ATe = 125, RTe = ? 1/3
RAl 27 = RTe 125
5 3 = ⇒ RTe = RAl 3 5
6. (a) :
2lD . a
The Boolean expression of this arrangement is Y = A + B = A⋅B = A⋅B Thus, the combination represents AND gate. 7. (d) : In SHM, velocities of a particle at distances x1 and x2 from mean position are given by V12 = w2(a2 − x12 ) V22
2
= w (a
2
− x22 ) Physics for you | june ‘15
…(i) …(ii) 15
From equations (i) and (ii), we get V12
− V22
w=
=w
2
(x22
V12 − V22 x22 − x12
− x12 )
\ T = 2p
x22 − x12
V12 − V22
8. (a) : Given combination is equivalent to three lenses. In which two are plano-convex with refractive index 1.5 and one is concave lens of refractive index 1.7.
10. (d) : According to question, velocity of unit mass varies as v(x) = bx–2n …(i) dv −2n − 1 …(ii) = − 2nbx dx Acceleration of the particle is given by dv dv dx dv a= = × = ×v dt dx dt dx Using equation (i) and (ii), we get a = (–2nbx–2n – 1) × (bx–2n) = –2nb2 x–4n – 1 11. (a) : According to question, electric field varies as E = Ar Here r is the radial distance.
Using lens maker formula, 1 1 1 = (m − 1) − f R1 R2 For plano-convex lens R1 = ∞, R2 = –20 cm, 1 1 1 0. 5 1 1 \ = = (1.5 − 1) − = = f1 f2 ∞ −20 20 40 So, f1 = f2 = 40 cm For concave lens, m = 1.7, R1 = –20 cm, R2 = 20 cm 1 1 1 \ = (1.7 − 1) − f3 −20 20 7 −2 = 0. 7 × = − 20 100 100 So, f3 = − cm 7 Equivalent focal length (f eq) of the system is given by 1 1 1 1 1 1 1 = + + = + + feq f1 f3 f2 40 −100 / 7 40 1 7 2 1 − =− =− 20 100 100 50 \ feq = – 50 cm =
e 9. (b) : Current in the orbit, I = T e we (2pn)e I= = = = ne (2p/w) 2p 2p Magnetic field at centre of current carrying circular coil is given by m I m ne B= 0 = 0 2r 2r 16
Physics for you | june ‘15
At r = a, E = Aa …(i) Net flux emitted from a spherical surface of radius a is q φ net = en e0 ⇒
(Aa) × (4pa2) =
\ q = 4pe0Aa3
q e0
[Using equation (i)]
12. (d) : Energy of radiation, E = hu = Also, its momentum p =
hC l
h E = = pi l C
E C So, momentum transferred to the surface E E 2E = pi − pr = − − = C C C pr = − pi = −
13. (c) : The current flowing in the different branches of circuit is indicated in the figure.
VA = IR
2I 3 VB = × R = IR 3 2 I VC = × 3R = IR 3 Thus, VA = VB = VC 14. (b) : Given situation is shown in figure. N1 = Normal reaction on A N2 = Normal reaction on B W = Weight of the rod In vertical equilibrium, N1 + N2 = W …(i) Torque balance about centre of mass of the rod, N1x = N2(d – x) Putting value of N2 from equation (i) N1x = (W – N1)(d – x) ⇒ N1x = Wd – Wx – N1d + N1x ⇒ N1d = W(d – x) W (d − x) \ N1 = d 15. (a) : Given situation is shown in the figure.
Parallel wires 1 and 3 are semi-infinite, so magnetic field at O due to them m I ^ B1 = B3 = − 0 k 4pR Magnetic field at O due to semi-circular arc in YZ-plane is given by m I^ B2 = − 0 i 4R Net magnetic field at point O is given by B = B1 + B2 + B3 m I ^ m I^ m I ^ =− 0 k− 0 i − 0 k 4pR 4R 4pR m I ^ ^ = − 0 (p i + 2 k) 4pR 16. (a) : By Bernoulli’s theorem, 1 1 P1 + rv12 = P2 + r v22 2 2 inside
outside
Assuming that the roof width is very small Pressure difference, 1 P1 − P2 = r(v22 − v12 ) 2 Here, r = 1.2 kg m–3, v2 = 40 m s–1, v1 = 0,
A = 250 m2 1 P1 − P2 = × 1.2 (402 − 02 ) 2 1 = × 1.2 × 1600 = 960 N m–2 2 Force acting on the roof F = (P1 – P2) × A = 960 × 250 = 2.4 × 105 N upwards 17. (c): For double slit experiment, d = 1 mm = 1 × 10–3 m, D = 1 m, l = 500 × 10–9 m Dl Fringe width b = d 2lD Width of central maxima in a single slit = a As per question, width of central maxima of single slit pattern = width of 10 maxima of double slit pattern 2lD lD = 10 d a 2d 2 × 10−3 a= = = 0.2 × 10−3 m = 0.2 mm 10 10 18. (a) : According to law of conservation of angular momentum mvr = mv′r′ R v0 R0 = v 0 ; v = 2v0 2 1 2 K 0 2 mv0 v0 2 \ = = K 1 2 v mv 2 2 K v = = (2)2 or K 0 v0 K = 4K0 = 2mv 02
…(i)
(Using (i))
19. (d) : Gravitational force of attraction between sun and planet provides centripetal force for the orbit of planet. GMm mv 2 \ = r r2 GM …(i) v2 = r Time period of the planet is given by 2pr 4p2r 2 , T2 = 2 v v 2 2 4p r T2 = [Using equation (i)] GM r 4p2r 3 T2 = GM T=
Physics for you | june ‘15
…(ii) 17
According to question, T2 = Kr3 Comparing equations (ii) and (iii), we get K=
n C p = + 1 R 2
…(iii)
n +1 R n + 2 2 2 \ γ =1 + γ= = = n (n/2)R Cv n
4p2 \ GMK = 4p2 GM
Cp
20. (b) : Net moment of inertia of the system, I = I1 + I2 + I3 The moment of inertia of a shell about its diameter, 2 I1 = mr 2 3 The moment of inertia of a shell about its tangent is given by 2 5 I2 = I3 = I1 + mr 2 = mr 2 + mr 2 = mr 2 3 3 2
12mr 5 2 = 4mr 2 \ I = 2 × mr 2 + mr 2 = 3 3 3 21. (d) : Given situation is shown in the figure.
23. (b) : Diode is forward bias for positive voltage i.e. V > 0, so output across RL is given by
24. (d) : According to Einstein’s equation hc hc eVs = − l l0 where Vs = Stopping potential l = Incident wavelength l0 = Threshold wavelength or Vs =
photoelectric
hc 1 1 − e l l0
For the first case hc 1 1 3V0 = − e l l0 For the second case Velocity of ship A vA = 10 km h–1 towards west Velocity of ship B vB = 10 km h–1 towards north OS = 100 km OP = shortest distance Relative velocity between A and B is
hc 1 1 − e 2 l l0 Divide eqn. (i) by (ii), we get V0 =
1 1 l − l 0 3= 1 1 2 l − l 0
v AB = v 2A + v B2 = 10 2 km h −1 cos 45° =
1 1 1 1 = − 3 − 2 l l0 l l0 3 3 1 1 − = − 2 l l0 l l0 1 2 = or l0 = 4 l 2 l l0
OP 1 OP ; = OS 2 100
100 100 2 = = 50 2 km 2 2 The time after which distance between them equals to OP is given by OP =
t=
OP 50 2 = ⇒ t = 5h vAB 10 2
22. (a) : For n degrees of freedom, C = n R v 2 Also, Cp – Cv = R n C p = Cv + R = R + R 2 18
25. (b) :
Physics for you | june ‘15
q = CV ⇒ V = q/C Due to dielectric insertion, new capacitance C2 = CK
...(i)
...(ii)
Initial energy stored in capacitor, U1 =
q2 2C
1+
q2 Final energy stored in capacitor, U 2 = 2KC Change in energy stored, DU = U2 – U1
Q1 = 90 J So, 90 J heat is absorbed at lower temperature.
2
q 1 1 2 1 − 1 = CV − 1 K 2C K 2 New potential difference between plates q V V′ = = CK K 26. (c): Constant power acting on the particle of mass m is k watt. or P = k dW = k ; dW = kdt dt W t DU =
Integrating both sides,
∫ dW = ∫ k dt 0
0
⇒ W = kt Using work energy theorem, 1 1 W = mv 2 − m(0)2 2 2 1 2 kt = mv [Using equation (i)] 2
…(i)
2kt m dv Acceleration of the particle, a = dt 1 2k 1 k a= = 2 m t 2mt mk mk −1/2 Force on the particle, F = ma = = t 2t 2 27. (a) : For Carnot engine, v=
T 1 T Efficiency, η = 1 − 1 ; = 1 − 1 T2 10 T2 T1 1 9 =1− = 10 10 T2
…(i)
or
=
5nR PBVB PAVA − [ PV = nRT ] 2 nR nR
5 5 = (PBVB − PAVA ) = (2 × 103 × 6 − 5 × 103 × 4) 2 2 5 3 = ( − 8 × 10 ) = − 20 kJ 2 29. (c): Here, m = 10 kg, vi = 10 m s–1 Initial kinetic energy of the block is 1 1 K i = mvi2 = × (10 kg ) × (10 m s −1 )2 = 500 J 2 2 Work done by retarding force 30 x2 30 x2 W = ∫ Fr dx = ∫ −0.1 xdx = − 0.1 2 20 20 x 1
900 − 400 = − 0. 1 = −25 J 2 According to work-energy theorem, W = Kf – Ki Kf = W + Ki = – 25 J + 500 J = 475 J 30. (d) : Energy of electron in He+ 3rd orbit 4 Z2 E3 = − 13.6 × 2 eV = − 13.6 × eV 9 n = − 13.6 ×
Q2 T2 = Q1 T1
Q1 + W T2 = Q1 T1
Q1 + 10 10 = Q1 9
28. (d) : We know, DU = nCv DT 5R 5R = n (TB − TA ) [for diatomic gas,Cv = ] 2 2
4 × 1.6 × 10−19 J 9.7 × 10−19 J 9 As per Bohr’s model, Kinetic energy of electron in the 3rd orbit = – E3 1 \ 9.7 × 10−19 = mev 2 2
For refrigerator, \
10 10 10 10 1 = ; = −1= Q1 9 Q1 9 9
[Using equation (i)]
v=
2 × 9.7 × 10−19 9.1 × 10−31
= 1.46 × 106 m s −1
Physics for you | june ‘15
19
31. (c): Case I : P = VrmsIrms A cot = 2
V = Vrms × rms R V2 2 = PR P = rms ⇒ Vrms R
...(i)
Case II : Power drawn in LR circuit V R P ′ = Vrms Irms cos φ = Vrms × rms × Z Z R R 2 = Vrms 2 = PR × 2 Z Z
A m = cot 2
A+d A sin 2 2 = A A sin sin 2 2
cos
A d p A A d p A sin − = sin + ; − = + 2 2 2 2 2 2 2 2 \ d = p – 2A = 180° – 2A
[Using eqn (i)] 2
R P′ = P 2 Z 32. (a) : Given situation is shown in the figure. Here, N = m1g f = mkN = mkm1g
A+d sin 2 A sin 2
…(i)
34. (d) : According to Wein’s displacement law lmT = constant …(i) For star P , intensity of violet colour is maximum For star Q, intensity of red colour is maximum. For star R, intensity of green colour is maximum. Also, lr > lg > lv Using equation (i), Tr < Tg < Tv TQ < TR < TP 35. (b) : Here, PQ = RS = PR =QS = a
Let a be the acceleration of blocks. Equation of motion for A and B T – f = m1a m2g – T = m2a Adding equation (ii) and (iii), we get m2g – f = (m1 + m2)a a=
…(ii) …(iii)
m2 g − f m1 + m2
Put this value of a in equation (iii) (m g − f ) T = m2 g − m2 2 m1 + m2 m1m2 g + m1m2mk g m1m2(1 + mk )g = = m1 + m2 m1 + m2 (Using (i)) A+d sin 2 33. (d) : As m = A sin 2 20
Physics for you | june ‘15
Emf induced in the frame e = B1(PQ)V – B2(RS)V =
m0 I m0 I aV aV − 2p(x + a/2) 2p(x − a/2)
m I 2 2 aV = 0 − 2p (2x − a) (2x + a) m I 2a = 0 ×2 aV 2p (2x − a)(2x + a) 1 \ e∝ (2x − a)(2x + a) 36. (a) 37. (a) : Here, KP > KQ Case (a) : Elongation (x) in each spring is same. 1 1 WP = K P x 2 , WQ = K Q x 2 2 2 \ WP > WQ
Case (b) : Force of elongation is same. So, x1 =
400 − 20 = DQAC − (2 × 104 × 2 × 10−3 +
F F and x2 = KP KQ
1 1 F2 KP x12 = 2 2 KP 1 1 F2 WQ = KQ x22 = 2 2 KQ
1 × 2 × 10−3 × 4 × 104 ) 2 380 = DQAC – (40 + 40), DQAC = 380 + 80 = 460 J
WP =
\ WP < WQ
38. (a) : Total initial energy of two particles 1 1 = m1u12 + m2u22 2 2 Total final energy of two particles 1 1 = m2v22 + m1v12 + e 2 2 Using energy conservation principle, 1 1 1 1 m u2 + m u2 = m v 2 + m v 2 + e 2 11 2 2 2 2 11 2 2 2 1 1 1 1 \ m1u12 + m2u22 − e = m1v12 + m2v22 2 2 2 2 39. (a) : Depth of ocean d = 2700 m Density of water, r = 103 kg m–3 Compressibility of water, K = 45.4 × 10–11 Pa–1 DV =? V Excess pressure at the bottom, DP = rgd = 103 × 10 × 2700 = 27 × 106 Pa DP We know, B = (DV /V ) 1 D V D P = K . DP K = = V B B = 45.4 × 10–11 × 27 × 106 = 1.2 × 10–2
40. (a) : As initial and final points are same so DUABC = DUAC AB is isochoric process. DWAB = 0 DQAB = DUAB = 400 J BC is isobaric process. DQBC = DUBC + DWBC 100 = DUBC + 6 × 104 (4 × 10–3 – 2 × 10–3) 100 = DUBC + 12 × 10 DUBC = 100 – 120 = – 20 J As, DUABC = DUAC DUAB + DUBC = DQAC – DWAC
41. (a) : For closed organ pipe, fundamental frequency is given by v uc = 4l For open organ pipe, fundamental frequency is given by v uo = 2l ′ 2nd overtone of open organ pipe 3v u′ = 3uo ; u′ = 2l ′ According to question, uc = u′ v 3v = 4l 2l ′ l′ = 6l Here, l = 20 cm, l′ = ? \ l′ = 6 × 20 = 120 cm
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42. (a) : Here, y1 = asinwt p y2 = b cos wt = b sin wt + 2
flow of electrons will be uniform so current will be constant. 45. (c): Required potential gradient = 1 mV cm–1 1 = V m −1 10 Length of potentiometer wire, l = 4 m
Hence, resultant motion is SHM with amplitude a 2 + b2 . 43. (a) : Let S = kEaV bTc where k is a dimensionless constant. Writing the dimensions on both sides, we get [M1L0T–2] = [ML2T–2]a[LT–1]b [T]c
= [M a L2a + b T−2a − b + c ] Applying principle of homogeneity of dimensions, we get, a = 1 ... (i) 2a + b = 0 ... (ii) – 2a – b + c = – 2 ... (iii) Adding (ii) and (iii), we get c=–2 From (ii), b = – 2a = – 2 \ S = kEV–2 T–2 or [S] = [E V–2T–2]
44. (d) : The area of cross section of conductor is non uniform so current density will be different but the
So potential difference across potentiometer wire 1 …(i) = × 4 = 0. 4 V 10 In the circuit, potential difference across 8 W =I×8=
2 ×8 8+R
…(ii)
Using equation (i) and (ii), we get 2 ×8 8+R 4 16 = , 8 + R = 40 10 8 + R \ R = 32 W 0. 4 =
nn
AIPMT keys leaked, students get answers via bluetooth vests Haryana police have nabbed four people, including two dentists and an MBBS student, from Rohtak for allegedly passing on answer keys to students using vests with SIM card units and bluetooth-enabled earpieces during the All India Pre Medical Test (AIPMT). Apart from probing how 90 answer keys to the highly competitive all-India test were leaked, police are also investigating at least nine candidates who allegedly paid the gang around Rs 15-20 lakh for the “help” in their bid to become doctors. Police said the gang may have spread its reach to other states too, particularly Bihar and Rajasthan. They added that the accused claimed they had purchased the “engineered vests” from a shop in new Delhi. Of the four arrested, police have identified two as BDS doctors Sanchit and Bhupender, one as second-year MBBS student Ravi and the fourth as Rajesh. The alleged “kingpin” of this racket, Roop Singh Dangi, is on the run, police said. Another MBBS doctor – his identity has been withheld – is also under the scanner for acting as a “mediator” between the accused and the students. Shrikant jadhav, Inspector General of Police (Rohtak range), said the arrests followed a tip-off. “We alerted the examination authorities and ordered a thorough frisking of every student. Also, we received concrete information about the four accused who were staying in a hotel in Panipat. We tracked their mobile phones to the jhajjar bypass in Rohtak and caught them,” jadhav told The Indian express. elaborating on the “special” vests, one of the investigating officers said, “They had sim card units linked to earpieces via bluetooth. Specially configured phones were also supplied to some students. Soon after the exam began, the
accused started sending answer keys to the nine students from whom they had allegedly taken Rs 15-20 lakh each. For students with phones, answer keys were sent through WhatsApp, and for those using the earpieces, they were passed on through phone calls.” Police said the “engineered vests” and keys to the 90 questions were recovered from the four accused who were produced in a Rohtak court – they were sent to police custody for four days for further interrogation. ”So far, we have received a list of nine students who appeared in the AIPMT exams by allegedly paying money to the accused. Raids are being conducted to nab all those involved,” jadhav said. ”We have also sent teams to nab those who sold these vests to the accused. each vest was purchased by the accused at a cost of approximately Rs 9,000 each,” he added. Police also suspect that at least one of the accused – Ravi, the MBBS student – may have resorted to similar means to pass his own medical entrance test. ”He undertook PMT coaching from an institute in Kota, Rajasthan in 2005-06 and got through the exam in 2007. But after eight years, he is still in the second year. It appears from interrogation that he might have got through the exam using unfair means,” the investigating officer said. As for Dangi, the alleged “kingpin”, police said he is a Meham resident who operates from Alwar in Rajasthan. They believe that he allegedly procured and supplied the answer keys to the four accused. During their interrogation, police said, the accused also claimed that the shop from where they purchased the vests had sold 700 such units in Bihar. “There could be a possibility that the magnitude of this racket is much larger in Bihar and Rajasthan,” jadhav said. Courtesy : The Indian Express
22
Physics for you | june ‘15
chapterwise McQ’s for practice
Useful for All National and State Level Medical/Engg. Entrance Exams units and MeasureMent
1. Which of the following units denotes the dimensions [ML2Q–2], where Q denotes the electric charge? (a) weber (Wb) (b) Wb/m2 (c) henry (H) (d) H/m2 2. In an experiment of simple pendulum, the errors in the measurement of length of the pendulum (L) and time period (T) T) are 3% and 2% respectively. The T L maximum percentage error in the value of is T2 (a) 5% (b) 7% (c) 8% (d) 1% a a − t2 in the equation P = , b bx where P is pressure, x is distance and t is time are (a) [M2LT–3] (b) [ML0T–2] 3 –1 (c) [ML T ] (d) [MLT–3]
3. The dimensions of
4. In the following equation, x, x t and F represent displacement, time and force respectively, 1 F = a + bt + + A sin(wt + f) c + d⋅x The dimensional formula for A·d d is –1 –1 –1 (a) [T ] (b) [L ] (c) [M ] (d) [TL–1] 5. What is the number of significant figures in 0.310 × 103? (a) 2 (b) 3 (c) 4 (d) 5 6. The study of the earth’s surface is normally performed with (a) rectangular cartesian co-ordinates (b) gaussian system (c) cartesian co-ordinates, but spherical (d) none of these 7. Which one of the following is dimensionally incorrect?
Capacitance C = [M–1L–2T4A2] Magnetic field induction B = [ML0T–2A–1] Coefficient of self-induction L = [ML2T–2A–1] Specific resistance r = [M L3T–3A–2] DV , where e0 is the 8. A quantity X is given by e0 L Dt permittivity of free space, L is length, DV is potential (a) (b) (c) (d)
difference and Dt is time interval. The dimensional formula for X is the same as that of (a) resistance (b) charge (c) voltage (d) current 9. The moment of inertia of a body rotating about a given axis is 12.0 kg m2 in the SI system. What is the value of the moment of inertia in a system of units in which the unit of length is 5 cm and the unit of mass is 10 g? (a) 2.4 × 103 (b) 6.0 × 103 5 (c) 5.4 × 10 (d) 4.8 × 105 10. A wire has a mass (0.3 ± 0.003) g, radius (0.5 ± 0.005) mm and length (6 ± 0.06) cm. The maximum percentage error in the measurement of its density is (a) 1% (b) 2% (c) 3% (d) 4% 11. Match List I with List II and select the correct answer : List I List II A. spring constant 1. [M1L2T–2] B. pascal 2. [M0L0T–1] C. hertz 3. [M1L0T–2] D. joule 4. [M1L–1T–2] A B C D (a) 3 4 2 1 (b) 4 3 1 2 (c) 4 3 2 1 (d) 3 4 1 2 Physics for you | june ‘15
23
12. Which of the following statements is incorrect regarding significant figures? (a) All the non-zero digits are significant. (b) All the zeros between two non-zero digits are significant. (c) Greater the number of significant figures in a measurement, smaller is the percentage error. (d) The power of 10 is counted while counting the number of significant figures. 13. The length and breadth of a rectangular sheet are 16.2 cm and 10.1 cm, respectively. The area of the sheet in appropriate significant figures and error is (a) 164 ± 3 cm2 (b) 163.62 ± 2.6 cm2 2 (c) 163.6 ± 2.6 cm (d) 163.62 ± 3 cm2 14. The density of a material in CGS system of units is 4 g cm–3. In a system of units in which unit of length is 10 cm and unit of mass is 100 g, the value of density of material will be (a) 0.04 (b) 0.4 (c) 40 (d) 400 15. Distance Z travelled by a particle is defined by Z = a + bt + gt2. Dimensions of g are (a) [LT–1] (b) [L–1T] –2 (c) [LT ] (d) [LT2] KineMatics
16. A cricketer can throw a ball to a maximum horizontal distance of 100 m. With the same speed how much high above the ground can the cricketer throw the same ball? (a) 50 m (b) 100 m (c) 150 m (d) 200 m 17. A particle moving along the x axis has position given by x = (24t – 2.0t3) m, where t is measured in s. What is the magnitude of the acceleration of the particle at the instant when its velocity is zero? (a) 24 m s–2 (b) zero –2 (c) 12 m s (d) 48 m s–2 18. 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 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 19. Six vectors, a through f have the magnitudes and directions indicated in the figure. Which of the following statements is true? 24
Physics for you | june ‘15
(a) b + c = (b) d + c = (c) d + e = (d) b + e =
f f f f
b
a
d
c
f
e
20. A particle is projected vertically upwards from a point A on the ground. It takes time t1 to reach a point B, but it still continues to move up. If it takes further time t2 to reach the ground from point B. Then height of point B from the ground is 1 2 (a) g (t1 + t2 ) (b) gt1t2 2 1 1 (c) g (t1 + t2 )2 (d) gt1t2 2 8 21. A rectangular vessel when full of water, takes 10 min to be emptied through an orifice in its bottom. How much time will it take to be emptied when half filled with water? (a) 9 min (b) 7 min (c) 5 min (d) 3 min 22. Time taken by the projectile to reach u from A to B is t, then 60° the distance AB is 30° A equal to (a) 2ut (b) 3 ut (c)
3 ut 2
(d)
B
ut 3
23. A ball A is thrown up vertically with a speed u and at the same instant another ball B is released from a height h. At time t, the speed of A relative to B is (a) u (b) 2u (c) u – gt
(d)
u2 − gt
24. A car covers the first one-third of a distance x at a speed of 10 km h–1, the second one-third at a speed of 20 km h–1 and the last one-third at a speed of 60 km h–1. Find the average speed of the car over the entire distance x. (a) 10 km h–1 (b) 12 km h–1 –1 (c) 18 km h (d) 20 km h–1 25. The co-ordinates of a moving particle are x = at2, y = bt2 where a and b are constants. The velocity of the particle at any moment is (a) 2t a2 + b2
(b) 2t a + b
(c) 2t a2 − b2
(d) 2 a2 + b2
26. The speed of a projectile when it is at its greatest height is 2 / 5 times its speed at half the maximum height. What is its angle of projection? (a) 30° (b) 60° (c) 45° (d) 0° 27. A passenger is walking on an escalator at a speed of 6 km/h relative to escalator. The escalator is moving at 3 km/h relative to ground and has a total length of 120 m. The time taken by him to reach the end of the escalator is (a) 16 s (b) 48 s (c) 32 s (d) 80 s 28. A body is projected such that its kinetic energy at the top is (3/4)th of its initial kinetic energy. What is the angle of projection with the horizontal? (a) 30° (b) 60° (c) 45° (d) 120° 29. A particle is moving on circular path as shown in the figure. Then displacement from P1 to P2 is (a) 2r cos
q 2
(c) 2r sinq
O
r r
P2 P1
q 2 q (d) 2r sin 2
(b) 2r tan
30. A particle moves in x-y plane. The position vector 2 of particle at any time t is r = {(2t )i + (2t )j} m. The rate of change of q at time t = 2 s (where q is the angle which its velocity vector makes with positive x-axis) is 2 1 rad s −1 rad s −1 (a) (b) 17 14 4 6 rad s −1 rad s −1 (c) (d) 7 5 solutions
1. (c) : [ML2Q–2] = [ML2T–2A–2] [Wb] = [ML2T–2A–1] Wb −2 −1 2 =[M T A ] m [henry] = [ML2 T–2 A–2] H −2 −2 2 = [MT A ] m
ML2 Obviously henry (H) has dimensions . Q2
2. (b) : Time period of simple pendulum is T = 2π
L g
Squaring both sides, we get L \ T 2 = 4π2 g or g = 4 π2
L
T2 The maximum percentage error in g is
...(i)
Dg DL DT × 100 = × 100 + 2 × 100 g L T = 3% + 2 × 2% = 7% From (i), we get g L = 2 T 4 π2 L The maximum percentage error in is T2 L D T 2 × 100 = Dg × 100 = 7% L g 2 T 3. (b) : P =
a − t2 bx
[a] = [T2], as t2 is subtracted from a. From, P =
a − t2 t2 = bx bx
t2 [T2 ] [b] = = = [M–1L0T4] Px [ML−1T −2 ][L] \
[T2 ] a = = [ML0T–2] b [M −1L0 T 4 ]
1 + A sin(wt + f) c + d⋅x As sin(wt + f) is dimensionless, therefore A has dimensions of force. \ [A] = [F] = [MLT–2] As each term on RHS represents force 1 \ =F c + d⋅x 1 =F c 1 1 \ [c] = = = [M −1L−1T2 ] [F ] [MLT−2 ]
4. (b) : F = a + bt +
Physics for you | june ‘15
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As c is added to d·x, therefore dimensions of c are same that of d·x. \ [d·x] = [c]
1
[c] [M −1L−1T2 ] = = [M −1L−2 T2 ] [x] [L] The dimensional formula for A·d is [A·d] = [MLT–2][M–1L–2T2] = [L–1] or
1
5. (b) 6. (c) : Spherical cartesian co-ordinates are used with latitudes and longitudes. Carte also means map in French. This is derived from Descrate the great mathematician. 7. (c) : [C] = [B] =
a
12. 13.
...(i)
(Using (i))
Physics for you | june ‘15
(πr 2 )L
F [MLT −2 ] = = [M1L0 T−2 ] [L] x pascal = unit of pressure F [MLT −2 ] = = = [M1L−1T−2 ] A [L2 ] 1 hertz = unit of frequency = T = [M0L0T–1] joule = unit of work = force × distance =[MLT–2] [L] = [M1L2T–2]. (d) : The power of 10 is irrelevant to the determination of significant figures. (a) : Let length and breadth of a rectangular sheet are measured by using a metre scale as 16.2 cm and 10.1 cm respectively. Each measurement has three significant figures. \ Length l can be written as l = 16.2 ± 0.1 cm = 16.2 cm ± 0.6% Similarly, the breadth b can be written as b = 10.1 ± 0.1 cm = 10.1 cm ± 1% Area of the sheet, A = l × b = 163.62 cm2 ± 1.6% =163.62 ± 2.6 cm2 Therefore, as per rule, area will have only three significant figures and error will have only one significant figure. Rounding off, we get A = 164 ± 3 cm2 (c) : As n1u1 = n2u2 100 g g 4 = n2 ⇒ n2 = 40 cm3 (10 cm)3 (c) : By homogeneity of dimensions of LHS and RHS, −2 Distance (LHS) = [g] × [T]2 \ [ g] = [LT ]
11. (a) : Spring constant =
e0 A C = L
b
Dm 0.003 Dr 0.005 DL 0.06 = , = , = m 0. 3 r 0. 5 L 6 m
= 1 + 2 + 1 = 4%
[R][ A] [ML2T−3A −2 ][L2 ] = = [ML3T−3A −2 ] [L] [L]
c
M L T 9. (d) : n2 = n1 1 1 1 M2 L2 T2 Dimensional formula of moment of inertia = [ML2T0] \ a = 1, b = 2, c = 0 Here, n1 = 12.0, M1 = 1 kg, M2 = 10 g L1 = 1 m, L2 = 5 cm, T1 = 1 s, T2 = 1 s
2
Dr Dm 2Dr DL \ r × 100 = m + r + L × 100 0.003 2 × 0.005 0.06 = + + × 100 0. 5 6 0. 3
[F ] [MLT−2 ] = = [ML0T−2 A−1] [I ][l] [A][L]
Choice (c) is dimensionally wrong.
26
As r =
[q ] [AT] = = [M −1L−2T4 A2 ] − 2 2 [W ] [ML T ]
Dq 8. (d) : As, C = DV A Dq or e0 = L DV (Dq)L or e0 = A(DV ) DV X = e0 L (Given) Dt (Dq)L DV \ X= L A(DV ) Dt But [A] = [L]2 Dq \ X= = current Dt
(Using (i))
10. (d) : Here,
2
W q [t ] ML2T−2 [T] e [L] = = = AT [A] [i] di dt = [ML2T−2 A −2 ] [r] =
0
1000 g 100 cm = 12 × ×1 10 g 5 cm = 12 × 100 × 400 = 4.8 × 105
[d] =
2
2
1 kg 1 m 1 s n2 = 12.0 10 g 5 cm 1 s
...(i)
14.
15.
16. (a) : Here, Rmax = 100 m Rmax = of ball)
Therefore, initial velocity of the particle is
2
u (where u is the velocity of the projection g
or u2 = 100g
...(i)
Using v2 – u2 = 2as (0)2 – 100g = 2 (–g) (H) or H = 50 m
[Using (i)]
d2 x dt
2
=
Therefore, height of point B from the ground is 1 1 t +t h = ut1 − gt12 = g 1 2 t1 − gt12 2 2 2
d (24 − 6t 2 ) m s −2 dt
= –12t m s–2
10 min t \ t2 = 1 = ≈ 7 min 2 2
For v = 0, we get 24 – 6t2 = 0 or t = 2 s Hence, at t = 2 s, the acceleration will be a = –12(2) m s–2 = –24 m s–2 Its magnitude is 24 m s–2.
22. (d) : Refer the figure below. Horizontal component of velocity at A. u
18. (b) : Let d be distance of the target from the point of projection. \
u sin(2 × 75°)
or and or
2
2
g
A
= d − 10
u = d − 10 2g u2 sin(2 × 45°) g
...(i) = d + 10
u2 = d + 10 g
...(ii)
Divide (i) by (ii), we get d − 10
1 = or d = 30 m d + 10 2 19. (c) : As per the laws of vector addition, d +e = f This is as shown in adjacent figure.
(Using (i))
or
dx d = (24t − 2.0t 3 ) m s −1 dt dt = 24 – 6t2 m s–1
Acceleration, a =
...(i)
t2 t t 1 h = g 1 + 1 2 − gt12 or h = 1 gt1t2 2 2 2 2 21. (b) : If A0 is the area of orifice at the bottom below the free surface and A that of vessel, time t taken to empty the tank, A 2H \ t1 = H1 = H1 t= t2 H2 H1 / 2 A0 g
17. (a) : Given : x = 24t – 2.0t3 m Velocity, v =
t +t u= g 1 2 2
e f
d
20. (d) : Time taken for the particle to reach the highest t +t point is 1 2 . 2 As v = u – gt At highest point, v = 0
uH = u cos60° =
B
60° 30°
C
u ut \ AC = uH × t = 2 2
AB = AC sec 30° =
ut ut 2 × = 2 3 3
23. (a) : At time t, Velocity of A, vA = u – gt (upwards) Velocity of B, vB = gt (downwards) = –gt (upwards) Relative velocity of A with respect to B is vAB = vA – vB = (u – gt) – (–gt) = u
B uB = 0 h uA = u A
24. (c) : For first one-third of distance x Distance covered = km 3 speed = 10 km h–1. The time taken for the journey, x /3 x h= h 10 30 For the next one-third of distance : x Distance covered = km. 3 t1 =
Physics for you | june ‘15
27
Speed = 20 km h–1 The time taken for travel is x /3 x t2 = h= h 20 60 For the last one-third of distance : x Distance covered = km. 3 –1 Speed is 60 km h The time taken for travel is x /3 x t3 = h= h 60 180 total distance x = \ Average Speed = x x x total time + + 30 60 180 180 x = = 18 km h–1 10 x dx dy 25. (a) : v x = = 2at ; v y = = 2bt dt dt \
2 2 v = v 2x + v 2y = 4a2t 2 + 4b2t 2 = 2t a + b
26. (b) : Maximum height, 2
2
2
2
x
H and v 2y = u2 sin2 q − 2 g × = u2 sin2q − gH 2
u2 sin2 q u2 sin2 q = (Using (i)) 2 2 1/2 H \ Net velocity at height = v 2x + v 2y 2 As per question, 1/2 2 2 2 2 2 v x + v 2y = v H or v + v 2y = v H 5 5 x 2 2 u2 2 or sin2 q = u2 cos2 q u cos q + 5 2 or sin q = 3 cos q or sin2q = 3cos2q v 2y = u2 sin2 q −
(
(
)
or tan q = 3 = tan 60°
)
(
)
or q = 60° 27. (b) : Velocity of the passenger with respect to ground vPG = vPE + vEG = 6 + 3 = 9 km h–1 120 m x 240 s = 48 s Time taken t = = = 5 v PG 5 −1 9× m s 18 28
According to the given problem 1 3 1 m(u cos q)2 = × mu2 2 4 2 3 3 2 or cos q = or cos q = = cos 30° 4 2 or q = 30° 29. (d) :
Physics for you | june ‘15
P2 r x P1 r
O
According to cosine formula, cos q =
u sin q u sin q ...(i) or gH = 2g 2 Velocity at highest point, vH = u cosq Let vx, vy be the horizontal and vertical velocity of H projectile at height . Then 2 v = ucosq H=
28. (a) :
r2 + r2 − x2 2r 2
or 2r 2 cos q = r 2 + r 2 − x 2 or x2 = 2r2 – 2r2cosq = 2r2 [1 – cosq] q = 2r 2 2 sin2 2 Displacement from P1 to P2 is q x = 2r sin 2 ^ ^ 30. (a) : Given, r = {(2t ) i + (2t 2 ) j } m Comparing it with standard equation of position ^
^
vector, r = x i + y j , we get x = 2t and y = 2t2 dy dx = 4t = 2 and v y = ⇒ vx = dt dt v y 4t \ tan q = = = 2t vx 2 Differentiating with respect to time we get, dq =2 dt dq (1 + tan2 q) = 2 dt dq 2 = dt 1 + 4t 2 (sec2 q) or or
or
(1 + 4t 2 )
dq =2 dt
2 2 dq at t = 2 s, = rad s −1 = 2 dt 1 + 4(2) 17
R
3. (b) :
O
Solution Set-22
1. (b) : The woman experiences three forces; mg, her weight acting vertically downwards; N1, reaction due to her weight; N2, horizontal reaction which provides the centripetal acceleration.
G
B A
R
W
a/2 a = As sin λ sin(p − q)
⇒ sin q = 2 sin λ \ q = sin −1(2 sin λ)
4. (b) : At NTP, temperature = 273 K and pressure = 105 N m–2 gRT RT P v= , = M M ρ
From Newton’s second law,
mv 2 ∑ Fx = N 2 = r SFy = N1 – mg = 0 v = (2pr)u (where u is frequency) = (2p × 9) (6/60) = 1.8p m s–1 Therefore, 2
(50)(1.8p) = 178 N 9 N1 = mg = 490 N The magnitude of her weight is the magnitude of the resultant force exerted on her by the chair. N2 =
N = N12 + N 22 = 4902 + 1782 = 521 N 2. (c): Let T be the time of flight and u be the initial velocity of the stone then
u A
and T =
u2 sin 2q g
...(i)
2u sin q g
...(ii)
Eliminating u, we get tan q =
10 × 102 1 gT 2 = = 4R 4 × 250 3 3
\ q = 30°
v=
gP ρ
⇒ g=
v 2ρ (330)2 × 1.3 = = 1. 4 P 105
and g = 1 + (2/f) ⇒ f = 2/(g – 1) ⇒ f = 5 5. (d) : Along the vertical direction, Net impulse due to normal = change in momentum ...(i) ∫ Ndt = m (v/2) + mv cos q (where N is the normal by the floor on the ball) Along the horizontal direction, Friction force, f = mN Let horizontal velocity of the ball after collision = v′ Net impulse due to friction = change in momentum in horizontal direction −m ∫ Ndt = mv ′ − mv sin q
⇒ –m[m(v/2) + mv cosq] = mv′ – mv sinq (Using eqn. (i)) \ v′ = v sin q – m(v/2) + v cos q} 6. (c):
mg 2 mg T ′ = Tcosq = cos q 2sin q mg \ T′= 2 tan q Tsinq =
B 2R
2R =
⇒
7. (c): Tension T3 required to move third block = mmg. Tension T2 to move 2nd block = mmg + 2T3 = 3mmg. Force F required to move the first block = 2T2 + mmg = 7mmg. physics for you | june ‘15
29
8. (d) : Change in position of CM in vertical direction cos 30° ∆h = l +l 2
d2 y 2
B
dy 2 1 + dx R= d2 y
lcos30= l 3/2
60º
60º l
1 kg
JT
JT
30º
(1 + 1)3/2 = 10 (2)3/2 1 10
v2 36 9 2 aR = T = = units R 20 2 10
l 3 Work done by gravity = mg ∆ h = mg + l 4 3 +4 = mgl 4 1 kg
=
= 10 2 × 2 × 2 = 20 2 Normal acceleration
F
9. (b) :
3/2
dx 2
l
C
1 10
dx Radius of curvature at the given point is
A
CM
=
solution of may 2015 crossword
10 cos30º N s 10 N s
30º
Here 10cos30° – JT = JT [Both the masses will move with same velocity along the string] 5 ⇒ JT = 5 cos 30° = 3N s 2 10. (b) : Equation of path, x2 20 x dy x dx v = , vy = 10 x dt 10 dt At x = 10 vy = vx y=
Total v =
v 2x
+ v 2y ,
6=
36 = = 18 , v x = 3 2 2 vy = 3 2
2v 2x
v 2x
Now since the magnitude of velocity is 6 m s–1 constant, tangential acceleration will be zero. It has only normal acceleration. Now x2 y= 20 dy 2x x \ dy = = dx x =10 = 1 dx 20 10 30
physics for you | june ‘15
Guneet Kaur Harsh Gupta rizwan Khan
Winners (May 2015)
solution senders (April 2015) Aditya srivastava Chandra shekhar Panigrahi rajneesh Kumar
Solution Senders of Physics Musing sEt-22 1. Swayangdipta Bera 2. Debajyoti Dash (West Bengal) 3. Sayantan Bhanja (West Bengal) sEt-21 1. Komal Khatri (Pune) 2. Girish Ranjan (Bihar)
1.
2.
A wave is represented by y(x, t) = a sin(kx – wt + f). The phase of the wave is (a) f (b) kx – wt (c) wt + f (d) kx – wt + f Name of units of some physical quantities are given in List I and their dimensional formulae are given in List II. Match List I with List II and select the correct answer. List I Pa s
P. Q. N m K–1 R. J kg–1K–1 S. W m–1 K–1
3.
4.
5.
1. 2. 3. 4.
List II [M0L2T–2K–1] [MLT–3K–1] [ML–1T–1] [ML2T–2K–1]
P Q R S (a) 4 3 1 2 (b) 3 2 4 1 (c) 3 1 4 2 (d) 3 4 1 2 The apparent weight of a person in a lift moving downwards is half his apparent weight in the same lift moving upwards with the same acceleration. The acceleration of the lift is g g g (a) g (b) (c) (d) 2 3 4 The ratio of the angular velocities of the earth about its own axis and the hour hand of a watch is (a) 1 : 2 (b) 2 : 1 (c) 1 : 12 (d) 12 : 1 Consider the following statements : Nuclear force is 1. charge independent 2. long range 3. central Which of the above statements is/are correct? (a) 1 only (b) 1 and 2 (c) 2 and 3 (d) 1 and 3
6.
The magnetic susceptibility of a material of a rod is 299. The permeability of the material of the rod is (m0 = 4p × 10–7 H m–1) (a) 3771 × 10–5 H m–1 (b) 3771 × 10–6 H m–1 (c) 3771 × 10–7 H m–1 (d) 3771 × 10–8 H m–1
7.
A particle motion on a space curve is governed by x = 2sin t, y = 3cost and z = 5 sin t. The speed of the particle at any instant is (a) 3 2 sint (b) 3 cos 2t (c) 3 sin 2t
8.
9.
(d) independent of time
A TV transmitting antenna is 128 m tall. If the receiving antenna is at the ground level, the maximum distance between them for satisfactory communication in LOS mode is (Radius of the earth = 6.4 × 106 m) 128 (a) 64 10 km (b) km 10 64 (c) 128 10 km (d) km 10 A battery of emf 3 V and internal resistance 0.2 W is being charged with a current of 5 A. What is the potential difference between the terminals of the battery? (a) 2 V (b) 3 V (c) 3.5 V (d) 4 V
10. When NaCl is added to water, the surface tension of
water (a) increases (b) decreases (c) remains constant (d) nothing can be said
11. A person walks on a straight road from his house
to a market 2.5 km away with a speed of 5 km h–1. Finding the market closed, he instantly turns back and reaches his house with a speed of 7.5 km h–1. The average speed of the person is physics for you | june ‘15
31
5 14 (b) m s −1 m s −1 3 3 1 5 (c) m s −1 (d) m s −1 3 6 12. The half life of a radioactive element is 10 h. The fraction of initial activity of the element that will remain after 40 h is 1 1 1 1 (a) (b) (c) (d) 4 16 8 2 13. Light travels from air to water, from water to glass and then again from glass to air. If x represents refractive index of water with respect to air, y represents refractive index of glass with respect to water and z represents refractive index of air with respect to glass, then which one of the following is correct? (a) xy = z (b) yz = x (c) zx = y (d) xyz = 1
18. The successive resonance frequencies in an open
14. Which of the following parameters is the same for
wave-particle duality? (a) Wave-particle duality holds for matter particles but not for light. (b) Wave-particle duality holds for light but not for matter particles. (c) Wave-particle duality holds for electrons but not for protons. (d) Wave-particle duality holds for light as well as for matter particles.
(a)
molecules of all gases at a given temperature? (a) Mass (b) Speed (c) Momentum (d) Kinetic energy
15. In a meter bridge experiment, the length AB of the
wire is 1 m. The resistors X and Y have values 5 W and 2 W respectively. When a shunt resistance S is connected to X, the balancing point is found to be 0.625 m from A. Then, the resistance of the shunt is
organ pipe are 1944 Hz and 2600 Hz. If the speed of sound in air is 328 m s–1, then the length of the pipe is (a) 0.40 m (b) 0.04 m (c) 0.50 m (d) 0.25 m
19. A machine gun fires 240 bullets per minute with a
velocity of 600 m s–1. If the mass of each bullet is 10 g, the power of the gun is (a) 7.2 kW (b) 72 kW (c) 3.6 kW (d) 36 kW
20. The sensitivity of a galvanometer that measures
1 times by using shunt 40 resistance of 10 W. Then, the value of the resistance of the galvanometer is (a) 400 W (b) 410 W (c) 30 W (d) 390 W current is decreased by
21. Which one of the following is correct about
22. Work done to increase the temperature of
one mole of an ideal gas by 30°C, if it is expanding under the condition V ∝ T 2/3 is (R = 8.31 J mol–1K–1) (a) 116.2 J (b) 136.2 J (c) 166.2 J (d) 186.2 J
(a) 5 W
(b) 10 W
(c) 7.5 W (d) 12.5 W
16. The equation of trajectory of a projectile is
5 y = 10 x − x 2 m. The range of the projectile is 9
(a) 36 m (b) 24 m (c) 18 m (d) 9 m 17. A ray of light travels from an optically denser
medium towards a rarer medium. The critical angle for the two media is C. The maximum possible angle of deviation of the ray is p (a) − C (b) p – 2C 2 p (c) 2C (d) + C 2
32
physics for you | june ‘15
23. Electric charges A and B attract each other. Electric
charges B and C also attract each other. But electric charges C and D repel each other. If A and D are held close together then which one of the following is correct? (a) They cannot affect each other. (b) They attract each other. (c) They repel each other. (d) Cannot be predicted due to insufficient data.
24. In a transistor if a varies between
then the value of b lies between (a) 1–10 (b) 0.95–0.99 (c) 20–100 (d) 200–300
20 100 and , 21 101
25. The minimum force required to move a body up
an inclined plane is three times the minimum force required to prevent it from sliding down the plane. If the coefficient of friction between the body and 1 the inclined plane is , the angle of the inclined 2 3 plane is (a) 60°
(b) 45°
(c) 30°
(d) 15°
26. The threshold frequency of the metal of the cathode
in a photoelectric cell is 1 × 1015 Hz. When a certain beam of light is incident on the cathode, it is found that a stopping potential 4.144 V is required to reduce the current to zero. The frequency of the incident radiation is (h = 6.63 × 10–34 J s) (a) 2.5 × 1015 Hz (b) 2 × 1015 Hz 15 (c) 4.144 × 10 Hz (d) 3 × 1016 Hz
27. A 50 mF capacitor is connected to an ac source
V = 220 sin50t where V is in volt and t is in second. The rms current is 0.55 (a) 0.55 A (b) A 2 2 (c) (d) 2 A A 0.55
31. If the ratio of maximum and minimum intensities
of an interference pattern is 36 : 1, then the ratio of amplitudes of the two interfering waves will be (a) 3 : 7 (b) 7 : 4 (c) 4 : 7 (d) 7 : 5
32. A wire of length 6.28 m is bent into a circular coil
of 2 turns. If a current of 0.5 A exists in the coil, the magnetic moment of the coil is 1 A m2 4
(a) p A m2
(b)
(c) p A m2
(d) 4p A m2
4
33. The electric field for an electromagnetic wave in
^ free space is E = 30 cos(kz − 6 × 108 t ) i Vm −1 . The magnitude of wave vector is (a) 2 rad m–1 (b) 3 rad m–1 (c) 4 rad m–1 (d) 6 rad m–1
34. Two parallel plane sheets 1 and 2 carry uniform
charge densities s1 and s2(s1 > s2) as shown in the figure. The magnitude of the resultant electric field in the region marked I is 1
2
28. Of the following, NAND gate is
(a) (b) (c) (d) 29. On a temperature scale Y, water freezes at –160°Y
and boils at –50°Y. On this Y scale, a temperature of 340 K is (a) –106.3°Y (b) –96.3°Y (c) –86.3°Y (d) –76.3°Y
30. A satellite is revolving very close to a planet of
density r. The period of revolution of the satellite is (a)
3pr G
(b)
3p 2rG
(c)
3p rG
(d)
3pG r
(a)
s1 2e 0
(b)
s2 2e 0
(c)
s1 + s 2 2e 0
(d)
s1 − s 2 2e 0
35. In Millikan’s oil drop experiment, a charged oil
drop of mass 3.2 × 10–14 kg is held stationary between two parallel plates 6 mm apart, by applying a potential difference of 1200 V between them. How many electrons does the oil drop carry ?(g = 10 m s–2) (a) 7 (b) 8 (c) 9 (d) 10 36. Two long wires each parallel to the z-axis and each carrying current I, are at (0, 0) and (a, b). The force per unit length of each wire is m0 I 2 (a + b) m0 I 2 (a) (b) 2 p(a2 + b2 ) 2 p(a2 + b2 ) 2 m0 I m0 I 2 (c) (d) 2 p(a2 + b2 )3/2 2 p(a2 + b2 )1/2 physics for you | june ‘15
33
37. Two blocks of masses 1 kg and
2 kg are connected by a metal wire going over a smooth pulley as shown in the adjacent figure. The breaking stress of the metal is 40 × 106 N m −2. What should be 3p the minimum radius of wire used if it should not break? (g = 10 m s–2) (a) 0.5 mm (b) 1 mm (c) 1.5 mm (d) 2 mm 38. A rectangular coil is rotating in a uniform magnetic
field B. The emf induced in the coil is maximum when the plane of the coil (a) is parallel to B (b) makes an angle 30° with B (c) makes an angle 45° with B (d) is perpendicular to B
39. A steady dc current is flowing through a cylindrical
conductor. Which of the following statements is/are correct? 1. The electric field at the axis of the conductor is zero. 2. The magnetic field at the axis of the conductor is zero. Select the correct answer using the code given below : (a) 1 only (b) 2 only (c) both 1 and 2 (d) neither 1 nor 2
40. The ratio between kinetic and potential energies of
a body executing simple harmonic motion, when 1 it is at a distance of of its amplitude from the N mean position is (a) N2 + 1 (c) N2
1 N2 (d) N2 – 1 (b)
41. PQR is a right angled triangular plate of uniform
thickness as shown in the figure. If I1, I2 and I3 are moments of inertia about PQ, QR and PR axes respectively, then
34
physics for you | june ‘15
(a) I3 < I2 < I1 (c) I2 > I1 > I3
(b) I1 = I2 = I3 (d) I3 > I1 > I2
42. A tank of height 5 m is full of water. There is a hole
of cross-sectional area 1 cm2 in its bottom. The volume of water that will come out from this hole per second is (g = 10 m s–2) (a) 10–3 m3 s–1 (b) 10–4 m3s–1 3 –1 (c) 10 m s (d) 10–2 m3s–1
43. The temperature of a perfect black body is 727°C
and its area is 0.1 m2. If Stefan’s constant is 5.67 × 10–8 W m–2 K–4, then heat radiated by it in 0.3 minutes is (a) 1701 J (b) 17010 J (c) 102060 J (d) 1020 J
44. The plates in a parallel plate capacitor are separated
by a distance d with air as the medium between the plates. In order to increase the capacity by 66%, a dielectric slab of dielectric constant 5 is introduced between the plates. What is the thickness of the dielectric slab? d d (a) (b) 4 2 5d (c) (d) d 8 45. A simple pendulum is executing SHM with a period of 6 s between two extreme positions B and C about a point O. If the length of the arc BC is 10 cm, how long will the pendulum take to move from position C to a position D towards O exactly midway between C and O ? (a) 0.5 s (b) 1 s (c) 1.5 s (d) 3 s soLuTioNs 1. (d) : The argument of the sine function is called the phase. Thus the phase of the wave is kx – wt + f. 2. (d) : P. Dimensional formula of Pa s = [ML–1T–2][T] = [ML–1T–1] Q. Dimensional formula of N m K–1 = [MLT–2][L][K–1] = [ML2T–2K–1] R. Dimensional formula of J kg–1 K–1 = [ML2 T–2][M–1][K–1] = [M0L2T–2K–1] S. Dimensional formula of W m–1 K–1 = [ML2 T–3][L–1][K–1] = [MLT–3K–1] Thus P – 3, Q – 4, R – 1, S – 2
3. (d) : Let a be acceleration of the lift. When the lift is moving downwards with acceleration a, then the apparent weight of the person inside it, Wapp = m(g – a) where m is the mass of the person. When it is moving upwards with same acceleration a, then his apparent weight is W′app = m(g + a) 1 As Wapp = Wapp ′ (given) 2 1 \ m( g − a) = m( g + a) 2 2(g – a) = g + a or 2g – 2a = g + a g 3a = g or a = 3 4. (a) : The earth completes one rotation about its own axis in 24 h. Its angular velocity is 2p rad w1 = 24 h The hour hand completes one rotation in 12 h. Its angular velocity is 2 p rad w2 = 12 h Their corresponding ratio is 2 p rad 24 h w1 12 1 = = = w2 2 p rad 24 2 12 h 5. (a) : Nuclear force is charge independent, short range and non-central. 6. (c) : The permeability (m), permeability of vacuum (m0) and magnetic susceptibility (c) are related as m = m0(1 + c) Here, m0 = 4p × 10–7 H m–1, c = 299 \ m = 4p × 10–7 H m–1(1 + 299) 22 =4× × 10–7 × 300 H m–1 7 = 3771 × 10–7 H m–1 7. (d) : Here, x = 2sint, y = 3cost and z = 5 sint dx d \ vx = = (2sint) = 2cost dt dt dy d vy = = (3 cost) = –3sint dt dt dz d = ( 5 sin t ) = 5 cos t and vz = dt dt The speed of the particle at any instant is
v=
v 2x + v 2y + v z2
=
(2 cos t )2 + (−3 sin t )2 + ( 5 cos t )2
=
4 cos2t + 9 sin2 t + 5 cos2t
=
4 cos2 t + 5 sin2 t + 4 sin2 t + 5 cos2 t
=
5 sin2 t + 5 cos2 t + 4 sin2 t + 4 cos2 t
=
5(sin2 t + 5 cos2 t ) + 4 (sin2 t + cos2 t )
= 5+4 ( sin2 q + cos2 q = 1) =3 Hence it is independent of time. 8. (b) : Here, Radius of the earth, R = 6.4 × 106 m Height of transmitting antenna, hT = 128 m The maximum distance (dM) between the transmitting and receiving antennas for satisfactory communication in LOS mode is dM = 2RhT + 2RhR where hR is the height of the receiving antenna. As the receiving antenna is at the ground level so, hR = 0. \
dM =
2RhT = 2(6.4 × 106 m)(128 m) 6 2 = 12.8 × 128 × 10 m
128 × 128 × 106 2 m 10 128 × 103 128 m = km = 10 10 9. (d) : The potential difference between the terminals of the battery during charging is V = e + Ir Here, e = 3 V, r = 0.2 W, I = 5 A \ V = (3 V) + (5 A)(0.2 W) = 3 V + 1 V = 4 V =
10. (a) : As NaCl is completely soluble in water, so when NaCl is added to water, the surface tension of water increases. 11. (b) : Time taken by person to reach the market is 2.5 km 1 = h t1 = 5 km h −1 2 and time taken by him to return back to his house is 2.5 km 1 = h t2 = −1 3 7.5 km h 1 1 5 \ Total time taken = t1 + t2 = h + h = h 2 3 6 Total distance travelled = 2.5 km + 2.5 km = 5 km The average speed of the person is physics for you | june ‘15
35
Total distance travelled 5 km = Total time taken 5 h 6 5 5 = 6 km h–1 = 6 × m s–1 = m s–1 3 18 12. (b) : If A0 be initial activity of the element, then fraction of initial activity after time t is vav =
t /T
A 1 1/2 = A0 2 where T1/2 is the half life. Here, t = 40 h, T1/2 = 10 h \
A 1 = A0 2
40 h /10 h
4
1 1 = = 2 16
13. (d) : As amw × wmg × gma = 1 where the subscripts a, w and g represent air, water and glass respectively. Here, amw = x, wmg = y and gma = z xyz = 1 \ 14. (d) 15. (b) :
where u is the velocity of projection and q is the angle of projection with the horizontal. We get tanq = 10 ...(i) g 5 = and ...(ii) 2u2cos2q 9 Dividing eqn. (i) by eqn. (ii), we get tan q 10 = g 5 9 2u2cos2q sin q cos q = 90 g 5 2 2 2u cos q 2u2sin q cos q 90 = g 5 The range of the projectile is u 2sin 2q R= g =
2u 2 sin q cos q g
( sin 2q = 2 sin q cos q)
= 18 m 17. (b) When the resistance S is connected in parallel with X, the balance point is obtained at 0.625 m (= 62.5 cm) from A. XS 62.5 cm 62.5 cm X +S = = \ Y (100 − 62.5) cm 37.5 cm (5 W)S (5 W + S) 5 = 2W 3 (5 W)S 5 = 2 W(5 W + S) 3 (15 W)S = (10 W)(5 W + S) (15 W)S = (10 W)(5 W) + (10 W)S (15 W)S – (10 W)S =(10 W)(5 W) (5 W)S = (10 W)(5 W) (10 W)(5 W) S= = 10 W (5 W) 5 16. (c) : Comparing the equation y = 10x − x 2 with 9 the equation of trajectory of a projectile y = x tan q − 36
gx 2 2u2cos2q
physics for you | june ‘15
...(iii)
(using (iii))
18. (d) : Let L be the length of the open pipe. The resonance frequencies of vibration in the open pipe are nv un = ; n = 1, 2, 3, ........ 2L where v is the speed of sound in air. And the difference between successive frequencies is v v ( Dn = 1) Dun = Dn = 2L 2L or
L=
v 2Dun
Here, v = 328 m s–1 Dun = 2600 Hz – 1944 Hz = 656 Hz 328 m s −1 1 \ L= = m = 0.25 m 2(656 Hz) 4 19. (a) : Here, Mass of the bullet, m = 10 g = 10 × 10–3 kg Velocity of the bullet, v = 600 m s–1 The number of bullets fired per second is 240 n= =4 60
The power of the gun is 1 1 P = nmv 2 = × 4 × 10 × 10–3 × (600)2 W 2 2 = 72 ×102 W = 7.2 kW 20. (d) : Let the resistance of the galvanometer be G. I G As shunt resistance, S = g (I − Ig ) (I − Ig )S \ G= Ig 1 Here, Ig = I , S = 10 W 40 I I − 40 (10 W) \ G= = (39)(10 W) = 390 W I 40 21. (d) 22. (c) : Work done, dW = PdV According to an ideal gas equation nRT PV = nRT or P = V dV \ dW = nRT V As V ∝ T2/3 (given) \ V = KT2/3 where K is a constant of proportionality. Differentiating both sides, we get 2 dV = K T −1/3dT 3 \
2 K T −1/3dT 3 dV 2 2 dT = T −1dT = = V 3 3 T KT 2/3
Putting this value in eqn. (i), we get 2 dT 2 dW = nRT = nRdT 3 T 3 \ W=
T2
T
2 2 2 2 nRdT = nR dT = nR(T2 − T1) ∫3 ∫ 3 T 3 T 1
1
Here, n = 1, R = 8.13 J mol–1 K–1 T2 – T1 = 30°C = 30 K 2 \ W = (1)(8.31 J mol −1K −1)(30 K) = 166.2 J 3 23. (c) 24. (c) : The a and b are related by the relation a b= 1− a
...(i)
20 21 20 20 b = 21 = 21 = 20 20 1 1− 21 21 100 , For a = 101 For a =
100 100 b = 101 = 101 = 100 100 1 1− 101 101 Thus the value of b lies between 20-100. 1 25. (c) : Here, m = 2 3 Let m be mass of the body and q be angle of the inclined plane. The minimum force required to move the body up the inclined plane is F1 = mgsinq + mmgcosq = mg(sinq + mcosq) and the minimum force required to prevent it from sliding down the plane is F2 = mgsinq – mmgcosq = mg(sinq – mcosq) As F1 = 3F2 (given) \ mg(sinq + mcosq) = 3mg(sinq – mcosq) sinq + mcosq = 3sinq –3mcosq 2sinq = 4mcosq 1 1 tanq = 2m = 2 = 2 3 3 1 q = tan −1 = 30° 3 26. (b) : Here, Threshold frequency, u0 = 1 × 1015 Hz Stopping potential, V0 = 4.144 V Planck’s constant, h = 6.63 × 10–34 J s According to Einstein’s photoelectric equation Kmax = hu – f0 where u is the frequency of incident radiation. or hu = Kmax + f0 But f0 = hu0 and Kmax = eV0 \ hu = hu0 + eV0 eV or u = u 0 + 0 h (1.6 × 10 −19 C)(4.144 V) 15 = 1 × 10 Hz + (6.63 × 10 −34 J s) 15 = 1 × 10 Hz + 1 × 1015 Hz = 2 × 1015 Hz physics for you | june ‘15
37
27. (b) : On comparing V = 220 sin50t with V = V0 sinwt, we get V0 = 220 V and w = 50 rad s–1 The capacitive reactance is 104 1 1 XC = = = W wC (50 rad s −1 )(50 × 10−6 C) 25 The rms current is V0 Vrms V ( Vrms = 0 ) Irms = = 2 XC XC 2 220
V 220 × 25 0.55 2 = 4 = 4 A= A 2 10 × 2 10 W 25 28. (d) : The symbol for NAND gate is
29. (c) :
If TY be temperature on the Y scale corresponding to 340 K on kelvin scale, then 340 K − 273 K TY − (−160°Y) = −50°Y − (−160°Y) 373 K − 273 K TY + 160°Y 67 K = 110°Y 100 K TY + 160°Y = 0.67(110°Y) TY + 160°Y = 73.7°Y TY = 73.7°Y – 160°Y = –86.3°Y 30. (c) : As the satellite is revolving very close to the planet, its period of revolution is R3 ...(i) GM where M is the mass of the planet, R is its radius and G is the universal gravitational constant. M 4 As r = or M = pR3r 4 3 3 pR 3 Putting this value of M in eqn.(i), we get T = 2p
T = 2p
38
R3 3 = 2p 4 4 prG G pR3r 3
physics for you | june ‘15
=
3(4 p2 ) 3p = 4 prG rG
31. (d) : Let A1 and A2 be amplitudes of the two interfering waves. Then 2 I max (A1 + A2 ) = I min (A1 − A2 )2 But \
I max 36 = (given) I min 1
(A1 + A2 )2 36 = (A1 − A2 )2 1 A1 + A2 6 = A1 − A2 1 A1 + A2 = 6(A1 – A2) A1 + A2 = 6A1 – 6A2 A 7 5A1 = 7A2 or 1 = A2 5
32. (a) : When a wire of length l is bent into a circular coil of N turns of radius r, then l l = N2pr or r = N 2p Here, l = 6.28 m, N = 2 6.28 m 1 \ r = = m = 0. 5 m 2(2 × 3.14) 2 The area of the coil is A = pr2 = p(0.5 m)2 = 0.25p m2 The magnetic moment of the coil, M = NIA = (2)(0.5 A)(0.25p m2) p = 0.25p A m2 = A m 2 4 33. (a) : Comparing the given equation with ^ E = E0 cos(kz − wt ) i we get, w = 6 × 108 rad s–1 The velocity of electromagnetic wave in free space is c = 3 × 108 m s–1 The magnitude of wave vector is 8 −1 w 6 × 10 rad s = 2 rad m −1 k= = 8 − 1 c 3 × 10 m s 34. (c) :
In the region I, the electric fields E1 and E2 due to sheets 1 and 2 act in the same direction.
\ The magnitude of the resultant electric field in the region I is s + s2 s s EI = E1 + E2 = 1 + 2 = 1 2e0 2e0 2e0 35. (d) : Here, Mass of the drop, m = 3.2 × 10–14 kg Distance between the plates, d = 6 mm = 6 × 10–3 m Potential difference between the plates, V = 1200 V The electric field between the plates is 1200 V V E= = d 6 × 10−3 m Let the charge on the drop be q. As the drop is held stationary, \ Upward force on drop = Weight of the drop due to electric field qE = mg V q = mg d mgd q= V
Tension Area of cross-section
To avoid breaking, this stress should not exceed the breaking stress. If the minimum radius needed to avoid breaking is r, then Breaking stress =
T pr 2
40 N 40 × 106 N m −2 = 3 2 3p pr 40 N 2 3 = 1 × 10−6 m2 r = 40 p× × 106 N m −2 3p
r = 10–3m = 1 mm e = NBAwsinwt
p i.e. 2 when the plane of the coil is parallel to the field. This is maximum when sinwt = 1 or wt =
(3.2 × 10 −14 kg)(10 m s −2 )(6 × 10 −3 m) 1200 V = 1.6 × 10–18 C The number of electrons the drop carries is −18 q 1.6 × 10 C n= = = 10 e 1.6 × 10 −19 C 36. (d) : The force per unit length of each wire is m0 I1I2 2 pd where I1 and I2 are the currents in the two wires respectively and d is the distance between them. f =
Here, I1 = I2 = I, d = a2 + b2 m0 I 2 2 p a 2 + b2
37. (b) : Here, m1 = 1 kg, m2 = 2 kg, g = 10 m s–2 Breaking stress =
=
38. (a) : The emf induced in the coil at any instant t is
=
\ f =
The stress in the wire is
40 × 106 N m −2 3p
The tension in the wire is −2
2m1m2 g 2(1 kg )(2 kg )(10 m s ) 40 T= = = N (m1 + m2 ) (1 kg + 2 kg) 3
39. (c) 40. (d) : The kinetic energy of the body executing simple harmonic motion at a distance x from the mean position is 1 K = mw2 ( A2 − x 2 ) 2 1 and the potential energy is U = mw2 x 2 2 where A is the amplitude, w is the angular frequency and m is the mass of the body. 2 1 A A At x = , K = mw2 A2 − N 2 N 1 A and U = mw2 N 2
2
Their corresponding ratio is
K = U
2 1 A mw2 A2 − N 2
1 A mw2 N 2
2
=
2 A2 A − 2 N A2
N2
physics for you | june ‘15
39
A2 2 (N − 1) 2 =N = N2 −1 A2 N2 41. (a) : The centre of mass of a triangular lamina lies at
the centroid. But the distance of a side of a triangle from its centroid is inversely proportional to the length of that side, so rPQ > rQR > rPR \ I1 > I2 > I3 or I3 < I2 < I1 42. (a) : Here,
Area of cross-section of the hole, A = 1 cm2 = 10–4 m2 Height of the tank, h = 5 m According to Torricelli’s theorem, Speed of water coming out of the hole is −2 −1 v = 2gh = 2(10 m s )( 5 m) = 10 m s Volume of water coming out of the hole per second is
Q = Av = (10–4 m2) (10 m s–1) = 10–3 m3 s–1 43. (c) : According to Stefan-Boltzmann law, Heat radiated by a perfect black body is Q = sAT4t where the symbols have their usual meanings. Here, s = 5.67 × 10–8 W m–2 K–4 A = 0.1 m2 T = 727°C = (727 + 273) K = 1000 K t = 0.3 min = 0.3 × 60 s = 18 s \ Q = (5.67 × 10–8 W m–2 K–4)(0.1 m2)(1000 K)4 (18 s) = 102060 J 44. (b) : The capacity of a parallel plate capacitor when
air in between the plates is eA ... (i) C= 0 d where A is the area of cross-section of each plate. When the dielectric slab of dielectric constant K (= 5) of thickness t is introduced between the plates, then its capacity becomes e0 A e0 A eA C′ = = 0 = 4 1 1 d −t 1− d − t 1 − d − t K 5 5 66 166 But C ′ = C + C= C (given) 100 100 40
physics for you | june ‘15
eA 166 C= 0 4 100 d− t 5 Dividing eqn. (i) by eqn. (ii), we get 4 100 d − 5 t = 166 d 664 100d = 166d − t 5 664 t = 66d 5 330 d t= d = 0.5d = 664 2 \
... (ii)
45. (b) : The situation is shown in figure.
In figure, O represents the mean position and B and C represent extreme positions and D is the midpoint between C and O. Since B and C are the extreme positions, therefore amplitude of the SHM oscillation is BC 10 cm A = OC = = = 5 cm 2 2 As D is the midpoint between C and O, OC A 5 \ CD =OD = = = cm 2 2 2 Since time is noted from extreme position, hence displacement x from the mean position at any time t is x = Acoswt Let t1 be the time taken by the pendulum to move from C to D. Then A At t = t1, x = 2 A = A cos wt1 \ 2 1 = cos wt1 2 p 2p 2p cos = cos t1 w = T T 3 t1 =
T 6s = =1s 6 6
nn
Solved PaPer 2015
Kerala PMT 1.
The wrong unit conversion among the following is (a) 1 angstrom = 10–10 m (b) 1 fermi = 10–15 m (c) 1 light year = 9.46 × 1015 m (d) 1 parsec = 3.08 × 1016 m (e) 1 astronomical unit = 1.496 × 10–11 m
2.
Choose the wrong statement. (a) The motion of an object along a straight line is a rectilinear motion. (b) The speed in general is less than the magnitude of the velocity. (c) The slope of the displacement-time graph gives the velocity of the body. (d) The area under the velocity-time graph gives the displacement of the body. (e) The negative slope of speed-time graph indicates a retarded motion.
3.
4.
The range of a projectile is R when the angle of projection is 40°. For the same velocity of projection and range, the other possible angle of projection is (a) 45° (b) 50° (c) 60° (d) 40° (e) 30°
5.
The scalar quantity among the following is (a) weight of body (b) temperature gradient (c) elementary area (d) magnetic field strength (e) electric potential
6.
Which one of the following motions on a smooth plane surface does not involve force? (a) Accelerated motion in a straight line. (b) Retarded motion in a straight line. (c) Motion with constant momentum along a straight line. (d) Motion along a straight line with varying velocity. (e) Motion in a circle with uniform speed.
7.
Pick out the wrong statement. (a) Newton's laws of motion hold good for both inertial and non-inertial frames. (b) During explosion, linear momentum is conserved. (c) Area under force-time graph gives the magnitude of impulse. (d) Force of friction is zero when no driving force is applied. (e) The apparent weight of a lift moving upwards with uniform velocity, equals its true weight.
8.
Two bodies of different masses are moving with same kinetic energy. Then the ratio of their momenta is equal to the ratio of their (a) masses (b) square of masses (c) square root of masses (d) cube root of masses (e) inverse of masses
The displacement of a particle as a function of time is shown in figure. It indicates that
(a) the velocity of the particle is constant throughout (b) the acceleration of the particle is constant throughout (c) the particle starts with a constant velocity and is accelerated (d) the particle starts from rest and is accelerated throughout (e) the motion is retarded and finally the particle stops
physics for you | june ‘15
41
9.
Two bodies of masses 1 kg and 2 kg moving with same velocities are stopped by the same force. Then the ratio of their stopping distances is (a) 1 : 2 (b) 2 : 1 (c) 2 : 1 (d) 1 : 2 (e) 1 : 3
10. If two circular discs A and B are of same mass but of radii r and 2r respectively, then the moment of inertia of A is (a) the same as that of B (b) twice that of B (c) four times that of B (d) half that of B (e) one-fourth that of B 11. Choose the wrong statement. (a) The centre of mass of a uniform circular ring is at its geometric centre. (b) Moment of inertia is a scalar quantity. (c) Radius of gyration is a vector quantity. (d) For same mass and radius, the moment of inertia of a ring is twice that of a uniform disc. (e) Force in translational motion is analogous to torque in rotational motion. 12. Orbital velocity of earth satellite does not depend on (a) mass of the earth (b) mass of the satellite (c) radius of the earth (d) acceleration due to gravity (e) its height from the surface of earth 13. Gravitational potential energy of a body of mass m at a height of h above the surface of earth (M = mass of earth, R = radius of earth) is GMm GMm (a) (b) ( R + h) h Gm −GM (c) (d) ( R + h) ( R + h) (e) −
GMm ( R + h)
14. A boat carrying a few number of big stones floats in a water tank. If the stones are unloaded into water, the water level (a) rises till half the number of stones are unloaded and then begins to fall (b) remains unchanged (c) rises 42
physics for you | june ‘15
(d) falls till half the number of stones are unloaded and then begins to rise (e) falls 15. Two wires of same length and same material but of radii r and 2r are stretched by forces F and f respectively to produce equal elongation. The ratio F to f is (a) 1 : 1 (b) 1 : 2 (c) 2 : 1 (d) 1 : 4 (e) 4 : 1 16. Choose the correct statement. (a) Ter mina l velo cit ies of rain drops are proportional to square of their radii. (b) Water proof agents decrease the angle of contact between water and fibres. (c) Detergents increase the surface tension of water. (d) Hydraulic machines work on the principle of Torricelli's law. (e) Venturimeter measures the flow speed of compressible fluids. 17. If DU represents the increase in internal energy and W the work done by the thermodynamic system, then (a) DU = –W is an isothermal process (b) DU = W is an isothermal process (c) DU = –W is an adiabatic process (d) DU = W is an adiabatic process (e) DU = W is an isochoric process 18. If the energy input to a Carnot engine is thrice the work it performs then, the fraction of energy rejected to the sink is 1 1 2 2 (a) (b) (c) (d) 4 3 5 3 3 (e) 4 19. The ratio of rms speed of an ideal gas molecules at pressure P to that at pressure 2P is (a) 1 : 2 (b) 2 : 1 (c) 1 : 2 (d) 2 : 1 (e) 1 : 1 20. A pendulum of time period 2 s on earth is taken to another planet whose mass and diameter are twice that of earth. Then its time period on the planet is (in second) 1 1 (a) (b) 2 2 (c) (d) 2 2 2 (e)
2
21. The physical quantity which remains constant in simple harmonic motion is (a) kinetic energy (b) potential energy (c) restoring force (d) displacement (e) frequency 22. Sound waves (a) can be polarized (b) can exhibit diffraction (c) are transverse in nature (d) can travel in free space (e) travel slower in liquids than in air 23. If a closed organ pipe has the same third harmonic frequency as that of an open organ pipe, then their respective lengths are in the ratio (a) 1 : 1 (b) 1 : 2 (c) 1 : 4 (d) 3 : 4 (e) 4 : 5 24. A particle of mass 1.96 × 10–15 kg is kept in equilibrium between two horizontal metal plates having potential difference of 400 V separated apart by 0.02 m. Then the charge on the particle is (e = electronic charge) (a) 3e (b) 6e (c) 2e (d) 5e (e) 4e 25. Two small spherical shells A and B are given positive charge of 9 C and 4 C respectively and placed such that their centres are separated by 10 m. If P is a point in betweeen them where the electric field intensity is zero, then the distance of the point P from the centre of A is (a) 5 m (b) 6 m (c) 7 m (d) 8 m (e) 4 m 26. Identify the wrong statement. (a) Charge is a vector quantity. (b) Current is a scalar quantity. (c) Charge can be quantised. (d) Charge is additive in nature. (e) Charge is conserved. 27. When the rate of flow of charge through a metallic conductor of non uniform cross section is uniform, then the quantity that remains constant along the conductor is (a) current density (b) electric field (c) electric potential (d) drift velocity (e) current 28. The resistance of a carbon resistor of colour code red, red, green, silver is (in kW) (a) 2200 ± 5% (b) 2200 ± 10% (c) 220 ± 10% (d) 220 ± 5%
(e) 2200 ± 1% 29. The slope of the graph showing the variation of potential difference V on x-axis and current on y-axis gives conductor’s (a) resistance (b) resistivity (c) reciprocal of resistance (d) conductivity (e) impedance 30. Identify the mismatched pair. (a) Hard magnet - Alnico (b) Soft magnet - Soft iron (c) Bar magnet - Equivalent solenoid (d) Electromagnet - Electric bells (e) Permanent magnet - Loud speaker 31. When the temperature of a magnetic material decreases, the magnetization (a) decreases in a diamagnetic material (b) decreases in a paramagnetic material (c) decreases in a ferromagnetic material (d) remains the same in a ferromagnetic material (e) remains the same in a diamagnetic material 32. The magnetic field at the centre of a circular coil carrying current I ampere is B. If the coil is bent into smaller circular coil of n turns, its magnetic field at the centre is B′. The ratio between B′ and B is (a) 1 : 1 (b) n : 1 (c) n2 : 1 (d) 2n : 1 (e) (n + 1) : 1 33. The magnetic flux linked with a circuit of resistance R changes by Df in a time Dt. Then the total quantity of charge Q that passes at any point in the circuit during the time Dt is Df 1 Df (a) (b) R R Dt Df Df (c) R (d) Dt Dt (e)
Df
R2 34. In an LCR series resonant circuit, the capacitance is changed from C to 4C. For the same resonant frequency, the inductance should be changed from L to L L (a) 2L (b) (c) 4L (d) 2 4 L (e) 8 physics for you | june ‘15
43
35. Changing magnetic fields can set up current loops in nearby metal bodies and the currents are called as (a) eddy currents (b) flux currents (c) alternating currents (d) leakage currents (e) wattless currents 36. Ozone layer in the atmosphere absorbs (a) radio waves (b) infrared waves (c) ultraviolet rays (d) X-rays (e) microwaves 37. The magnifying power of a convex lens of focal length 10 cm when the image is formed at the near point is (a) 6 (b) 5.5 (c) 4 (d) 3.5 (e) 2 38. The waves that require a medium to travel is (a) infrared radiation (b) ultraviolet radiation (c) visible light (d) X-rays (e) ultrasound 39. In Young's double slit experiment, the locus of the point P lying in a plane with a constant path difference between the two interfering waves is (a) a hyperbola (b) a straight line (c) an ellipse (d) a parabola (e) a circle 40. The ratio of the respective de Broglie wavelengths associated with electrons accelerated from rest with the voltages 100 V, 200 V and 300 V is (a) 1 : 2 : 3 (b) 1 : 4 : 9 1 1 1 1 (c) 1 : (d) 1 : : : 2 3 2 3 (e) 3 : 2 : 1 41. A radioactive source of half-life 2 hours emits radiation of intensity which is 64 times the permissible safe level. The minimum time in hours after which it would be possible to work safely with the source is (a) 12 (b) 8 (c) 6 (d) 24 (e) 3 42. Nuclear fusion is not found in (a) thermonuclear reactor (b) hydrogen bomb (c) energy production in sun (d) atom bomb (e) energy production in stars 44
physics for you | june ‘15
43. The approximate ratio of nuclear mass densities of and 107 47 Ag nuclei is (a) 197 : 107 (b) 47 : 79 (c) 79 : 47 (d) 1 : 1 (e) 107 : 197 44. Identify the gate used in the following diagram. 197 79 Au
(a) AND (b) OR (c) NAND (d) NOR (e) either NAND or NOR 45. Acceptor level in p-type semiconductors lies (a) nearer to the conduction band (b) at the middle of conduction band and valence band (c) within the valence band (d) nearer to the valence band (e) within the conduction band 46. The frequencies that are reflected and transmitted at ionospheric layer respectively are (a) 3 kHz and 10 MHz (b) 10 MHz and 40 MHz (c) 10 MHz and 20 MHz (d) 35 MHz and 70 MHz (e) 100 Hz and 100 kHz 47. The gap between the frequency of the side bands in an amplitude modulated wave is (a) twice that of the carrier signal (b) twice that of the message signal (c) the same as that of the message signal (d) the same as that of the carrier signal (e) the sum or difference of the frequencies of carrier and message signal 48. A transmitting antenna at the top of a tower has a height of 20 m. For obtaining 40 km as the maximum distance between the transmitter and receiver for satisfactory communication in LOS mode, the height of receiving antenna should be (radius of the earth R = 64 × 105 m) (a) 30 m (b) 35 m (c) 40 m (d) 45 m (e) 50 m solutions 1. (e) : 1 astronomical unit = 1.496 × 1011 m All the other unit conversions are correct. 2. (b) : The speed in general is greater than the magnitude of the velocity. All the other statements are correct.
3. (e) :
In figure, the displacement-time graph is a parabola having negative slope and finally becomes parallel to time axis. So it indicates that motion is retarded and finally the particle stops. 4. (b) : For the same velocity of projection and range, the sum of the two angles of projection is 90°. As one angle of projection is 40°, so the other possible angle of projection is 90° – 40° = 50°. 5. (e) : Among the given quantities electric potential is a scalar quantity whereas all others are vector quantities. 6. (c) : Motion with constant momentum along a straight line implies p = constant. So, according to Newton's 2nd law dp d F= = (constant) = 0 dt dt 7. (a) : Newton's laws of motion hold good for inertial frame. All the other statements are correct. 8. (c) : The kinetic energy (K) and momentum (p) of a body are related as K=
p2 or p = 2mK 2m
10. (e) : Let m be the mass of each disc. The moment of inertia of disc A is 1 IA = mr2 2 and that of disc B is 1 1 IB = m(2r )2 = 4 mr2 = 4 IA 2 2 1 or IA = IB 4
vo =
2
2
gR2 GM = R+h R+h
GM g = 2 R
Thus it is independent of the mass of the satellite (m) but depends on the mass of the earth (M), radius of the earth (R), acceleration due to gravity (g) and height (h) of the satellite from the surface of earth. 13. (e) : Gravitational potential energy of a body of mass m at a height h above the surface of earth is GMm U =− ( R + h) 14. (e) : When the stones are unloaded into water, the water level falls because the volume of the water displaced by stones in water will be less than the volume of water displaced when stones are in the boat. 15. (d) : According to definition of Young's modulus Y=
But K1 = K2 (given) p m1 \ 1= p2 m2
Here, m1 = 1 kg and m2 = 2 kg ds 1 kg 1 = \ 1 = 2 kg 2 ds
(using (i))
11. (c) : Radius of gyration is a scalar quantity. 12. (b) : Orbital velocity of earth satellite is
where m is the mass of the body. 2m1K1 p1 m1K1 = = \ p2 m2 K 2 2m2 K 2
9. (a) : If two bodies of masses m1 and m2 moving with the same velocities are stopped by the same force, then the ratio of their stopping distances is ds m 1 = 1 ds m2
...(i)
F / A F / pr 2 = DL / L DL / L
\ Elongation produced in a wire is FL DL = 2 pr Y where L is the length of the wire, r is its radius and F is the stretching force. As both wires are of same length (L) and same material i.e., Y is same and produce equal elongations \ DL1 = DL2 FL fL = 2 pr Y p(2r )2 Y F f = 2 r (2r )2 F r2 r2 1 = = = f (2r )2 4r2 4 physics for you | june ‘15
45
16. (a) : (a) The terminal velocity of a rain drop is
Tp
2r2 (ρ − σ) g vt = 9η
Te
Thus vt ∝ r2 (b) Water proof agents increase the angle of contact between the water and fibres. (c) Detergents decrease the surface tension of water. (d) Hydraulic machines work on the Pascal's law. (e) Venturimeter measures the flow speed of incompressible fluids. 17. (c) : According to first law of thermodynamics DQ = DU + W In an adiabatic process, DQ = 0 \ 0 = DU + W or DU = –W 18. (d) : If Q1 is the energy input and Q2 is the energy rejected to the sink, then work done W = Q1 – Q2 Dividing by Q1 on both sides, we get Q Q W W = 1 − 2 or 2 = 1 − Q1 Q1 Q1 Q1 As Q1 = 3W (given) Q 1 2 W \ 2 = 1− = 1− = 3W 3 3 Q1
Thus the fraction of energy rejected to the sink is
2 . 3
19. (c) : Since temperature is not specified, vrms ∝ P So,
vrms(P ) vrms(2P )
=
P 1 = 2P 2
Note : If temperature remains constant, the rms speed of an ideal gas is independent of the pressure of the gas. So vrms(P) : vrms(2P) = 1 : 1 20. (b) : Let L be length of the pendulum. \ Its time period on earth is L ge and that on the planet is Te = 2p
Tp = 2p
L gp
...(i)
...(ii)
where g e and g p are the acceleration due to gravity on the surface of the earth and the planet respectively. Dividing eqn. (ii) by eqn. (i), we get 48
physics for you | june ‘15
=
ge gp
...(iii)
If Me, De be mass and diameter of earth and Mp, Dp be corresponding quantities of the planet, then GMe 4GMe ge = = 2 (De / 2) De2 and g p =
GM p
(Dp / 2)2
=
4GM p D2p
Substituting these values in eqn. (iii), we get, Tp Te
=
(4GMe / De2 )
(4GM p / D2p )
=
2 Me Dp M p De2
2 Me Dp or Tp = Te M p De2
But Mp = 2Me, Dp = 2De and Te = 2 s (given) \ Tp = (2 s)
Me (2De )2 =2 2 s 2 Me De2
21. (e) : The frequency remains constant in simple harmonic motion. 22. (b) : (a) Sound waves can't be polarized. (b) They can exhibit diffraction. (c) They are longitudinal in nature. (d) They can't travel in free space. (e) They travel faster in liquids than in air. 23. (b) : Let Lc and Lo be the lengths of the closed and the open organ pipes respectively. The frequency of third harmonic of the closed organ pipe is 3v u3c = 4 Lc and that of the open organ pipe is 3v u3o = 2Lc where v is the speed of the sound. As u3c = u3o (given) L 3v 3v 1 \ = or c = Lo 2 4 Lc 2Lo
24. (b) : Here, Mass of the particle, m = 1.96 × 10–15 kg Distance between the plates, d = 0.02 m Potential difference between the plates, V = 400 V The electric field between the plates is V 400 V E= = = 2 × 104 V m −1 d 0.02 m
Let the charge on the particle be q. As the particle is in equilibrium, \ Upward force on the particle due to electric field = Weight of the particle mg qE = mg or q = E
q=
(1.96 × 10−15 kg )(9.8 m s −2 )
(2 × 104 V m −1 ) = 9.6 × 10–19 C = 6 × 1.6 × 10–19 C = 6e ( e = 1.6 × 10–19 C) 25. (b) : Let the point P be at the distance x from the centre of A where the electric field intensity is zero.
\ At point P, EA = EB 1 9C 1 4C = 2 4 pε0 x 4 pε0 (10 m − x )2 9 4 = 2 x (10 m − x )2 3 2 = or 30 m – 3x = 2x x 10 m − x 30 m 5x = 30 m or x = =6 m 5 26. (a) : Charge is a scalar quantity. All the other statements are correct. 27. (e) : When the rate of flow of charge through a metallic conductor of non uniform cross section is uniform, then current remains constant along the conductor while current density, electric field, electrical potential and drift velocity are not constants and all vary inversely with area of cross section. 28. (b) : The number for red is 2 and that for green is 5. For silver, tolerance is 10%. \ The resistance of the resistor shown in the figure is
R = 22 × 105 W ± 10% = 2200 kW ± 10% 29. (c) : According to Ohm's law, Potential difference (V ) = Resistance (R) Current (I )
\ The slope of the given graph is I 1 = = = reciprocal of resistance V R 30. (e) : Electromagnet - Loudspeaker 31. (e) : The magnetization of a diamagnetic material is independent of the temperature. 32. (c) : Let r be the radius of the coil. µ I ... (i) \ B= 0 2r When the coil is bent into small circular coil of n turns of radius r′, then r n2pr′ = 2pr or r ′ = ... (ii) n µ nI µ nI n2 µ0 I \ B′ = 0 = 0 = 2r ′ 2(r / n) 2r
(using (ii))
Dividing eqn. (ii) by eqn. (i), we get B ′ n2 = 1 B
33. (a) : According to Faraday's law of induction, the magnitude of the induced emf in the circuit is Dφ |ε|= Dt As R is the resistance of the circuit, so induced current is | ε | Dφ I= = R RDt Dφ Dφ \ Q = I Dt = Dt = RDt R 34. (d) : The resonant frequency of an LCR series circuit is 1 ur = 2 p LC When the capacitance is changed to C′ (= 4C) and inductance changed to L′, the new resonant frequency becomes physics for you | june ‘15
49
ur′ =
1
2 p L ′C ′ But u′r = ur (given) 1 1 \ = 2 p L ′C ′ 2 p LC Squaring both sides, we get 1 1 LC LC L \ = L′ = = = L ′C ′ LC C ′ 4C 4
42. (d) : Atom bomb is based on nuclear fission whereas all others are based on nuclear fusion. 43. (d) : The nuclear mass density is independent of mass number (A). Thus the approximate ratio of nuclear 107 mass densities of 197 79 Au and 47 Ag nuclei is 1 : 1. 44. (a) :
The output is high only when both inputs A and 35. (a) : Changing magnetic fields can set up current B are high. So the logic gate is AND. loops in nearby metal (any conductor) bodies. They dissipate electrical energy as heat. Such 45. (d) : Acceptor level in p-type semiconductor lies currents are called eddy currents. nearer to the valence band. 36. (c) : Ozone layer in the atmosphere absorbs 46. (b) : The frequencies that are reflected and ultraviolet rays. transmitted at ionospheric layer respectively are 10 MHz and 40 MHz. 37. (d) : When the final image is formed at the near point, the magnifying power is 47. (b) : T he f re qu e nc y D spectrum of an m = 1+ amplitude modulated f wave is shown in the where D is the least distance of distinct vision and adjacent figure. f is the focal length of the convex lens. The gap between the frequency of the side bands Here, D = 25 cm, f = 10 cm (i.e. upper side band and lower side band) is called 25 cm \ m = 1+ = 1 + 2. 5 = 3. 5 bandwidth and it is given by 10 cm Bandwidth = uUSB – uLSB 38. (e) : Ultrasound are mechanical waves and they =(uc + um) – (uc – um) require a medium to travel whereas infrared = uc + um – uc + um = 2um radiation, ultraviolet radiation, visible and X-rays i.e. Bandwidth = twice of the frequency of the are all electromagnetic waves and they do not message signal. require a medium to travel. 48. (d) : The maximum distance between the transmitter 39. (a) : In Young's double slit experiment, the locus of and receiver for satisfactory communication in the point P lying in a plane with a constant path LOS mode is difference between the two interfering waves is a dM = 2RhT + 2RhR hyperbola. where hT and hR are the heights of transmitting 40. (c) : de Broglie wavelength associated with an electron and receiving antennas respectively. (mass m, charge e) accelerated from rest with a voltage Here, dM = 40 km = 40 × 103 m V volt is h hT = 20 m, R = 64 × 105 m 1 λ= or λ ∝ 2meV V \ 40 × 103 m = 2(64 × 105 m)(20 m) + 2(64 × 105 m)hR 1 1 1 \ λ1 : λ2 : λ3 = : : \ 40 × 103 m = 2(64 × 105 m)(20 m) + 2(64 × 105 m)hR V1 V2 V3 1 1 1 40 × 103 m = 16 × 103m + (128 × 105 m)hR = : : 100 V 200 V 300 V or (128 × 105 m)hR = 40 × 103 m − 16 × 103 m 1 1 = 1: : = 24 × 103 m 2 3 Squaring both sides, we get 41. (a) : Since the half-life is 2 hours, the intensity of (128 × 105 m)hR = (24 × 103 m)2 the radiation falls by a factor of 2 every 2 hours. In (24 × 103 m)2 12 hours, it will fall by a factor of (2)6 = 64. Thus, hR = = 45 m in 12 hours, the intensity attains the safe level. (128 × 105 m) nn 50
physics for you | june ‘15
CHAPTERWISE PRACTICE PAPER : ELECTROSTATICS 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. What is the nature of symmetry of electric field due
to an electric dipole ?
2. Is it possible to move a charge in an electric field
without doing any work? If so, how and if not, why? 3. A student plotted E – r curves for a point charge, a long charged straight wire and an electric dipole but failed to label them. Identify the curves and give reason for your choice.
4. Can a charge of 1 C be given to a spherical conductor
7. Two wire rings, each
having a radius R, are placed at a distance d apart with their axes coinciding as shown in the following figure. The charges on the two rings are +q and –q. Find the value of potential difference between the centres of the two rings.
8. Two dielectric slabs of dielectric constants K1 and
K2 are filled in between the two plates, each of area A1 of the parallel plate capacitor as shown in the figure. Find the net capacitance of the capacitor.
of 1 m radius? Why ? 5. Two point charges placed at a distance r apart in air exert a force F on each other. At what distance will these charges experience the same force F in a medium of dielectric constant er? section-b 6. n tiny drops, all of same size, are given equal charges.
If the drops coalesce to form a single bigger drop, then what will be the new potential of the drop? What is the surface charge density of the bigger drop?
9. Plot a graph depicting variation of potential energy
of an electric dipole placed in a uniform electric field with angle q between the direction of dipole moment and field. OR physics for you | june ‘15
51
A capacitor of capacitance C1 is charged to a potential V. On disconnecting with the battery, it is connected with an uncharged capacitor of capacitance C2. Find the ratio of total electrostatic potential energy before and after. 10. An electron is constrained to move along the axis
of the ring of charge q and radius a. Show that the electron can perform oscillations whose frequency
is given by ω =
qe
4pe0ma 3
.
section-c 11. Dedue the expression for the capacitance of a parallel
12.
13.
14.
15.
16.
52
plate capacitor when a dielectric slab is inserted between its plates. Assume the slab thickness is less than the plate separation. Two point charges + 4e and + e are fixed at a distance a apart. Where should a third point charge q be placed on the line joining the two charges that it may be in equilibrium? In which case the equilibrium will be stable and in which unstable ? A small sphere of radius r and charge q is enclosed by a spherical shell of radius R and charge Q. Show that if q is positive, charge q will necessarily flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge Q on the shell is. Two charged conducting spheres of radii a and b are connected to each other by a wire. What is the ratio of electric fields at the surfaces of the two spheres? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions. Derive an expression for the torque on an electric dipole placed in a uniform electric field. Hence define dipole moment. OR Two point charges +q and –q are placed distance d apart. What are the points at which the resultant electric field is parallel to the line joining the two charges? (a) Determine the electrostatic potential energy of a system consisting of two charges 7 mC and –2 mC (and with no external field) placed at (–9 cm, 0, 0) and (9 cm, 0, 0) respectively. (b) How much work is required to separate the two charges infinitely away from each other? (c) Suppose the same system of charges is now placed in an external electric field E = A(1/r2); A = 9 × 105 N C–1 m2. What would the electrostatic energy of the configuration be? physics for you | june ‘15
17. A charged particle, of charge 2 mC and mass
10 milligram, moving with a velocity of 1000 m s–1 enters a uniform electric field of strength 103 N C–1 directed perpendicular to its direction of motion. Find the velocity and displacement, of the particle after 10 s. 18. A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process? 19. A spherical capacitor has an inner sphere of radius 12 cm and outer sphere of radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of 2.5 mC. The space between the concentric spheres is filled with liquid of dielectric constant 32. (a) Determine the capacitance of the capacitor. (b) What is the potential of the inner sphere? (c) Compare the capacitance of this capacitor with that of an isolated sphere of radius 12 cm. Explain why the latter is much smaller. 20. Obtain the equivalent capacitor of the network in
figure. For a 300 V supply, determine the charge and voltage across each capacitor.
21. Consider three charges q1, q2, q3 each equal to q at
the vertices of an equilateral triangle of side l. What is the force on a charge Q (with the same sign as q) placed at the centroid of the triangle?
22. Obtain the formula for the electric field due to
a long thin wire of uniform linear charge density l without using Gauss’s law. section-d
23. Father of Devansh is a surgeon and has a video
monitor, Devansh observed that after the use the video monitor surface remain charged for long. He further heard his father instructing his assistant not to bring his finger close to the screen or touch the screen with contaminated gloves as the screen may become a source of bacteria. Devansh asked for the explanation from his father and shared his learning with his classmates during the physics period. His teacher and classmates had appreciations for him. (i) What values had Devansh shown? (ii) How did the father explain his observations?
section-e 24. Define dipole moment of an electric dipole. Show
mathematically that the electric field intensity due to a short dipole at a distance d along its axis is twice the intensity at the same distance along the equatorial axis. OR What is an electric dipole? An electric dipole is placed in a uniform electric field at some angle q. Deduce the expression for its potential energy. What is physical significance of electric dipole? 25. Draw a labelled diagram of Van de Graaff generator. State its working principle to show how by introducing a small charged sphere into a larger sphere, a large amount of charge can be transferred to the outer sphere. State the use of this machine and also point out its limitations. OR State Gauss’s theorem in electrostatics and express it mathematically. Using it, derive an expression for electric field at a point near a thin infinite plane sheet of electric charge. How does this electric field change for a uniformly thick sheet of charge? 26. Derive an expression for the energy density in a parallel plate capacitor. A parallel plate capacitor with air as dielectric is charged by a dc source to a potential V. Without disconnecting the capacitor from the source, air is replaced by another dielectric medium of dielectric constant 10. State with reason, how does (i) electric field between the plates and (ii) energy stored in the capacitor change? OR What are conductors? Explain the electrostatics of conductors. solutions 1. The electric field due to an electric dipole exhibits
cylindrical symmetry with axis as the axis of cylinder. 2. Yes, it is possible to move a charge in an electric field without doing any net work. If electric potential of initial and final points is same, i.e., VA = VB, then WAB = q(VB – VA) = 0 3. The curve 1 represents variation of electric field E with distance r normal to a long uniformly charged straight wire because here E ∝ 1/r and slope of curve 1 is the least. The curve 3 represents E – r curve for an electric dipole because here E ∝ 1/r3 and slope of curve 3 is maximum. Then, the curve 2 should represent E–r graph for a point charge, because here E ∝ 1/r2.
4. If a charge of 1 C is given to a spherical conductor
of radius 1 m, then field at a point just near its surface, Q 9 × 109 × 1 = = 9 × 109 V m −1 E= 4pe0R 2 (1)2 However, the dielectric strength of air present around the conductor is much less (about 3 × 106 V m–1). So, the charge will immediately leak. Thus, it is not possible to store a charge of 1 C in a spherical conductor of radius 1 m. 5. As, electric force between two point charges q1 and q2 in a medium = electric force between same charges in free space ke q1q2 qq r = ke 122 or er r ′ 2 = r 2 or r ′ = er r ′ 2 er r 6. Let each drop be having a radius r and charge q. Then, potential at the surface of each drop, q V= 4pe0r
and surface density of charge, q σ= 4pr 2 When n drops coalesce to form a single bigger drop of radius R, total volume remains unchanged. 4 4 Hence, pR 3 = n × pr 3 3 3 ⇒ R = (n)1/3r and total charge on the bigger drop, Q = nq \ Potential of bigger drop, Q nq V′ = = = (n)2/3V 4pe0R 4pe (n)1/3r 0
and new surface charge density, Q nq σ′ = = = (n)1/3σ 2 4pR 4pn2/3r 2 7. Electric potential at centre O1 of 1st ring,
V1 =
q 1 q + − 4pe0 R R 2 + d 2
and electric potential at the centre O2 of 2nd ring, q q 1 V2 = − + 4pe0 2 2 R R +d \ V1 – V2 = q q q 1 q − − − R 4pe0 R R2 + d 2 R2 + d 2 physics for you | june ‘15
53
=
q 1 1 − 2pe0 R R 2 + d 2
⇒
8. The arrangement shown in the figure is equivalent
to two capacitors each having a plate area A/2 and plate separation d. The two capacitors are filled with dielectrics of dielectric constants K1 and K2 respectively and are joined in parallel. K e A / 2 K1e0 A \ C1 = 1 0 = d 2d K e A and C2 = 2 0 2d \ Net capacitance of the capacitors, e A C = C1 + C2 = 0 (K1 + K 2 ) 2d 9. We know that potential energy of an electric dipole is given by U = –pE cosq Hence, (a) for q = 0°, U = –pE p (b) for q = , U = 0 2 (c) for q = p, U = +pE On the basis of this data, we plot the curve as shown in the given figure. It is a part of cosine curve.
1 C1V 2 Ui C + C2 2 = = 1 2 2 Uf C1 1 C1 V 2 (C1 + C2 )
10. Electric field at a point distant x from the ring of
charge q,
E = ke
2
qx
2 3/2
(a + x )
k qx = e3 a
(for x a, then Ea > Eb i.e. sphere with smaller radius produces more electric field on its surface. Hence, the charge density on the sharp and pointed ends of conductor is higher than on its flatter portions. 15. Consider an electric dipole consisting of charges +q and –q and of length 2a placed in a uniform electric field E making an angle q with it. It has a dipole moment of magnitude, p = q × 2a Force exerted on charge + q by field E = qE (along E ) Force exerted on charge – q by field E = −qE (opposite to E ) Fnet = + qE − qE = 0. Hence the net translating force on a dipole in a uniform electric field is zero. But the two equal and opposite forces act at different points of the dipole. They form a couple which exerts a torque. Torque = Either force × Perpendicular distance between the two forces t = qE × 2a sinq = (q × 2a)E sinq or
t = pE sinq
(p = 2aq)
As the direction of torque t is perpendicular to both p and E, so we can write t = p× E The direction of vector t is that in which a right handed screw would advance when rotated from p
to E. As shown in figure, the direction of vector t is perpendicular to both p and E , and points into the plane of paper. When the dipole is released, the torque t tends to align the dipole with the field E i.e., tends to reduce angle q to 0. When the dipole gets aligned with E , the torque t becomes zero. Clearly, the torque on the dipole will be maximum when the dipole is held perpendicular to E .Thus tmax = pE sin90° = pE. Dipole moment : As t = pEsinq If E = 1 unit, q = 90°,then t = p Hence dipole moment may be defined as the torque acting on an electric dipole, placed perpendicular to a uniform electric field of unit strength. OR As shown in figure, suppose the charges + q and –q are located at points A and B distance d apart. Let P be a point such that AP = r1 and BP = r2. Electric field at P due to charge +q is 1 q E1 = , 4pe0 r 2 1
along AP produced. Electric field at P due to charge –q is q 1 E2 = . 2, 4pe0 r 2
along PB produced. Draw PR||AB and PQ, PS ^ PR. If a and b are the angles made by AP and BP with AB, then Component of E1 along PR = E1 cosa Component of E1 along PQ = E1 sina Component of E2 along PR = E2 cosb Component of E2 along PS = E2 sinb The resultant field will be parallel to PR if the components E1 sina and E2 sinb are equal and opposite or zero. Hence (i) When E1 sina = E2 sinb q PS 1 q PS 1 . = . × 4pe0 r 2 r1 4pe0 r 2 r2 1
2
or r1 = r2 (ii) When E1 sina = E2 sinb = 0, we have sina = sinb = 0 i.e., a = 0° or 180° and b = 0° or 180° Thus the resultant intensity will be parallel to the physics for you | june ‘15
57
line joining A and B if (i) P lies on the perpendicular bisector of AB, or (ii) P lies on either side of AB i.e., a = b = 0° or 180° 16. (a) q1 = 7 mC = 7 × 10–6 C, q2 = –2 ×10–6 C, r = 18 cm = 0.18 m Electrostatic potential energy of the two charge system is 1 q1q2 . U= 4pe0 r 9 × 109 × 7 × 10 −6 × (−2) × 10 −6 = − 0. 7 J 0.18 (b) Work required to separate two charges infinitely away from each other, W = U2 – U1 = 0 – U = –(–0.7) = 0.7 J (c) Energy of the two charges in the external electric field = Energy of interaction of two charges with the external electric field + Mutual interaction energy of the two charges 1 q1q2 = q1V (r1) + q2V (r2 ) + 4pe0 r 2 =
A A A 1 q1q2 V = Er = + q2 + 2 r r1 r2 4pe0 r −2 mC 7 mC 5 −1 2 = + × 9 × 10 NC m − 0.7 J 0 . 09 m 0 . 09 m = q1
= (70 – 20) – 0.7 = 50 – 0.7 = 49.3 J 17. The velocity of the particle, normal to the direction of field vx = 1000 m s–1, is constant. The velocity of the particle, along the direction of field, after 10 s, is given by vy = uy + ayt qE y 2 × 10 −6 × 103 × 10 = 2000 m s −1 t= =0+ m 10 × 10 −6 The net velocity after 10 s,
v = v 2x + v 2y = (1000)2 + (2000)2 = 1000 5 m s −1 Displacement, along the x-axis, after 10 s, x = 1000 × 10 m = 10000 m Displacement along y-axis (in the direction of field) after 10 s, 1 1 qE y 2 y = uy t + a yt 2 = (0)t + t 2 2 m 1 2 × 10 −6 × 103 × (10)2 = 10000 m = × −6 2 10 × 10 Net displacement, r = x 2 + y 2 = (10000)2 + (10000)2 = 10000 2 m. 58
physics for you | june ‘15
18. Here, C1 = 600 pF, V1 = 200 V, C2 = 600 pF, V2 = 0
On connecting charged capacitor to uncharged capacitor, the common potential V across the capacitors is
−12 −12 C V + C2V2 600 × 10 × 200 + 600 × 10 × 0 V= 1 1 = C1 + C2 (600 + 600) × 10 −12
or V = 100 V Energy stored in capacitors before connection is 1 1 Ui = C1V12 + 0 = × 600 × 10 −12 × 2002 2 2 or Ui = 12 mJ Energy stored in capacitors after connection is 1 1 Uf = (C1 + C2 )V 2 = (600 + 600) × 10 −12 × 1002 2 2 or Uf = 6 mJ Hence the energy lost in the process is DU= Ui – Uf = (12 – 6) mJ or DU= 6 mJ. 19. Here, r1 = 13 cm, r2 = 12 cm, K = 32, Q = 2.5 mC (a) Capacitance of capacitor is C=
4pe0Kr1r2 1 × 32 13 × 10 −2 × 12 × 10 −2 × = r1 − r2 9 × 109 (13 − 12) × 10 −2
or C = 5.5 × 10–9 F (b) Electric potential of inner sphere is VB = VBB + VBA Q r1 − r2 1 Q Q = + − = 4pe0K r2 r1 4pe0K r1r2 9 × 109 13 − 12 10 −2 × 2.5 × 10 −6 × 32 13 × 12 10 −4 = 4.5 × 102 V. Capacitance of isolated sphere of radius 12 cm is =
(c)
1
× 12 × 10 −2 9 × 109 or C0 = 1.3 × 10–11 F Here C > C0, because a single conductor A can be charged to an electric potential till it reaches the breakdown value of surroundings. But when another earthed metallic conductor B is brought near it, negative charge induced on it decreases the electric potential on A, hence more charge can be stored on A. 20. Since C2 and C3 are in series so C × C3 C′ = 2 =100 pF C2 + C3 Now, C1 and C′ are in parallel so C′′ = C1 + C′ = 100 + 100 or C′′ = 200 pF C0 = 4pe0r =
AD AD = AB l 3l or AD = l cos 30° = 2 2 2 3l l As AO = AD = × or AO = ....(i) 3 3 2 3 l Similarly BO = CO = 3 Forces on charge +Q at O due to charges +q at A, B and C are Qq 1 1 3Qq = FOA = FOB = FOC = 2 4pe0 l 4pe0 l 2 3 Horizontal component of net force on +Q charge at O is Fx = FOB cos 30° – FOC cos 30° = 0 or Fx = 0 ...(ii) and vertical component of net force on +Q charge at O is Fy = FOB sin 30° + FOC sin 30° – FOA In D ABD, cos 30° =
Since C4 and C′′ are in series, so net capacitance of the network is 1 1 1 1 1 1+ 2 = + = + = C C ′′ C 4 200 100 200 200 or C = pF = 66.7 pF 3 Net charge stored on the combination is 200 Q = CV = × 10 −12 × 300 = 2 × 10–8 C 3 As C′′ and C4 are in series, so Q′′ = Q4 = Q or Q′′ = Q4 = 2 × 10–8 C Q ′′ 2 × 10 −8 C and hence V′′= = = 100 V C ′′ 200 × 10 −12 F Q4 2 × 10 −8 C = = 200 V C 4 100 × 10 −12 F C1 and C′ are in parallel, so V1 = V′ = V′′ or V1 = V′ = 100 V Hence, Q1 = C1V1 = 100 × 10–12 × 100 = 1 × 10–8 C and Q′ = C′V′ = 100 × 10–12 × 100 = 1 × 10–8 C C2 and C3 are in series, so Q2 = Q3 = Q′ = 1 × 10–8 C and V4 =
Hence,V2 =
Q2 1 × 10 −8 C = = 50 V C2 200 × 10 −12 F
Q 1 × 10 −8 C and V3 = 3 = = 50 V. C3 200 × 10 −12 F
21. Given situation is shown with DABC.
or Fy = FOA [2sin30° – 1]
[ FOB = FOC = FOA]
1 or Fy = FOA 2 × − 1 = FOA [1–1] 2 or Fy = 0 So, net force on charge +Q at O is
...(iii)
2 2 Fnet = Fx + Fy or Fnet = 0 22. Consider an infinite line of charge with uniform linear charge density l, as shown in figure. We wish to calculate its electric field at any point P at a distance y from it. The charge on small element dx of the line charge will be dq = ldx The electric field at the point P due to the charge element dq will be ldx 1 dq 1 . = . dE = 4pe0 r 2 4pe0 y 2 + x 2
physics for you | june ‘15
59
The field dE has two components : dEx = –dE sinq and dEy = dE cosq The negative sign in x-component indicates that dE x acts in the negative x-direction. Every charge element on the right has a corresponding charge element on the left. The x-components of two such charge elements will be equal and opposite and hence cancel out. The resultant field E gets contributions only from y-components and is given by E = Ey = ∫ dE y = =2
x =∞
∫
x =0
l = 2pe0
x =+∞
∫
dE cos q
x =−∞
ldx 1 . cos q 4pe0 y 2 + x 2
x =∞
∫
x =0
dx
y2 + x2
\
=
q = p /2
∫
cos q
l 2pe0 y
∫
OR
Refer point 1.4 (1, 6, 7), page-6, 7 (MTG Excel in Physics) 25. Refer point 1.11 (11 (ii)), page-17(MTG Excel in
Physics) OR
Physics)
y sec2 qdq
cos qdq =
q=0
l [sin q]0p/2 2pe0 y
l p sin − sin 0 2pe0 y 2 l E= 2pe0 y 23. (i) Inquisitiveness and sharing. =
(ii) The picture on the screen is produced by striking accelerated electrons on the positively charged screen, that is why Devansh observed charging on the screen. The charged screen also attracts airborne infected particles floating around. As many of the particles collected on the screen surface carry bacteria, the screen becomes contaminated with bacteria.
60
Physics)
26. Refer point 1.11 (9), page-16 (MTG Excel in
cos q 2 y (1 + tan 2 q)
q=0 q = p /2
24. Refer point 1.4 (2, 3, 4), page-6(MTG Excel in
Refer point 1.8 (2), page-11, 12 (MTG Excel in Physics)
Now x = y tanq dx = y sec2q dq l E= 2pe0
When a surgeon puts his finger with gloves near the screen the bacteria may shift to the gloves. To avoid the risk, surgeons are suggested not to bring finger near a video monitor.
physics for you | june ‘15
e A 1 Energy stored U = CV 2 where C = 0 2 d Now battery remains connected and dielectric of constant K = 10 introduced, the potential will remain the same, the capacity becomes C′= KC = 10C V (i) Electric field remains same as E = d 1 (ii) Energy changes as U′= C ′V 2 2 1 (10C )V 2 2 1 1 Change in energy, U′ – U = (10C )V 2 − CV 2 2 2 U′ =
DU =
9 CV2 2
OR
Refer point 1.9, page-13, 14 (MTG Excel in Physics) nn
Silvering of Lens
An object is kept infront of a lens whose one face has been mirrored and location of the image is asked, as below.
To solve such questions, we need to understand first of all that when thin optical devices (lens or mirror) are kept in contact, the optical power follows algebraic addition rule which simply means, a converging (lens or mirror) + another converging (lens or mirror) increases the converging nature whereas converging + diverging will be less of converging. Conventionally, the ability to converge is taken to be positive power and ability to diverge is negative power.
For mirror,
1 , fm where fm = focal length of mirror For lens, 1 Power, Pl = , where f l = focal length of lens fl Hence, if two thin lenses of focal lengths f1 and f2 are kept in contact we can replace the combination with a single lens of equivalent power.
Power, Pm = −
f1 f2
Peq = P1 + P2 ⇒
1 1 1 = + feq f1 f2
Here, we need to follow the sign convention of the respective lenses. Now, what happens if a lens is silvered? Any ray which is incident on the set up, has to pass through lens → mirror → lens, in the same order as indicated.
Note that for a lens, P and f have same sign whereas for mirror they follow opposite sign. Clearly following the definition of power stated above,
Hence, the incident ray passes through lens twice and once through mirror. \ Equivalent power of this combination, whose overall nature (behaviour) is that of a mirror can be written as Peq = Pl + Pm + Pl = 2Pl + Pm 1 2 1 1 1 2 ⇒ − = − ⇒ = − feq fl fm feq fm fl
Contributed By: Bishwajit Barnwal, Aakash Institute, Kolkata
62
physics for you | june ‘15
Feq
This formula can be used as a standard result for zfinding out equivalent focal length of the combination and then we can use mirror formula to trace the image formed. 1 1 1 + = v u feq Note : If focal length of lens has not been given, instead the radius of curvature of the two surfaces and refractive index of the material is given, first find out the focal length of the lens using lens makers formula 1 1 1 = (m − 1) − fl R1 R2 and if the second surface of radius of curvature R2 has been silvered, the focal length of the mirror becomes 1 R2 = fm 2 Apparent depth
Assumption : Until and unless specified, we would assume near normal incidence of incident rays, i.e., angle of incidence would be small, such that we can assume sini ≈ i. This is a very common phenomenon, in which objects kept in denser medium when viewed from rarer medium appear to be shifted closer. Let us see the physics involved behind this.
appear to be shifted even more far than the real depth as below :
Again since the object is in medium of refractive index m1, R R A= = m m12 1 m2
Clearly from both the results, we note that, depth increases when viewed from denser medium and decreases when viewed from rarer medium. Note : If the rarer medium is air, m1 = 1 whereas the other medium might be water, glass etc. whose refractive index, m2 = m (say). R when depth decreases \ A= m mR when depth increases Let us see some applications :
1. Two thin similar convex glass pieces are joined together front to front with its rear portion silvered such that a sharp image is formed 20 cm from the mirror. When the air between the glass pieces is replaced by water (m = 4/3), then find the location of image. Object O, is kept in medium of refractive index (R.I.) m2 and is viewed from a medium of refractive index m1. R is the real depth of the object whereas A is the apparent depth. \ m2sini = m1sinr ⇒ m2i ≈ m1r R R x x ⇒ A= = ⇒ m2 = m1 m2 m21 R A m1
m where, m21 = 2 = refractive index of 2nd medium m1 with respect to 1st medium. On similar lines, if the object was kept in rarer medium and viewed from denser medium, it would
Soln.: The first thing to note here is, what would happen with the thin convex glass pieces? Do they have any role to play? Imagine a thick slab,
The ray gets laterally displaced, inclination does not change since radius of curvature is identical. physics for you | june ‘15
63
So, if the glass piece is said to be thin, it means lateral displacement will be negligible.
So, basically, these glass pieces are used only to hold water, nothing else. In the first situation (without water), the entire set up is equivalent to only a concave mirror on which a ray travelling from infinity is incident. Hence they would be focussed at f =
R . 2
R = 20 cm ⇒ R = 40 cm 2 Now, we have a lens made of water. \
=4 3
R1 = +R
\ \
R2 = –R
1 4 1 1 2 = −1 − = fl 3 R − R 3R 2 4 1 1 2 = − = − ( − R ) 3R f eq fm fl =
−10 −10 1 = =− 3R 3 × 40 12
Let O, I ′ and I represent the object, image formed after 1st refraction, and final image respectively. Remember the rule, as many times there is a change in medium for a plane surface we can apply the concept of real and apparent depth, where we apply R mR for increment in depth, for decrement in m depth and that can be understood by taking a ray and see how it bends. If it bends towards normal, depth increases. If it bends away from normal, depth decreases. For 1st refraction (interface AB), R = x, A = mx 2nd refraction (interface CD), mx + t R = mx + t, A = m \ Shift in location of image, mx + t OI = ∆x = (x + t ) − m 1 ∆x = t 1 − m 3. A point object is kept in front of a glass slab of thickness 10 cm, at a distance of 16 cm. The rear face of slab is silvered and hence starts behaving as a mirror. The image of the object is formed 12 cm behind mirrored face. Find the refractive index of glass slab. Soln.:
\ feq = –12 cm, i.e., the set up behaves like a concave mirror of focal length 12 cm. Hence, rays get focussed at 12 cm from combination. 2. A glass slab of thickness t and refractive index m is kept in between a point object and observer. By what distance the object appears shifted?
Soln.:
To the object the mirror appears to be shifted by, 1 ∆x = 10 1 − m \ From this shifted mirror, object distance = image distance 1 10 ⇒ 16 + = 12 + 10 1 − m m ⇒
20 10 =6 ⇒ m= m 3 nn
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physics for you | june ‘15
Category I (Q1 to Q30)
Each question has one correct option and carries 1 mark, for each wrong answer 1/4 mark will be deducted. 1. An object is located 4 m from the first of two thin converging lenses of focal lengths 2 m and 1 m respectively. The lenses are separated by 3 m. The final image formed by the second lens is located from the source at a distance of
(a) 8.0 m (c) 6.0 m
3. The length of a metal wire is L1 when the tension is T1 and L2 when the tension is T2. The unstretched length of the wire is L +L (a) 1 2 (b) L1L2 2 T2L1 + T1L2 T L −T L (c) 2 1 1 2 (d) T2 + T1 T2 − T1 4. A hollow sphere of external radius R and thickness t ( t1). The rate of heat absorbed by the body is proportional to (a) t24 – t14 (b) (t24 + 273) – (t14 + 273) (c) t2 – t1 (d) t22 – t12 13. The work function of metals is in the range of 2 eV to 5 eV. Find which of the following wavelength of light cannot be used for photoelectric effect. (Consider, Planck constant = 4 × 10–15 eVs, velocity of light = 3 × 108 m/s) (a) 510 nm (b) 650 nm (c) 400 nm (d) 570 nm 14. A thin plastic sheet of refractive index 1.6 is used to cover one of the slits of a double slit arrangement. The central point on the screen is now occupied by what would have been the 7th bright fringe before the plastic was used. If the wavelength of light is 600 nm, what is the thickness (in mm) of the plastic? (a) 7 (b) 4 (c) 8 (d) 6 66
Physics for you | June ‘15
15. The length of an open organ pipe is twice the length of another closed organ pipe. The fundamental frequency of the open pipe is100 Hz. The frequency of the third harmonic of the closed pipe is (a) 100 Hz (b) 200 Hz (c) 300 Hz (d) 150 Hz 16. A 5 mF capacitor is connected in series with a 10 mF capacitor. When a 300 volt potential difference is applied across this combination, the total energy stored in the capacitors is (a) 15 J (b) 1.5 J (c) 0.15 J (d) 0.10 J 17. Two particles of mass m1 and m2, approach each other due to their mutual gravitational attraction only. Then (a) accelerations of both the particles are equal. (b) acceleration of the particle of mass m1 is proportional to m1. (c) acceleration of the particle of mass m1 is proportional to m2. (d) acceleration of the particle of mass m1 is inversely proportional to m1. 18. Three bodies of the same material and having masses m, m and 3m are at temperatures 40°C, 50°C and 60°C respectively. If the bodies are brought in thermal contact, the final temperature will be (a) 45°C (b) 54°C (c) 52°C (d) 48°C 19. A satellite has kinetic energy K, potential energy V and total energy E. Which of the following statements is true? (a) K = –V/2 (b) K = V/2 (c) E = K/2 (d) E = – K/2 20. The line AA′ is on a charged infinite conducting plane which is perpendicular to the plane of the paper. The plane has a surface density of charge s and B is a ball of mass m with a like charge of magnitude q. B is connected by a string from a point of the line AA′. The tangent of the angle (q) formed between the line AA′ and the string is qs qs (a) (b) 2e0mg 4pe0mg (c)
qs 2pe0mg
(d)
qs e0mg
21. The current I in the circuit shown is
(a) 1.33 A (b) zero
(c) 2.00 A (d) 1.00 A
22. The r.m.s. speed of oxygen is v at a particular temperature. If the temperature is doubled and oxygen molecules dissociate into oxygen atoms, the r.m.s. speed becomes (a) v (b) 2v (c) 2v (d) 4v 23. Two particles, A and B, having equal charges, after being accelerated through the same potential difference enter a region of uniform magnetic field and the particles describe circular paths of radii R1 and R2 respectively. The ratio of the masses of A and B is (a) R1/R2 (b) R1/R2 2
(c) (R1/R2)
2
(d) (R2/R1)
24. A large number of particles are placed around the origin, each at a distance R from the origin. The distance of the center of mass of the system from the origin is (a) = R (b) ≤ R (c) > R (d) ≥ R 25. A 20 cm long capillary tube is dipped vertically in water and the liquid rises upto 10 cm. If the entire system is kept in a freely falling platform, the length of water column in the tube will be (a) 5 cm (b) 10 cm (c) 15 cm (d) 20 cm 26. A train is moving with a uniform speed of 33 m/s and an observer is approaching the train with the same speed. If the train blows a whistle of frequency 1000 Hz and the velocity of sound is 333 m/s, then the apparent frequency of the sound that the observer hears is (a) 1220 Hz (b) 1099 Hz (c) 1110 Hz (d) 1200 Hz 27. A photon of wavelength 300 nm interacts with a stationary hydrogen atom in ground state. During the interaction, whole energy of the photon is transferred to the electron of the atom. State which possibility is correct. (Consider, Planck constant = 4 × 10–15 eVs, velocity of light = 3 × 108 m/s, ionization energy of hydrogen = 13.6 eV) (a) Electron will be knocked out of the atom (b) Electron will go to any excited state of the atom
(c) Electron will go only to first excited state of the atom (d) Electron will keep orbiting in the ground state of atom 28. Block B lying on a table weighs W. The coefficient of static friction between the block and the table is m. Assume that the cord between B and the knot is horizontal. The maximum weight of the block A for which the system will be stationary is
(a) W tan q m
(b) mW tanq
(c) mW 1 + tan2 q
(d) mW sinq
29. The inputs to the digital circuit are shown below. The output Y is
(a) A + B + C
(b) (A + B) C
(c) A + B + C
(d) A + B + C
30. Two particles A and B having different masses are projected from a tower with same speed. A is projected vertically upward and B vertically downward. On reaching the ground (a) velocity of A is greater than that of B. (b) velocity of B is greater than that of A. (c) both A and B attain the same velocity. (d) the particle with the larger mass attains higher velocity.
Category II (Q31 to Q35)
Each question has one correct option and carries 2 marks, for each wrong answer 1/2 mark will be deducted. 31. Two cells A and B of e.m.f. 2 V and 1.5 V respectively, are connected as shown in figure through an external resistance 10 W. The internal resistance of each cell is 5 W. The potential difference EA and EB across the terminals of the cells A and B respectively are Physics for you | June ‘15
67
(a) (b) (c) (d)
EA = 2.0 V, EB = 1.5 V EA = 2.125 V, EB = 1.375 V EA = 1.875 V, EB = 1.625 V EA = 1.875 V, EB = 1.375 V
32. A charge q is placed at one corner of a cube. The electric flux through any of the three faces adjacent to the charge is zero. The flux through any one of the other three faces is (a) q/3e0 (b) q/6e0 (c) q/12e0 (d) q/24e0 33. In the circuit shown below, the switch is kept in position ‘a’ for a long time and is then thrown to position ‘b’. The amplitude of the resulting oscillating current is given by
(a) E L/C (c) infinity
(b) E/R (d) E C / L
34. The pressure p, volume V and temperature T for a
AT − BT 2 , where A V and B are constants. The work done by the gas when the temperature changes from T1 to T2 while the pressure remains constant, is given by 2 2 (a) A(T2 − T1 ) + B (T2 − T1 ) certain gas are related by p =
(b)
A(T2 − T1) B(T22 − T12 ) − V2 − V1 V2 − V1
2 2 (c) A(T2 − T1 ) − B (T2 − T1 )
(d)
A(T2 − T22 ) V2 − V1
35. A cylinder of height h is filled with water and is kept on a block of height h/2. The level of water in the cylinder is kept constant. Four holes numbered 1, 2, 3 and 4 are at the side of the cylinder and at heights 0, h/4, h/2 and 3h/4 respectively. When all four holes are opened together, the hole from which water will reach farthest distance on the plane PQ is the hole no. 68
Physics for you | June ‘15
(a) 1 (c) 3
(b) 2 (d) 4
Category III (Q36 to Q40)
Each question has one or more correct option(s), choosing which will fetch maximum 2 marks on pro rata basis. However, choice of any wrong option(s) will fetch zero mark for the question. 36. Consider two particles of different masses. In which of the following situations the heavier of the two particles will have smaller de Broglie wavelength? (a) Both have a free fall through the same height. (b) Both move with the same kinetic energy. (c) Both move with the same linear momentum. (d) Both move with the same speed. 37. A circular disc rolls on a horizontal floor without slipping and the centre of the disc moves with a uniform velocity v. Which of the following values the velocity at a point on the rim of the disc can have? (a) v (b) –v (c) 2v (d) zero 38. A conducting loop in the form of a circle is placed in a uniform magnetic field with its plane perpendicular to the direction of the field. An e.m.f. will be induced in the loop if (a) it is translated parallel to itself. (b) it is rotated about one of its diameters. (c) it is rotated about its own axis which is parallel to the field. (d) the loop is deformed from the original shape. 39. Find the right condition(s) for Fraunhoffer diffraction due to a single slit. (a) Source is at infinite distance and the incident beam has converged at the slit. (b) Source is near to the slit and the incident beam is parallel. (c) Source is at infinity and the incident beam is parallel. (d) Source is near to the slit and the incident beam has converged at the slit.
40. Two charges +q and –q are placed at a distance ‘a’ in a uniform electric field. The dipole moment of ^
^
the combination is 2qa(cos q i + sin q j), where q is the angle between the direction of the field and the line joining the two charges. Which of the following statement(s) is/are correct? (a) The torque exerted by the field on the dipole vanishes. (b) the net force on the dipole vanishes. (c) The torque is independent of the choice of coordinates. (d) The net force is independent of ‘a’. solutions
1. (b) : For first lens, f1 = 2 m, u1 = –4 m, v1 = ? Using lens formula, 1 1 1 − = v1 u1 f1 1 1 1 1 1 1 1 = − = − = ; v1 −4 2 v1 2 4 4 v1 = 4 m For second lens, f2 = 1 m Object distance, u2 = |v1| – 3 = 4 – 3 = 1 m Image distance, v2 = ? Again using lens formula, 1 1 1 1 − = or = 1 + 1 = 2; v2 = 0.5 m v2 1 1 v2 So, distance of final image from the source point = 4 + 3 + v2 = 4 + 3 + 0.5 = 7.5 m 2. (b) : Given situation is shown in the figure. L = length of pendulum m = mass of the bob q = angle between pendulum and vertical v = speed of pendulum Let T be the tension in the string at given situation. v2 L Using Newton’s second law of motion, mv 2 Fnet = L mv 2 T − mg cos q = L mv 2 T = mg cos q + L
Acceleration of the bob =
3. (c) : Let the unstretched length of the wire be L. Area of cross-section of the wire = A (say) Stress Young’s modulus, Y = Strain Case I : Length of wire = L1 Tension in the wire = T1 Extension in the wire = (L1 – L) T1 /A T1L \ Y= ...(i) = (L1 − L)/L A(L1 − L) Case II : Length of wire = L2 Tension in wire = T2 Extension in wire = (L2 – L) T2 /A T2 L = \ Y= (L2 − L)/L A(L2 − L)
...(ii)
From eqns (i) and (ii), T1L T2 L = A(L1 − L) A(L2 − L) T1(L2 – L) = T2(L1 – L) T1L2 – T1L = T2L1 – LT2 L(T2 – T1) = T2L1 – T1L2 T L −T L \ L= 2 1 1 2 T2 − T1
4. (b) : Here, density of metal = r Mass of hollow sphere = (Surface area) × (thickness) × r = 4pR2 t r Sphere will float in water if 4 (4 pR2tr) g ≤ pR3 rw g 3 R rw t≤ 3 r \
t≤
R 3r
[ rw = 1 g cm–3]
rl rl × l rl 2 = = A A×l V Here, r and V are constant. \ R ∝ l2
5. (c) : As R =
\
R1 l12 = R2 l22
Given, l2 = 2l1 Hence,
R1 l12 1 = = R2 4l12 4 R2 : R1 = 4 : 1 Physics for you | June ‘15
69
6. (d) : In the circuit, diode D1 is reverse biased and offers infinite resistance, diode D2 is forward biased and offers 50 W resistance. Equivalent circuit is redrawn. Total resistance of the circuit, R = 50 + 50 + 150 = 250 W V = 10 V, I = ? V 10 I= = = 0.04 A R 250 So, current through the resistance 150 W is 0.04 A. 7. (b) : Assuming magnetic field (B), conductor length (l) and its velocity (v) are mutually perpendicular so, emf induced between the two ends of the conductor e = Blv Here, B = 0.1 T, l = 0.1 m, v = 15 m s–1, e = ? e = 0.1 × 0.1 × 15 = 0.15 V 8. (b) : Given situation is shown in the figure.
Angle of refraction, r = (90° – i) Using Snell’s law at the interface m1 sin i = m2 sin r 1 × sin i = m sin(90° – i) = m cos i tan i = m 9. (a) : Total angular momentum of two particles about the point O Lnet = LA + LB = rA × pA + rB × pB
Here, rA = 1.5 m j, rB = 2.8 m i pA = mv A = (6.5)(2.2 i ) = (14.3 i ) N s pB = mv B = (3.1)(3.6 j ) = (11.16 j ) N s \ Lnet = (1.5 j ) × (14.3 i ) + (2.8 i ) × (11.16 j ) = 21.45(−k ) + 31.248(k ) ≈ 9.8k Lnet = 9.8 kg m2s −1 70
Physics for you | June ‘15
10. (b) : Here, v A = (10 i ) m s −1
v B = (20 cos 60° i + 20 sin 60° j ) m s −1
3 1 = (20) i + (20) j = (10 i + 10 3 j ) m s −1 2 2 Relative velocity of B with respect to A is given by v BA = v B − v A = (10 i + 10 3 j ) − 10 i = 10 3 j m s −1 11. (c) : When light is refracted/reflected from a surface, then frequency of light does not change because it depends on the source of light. 12. (*) : Rate of heat radiation emitted by a body at temperature t1 °C (= (t1 + 273) K). u1 = e s A (t1 + 273)4 Rate of heat radiation absorbed by a body due to surrounding temperature t2°C (= (t2 + 273) K) u = e s A (t2 + 273)4 Net rate of heat absorbed by the body = u – u1 = e s A [(t2 + 273)4 – (t1 + 273)4] *None of the given options is correct. Option (a) would be correct if t1 and t2 were in kelvin. 13. (b) : Range of work function of metals = 2 – 5 eV hc = 4 × 10–15 eVs × 3 × 108 m s–1= 1200 eV-nm hc As, l = E 1200 eV - nm l min = = 240 nm 5 eV 1200 eV - nm l max = = 600 nm 2 eV Hence light of wavelength 650 nm cannot be used for photoelectric effect. 14. (d) : Here, m = 1.6, n = 7, l = 600 nm t=? According to question, shift in central point due to insertion of plastic sheet = 7b (m − 1)tD Dl or =7× d d \
t=
7 l 7 × 600 × 10−9 = m −1 1. 6 − 1
7 × 6 × 10−7 = 7 × 10−6 m 0. 6 t = 7 mm 15. (c) : Let length of open organ pipe = lo length of closed organ pipe = lc Also, lo = 2lc Fundamental frequency of the open pipe, uo = 100 Hz v Also, uo = = 100 2lo =
v v = 200; = 200 2lc lo v = 400 lc
Fundamental frequency of closed organ pipe v 400 uc = = = 100 Hz 4lc 4 So the frequency of third harmonic of the closed organ pipe = 3uc = 300 Hz. 16. (c) : Here, C1 = 5 mF C2 = 10 mF V = 300 V U=? Equivalent capacitance of the circuit, C × C2 5 × 10 10 10 C= 1 = = mF = × 10−6 F C1 + C2 5 + 10 3 3 1 1 10 U = CV 2 = × × 10−6 × (300)2 = 0.15 J 2 2 3 17. (c) : Given situation is shown in the figure. Gravitational force, Gm1m2 F= r2 Acceleration of particle of mass m1, F Gm2 a1 = = m1 r2 \ a1 ∝ m2 18. (b) : Here m1 = m2 = m, m3 = 3m s1 = s2 = s3 = s, T1 = 40° C, T2 = 50° C, T3 = 60° C Let T be the final temperature at the thermal equilibrium. Using principle of calorimetry, Heat gained by m1 and m2 = heat lost by m3 m1s1(T – T1) + m2s2(T – T2) = m3s3(T3 – T) ms(T – 40) + ms(T – 50) = 3ms(60 – T) T – 40 + T – 50 = 3(60 – T)
2T – 90 = 180 – 3T 5T = 270; \ T = 54°C 19. (a) : As potential energy of satellite GMm V =− r GMm Kinetic energy of satellite, K = 2r Total energy of satellite, E = K + V GMm GMm GMm E= − =− 2r r 2r V V or K + V = ; \ K = − 2 2 20. (d) : Electric field due to charged conducting sheet, s E= e0 Let T be the tension in the string. Ball is in equilibrium so T cos q = mg T sin q = qE T sin q qE \ = T cos q mg qE qs tan q = = mg e0mg
21. (a) : Given circuit can be redrawn as shown in figure.
Here, first two cells are in parallel so net emf, e 2 2 ∑r 2+2 =2 V = eeq = 1 1 1 + ∑r 2 2 1 1 1 req = ∑ = + = 1 W r 2 2 Current in the circuit, 2+2 4 I= = = 1.33 A 1+ 2 3 3RT M For oxygen molecules at temperature T, 3RT ...(i) vrms = v = M Now, temperature is doubled and oxygen molecules
22. (c) : As, vrms =
Physics for you | June ‘15
71
dissociate into oxygen atom (molar mass becomes M/2) then rms speed will be vrms ′ =
3R(2T ) 3RT =2 = 2v [using eqn. (i)] M (M /2)
23. (c) : Radius of circular orbit of a charged particle in a magnetic field is given by, mv p 2mK ...(i) r= = = qB qB qB Here, K = Kinetic energy of charged particle Charge q is accelerated through some potential difference V. So kinetic energy of charge, K = qV From eqn. (i), r=
2mqV qB
For given q, V and B, r ∝
m
2
m R r rA mA or A = A = 1 = mB rB R2 rB mB
25. (d) : Capillary rise in an accelerated lift used as a plateform, 2S cos q h= rr aeff
Here, aeff = 0 (for free falling plateform) \ h=∞ It means length of water column in the tube will be 20 cm.
26. (a) : According to Doppler’s effect of sound, apparent frequency heard by observer, v + v0 u= u0 v − v s Here, u0 = 1000 Hz, v0 = vs = 33 m s–1 v = 333 m s–1, u = ? 366 333 + 33 u= × 1000 = × 1000 = 1220 Hz 333 − 33 300 27. (d) : Here, wavelength of photon, l = 300 nm h = 4 × 10–15 eVs, c = 3 × 108 m s–1 Ionization energy of hydrogen = 13.6 eV Physics for you | June ‘15
=
= 4 eV For hydrogen atom, En = −
(300 × 10−9 m)
13.6
n2 First excitation energy = E2 – E1
eV
13.6 13.6 − − 22 12 = –3.4 + 13.6 = 10.2 eV Since energy of photon is less than 10.2 eV, so no excitation is possible, i.e., electron will keep orbiting in the ground state of atom. 28. (b) : Weight of block B = W =−
2
24. (b) : Suppose along arc AB there is a large number of particles. If arc AB → 0 the position of centre of mass will be at distance R from the origin. If arc length AB increases, centre of mass of the system starts moving down (< R).
72
hc l (4 × 10−15 eVs)(3 × 108 m s −1 )
Energy of photon, E =
Weight of block A = W′ T = tension in the string Normal on block B, N = W Friction force on block B, f = mN = mW System will be in equilibrium, if T cos q = f = mW T sin q = W′ Divide eqn (ii) by eqn (i), we get sin q W ′ = \ W′ = mW tan q cos q mW
(say)
...(i) ...(ii)
29. (c) :
Output of NAND gate, Y1 = AB Output of NOT gate, Y2 = C Output of OR gate, Y = Y1 + Y2 = AB + C = A+ B +C
( AB = A + B )
30. (c) 31. (c) : Suppose current I is I 2V 5 flowing through the circuit. A 10 Using Kirchhoff ’s voltage 1.5 V law in the circuit B 2 – 1.5 – 5I – 10I – 5I = 0 5 0. 5 1 0.5 = 20I; I = = A 20 40 Terminal potential difference across the cell A, 1 EA = 2 – I r = 2 − × 5 = 1.875 V 40 Terminal potential difference across the cell B, 1 EB = 1.5 + I r = 1.5 + × 5 = 1.625 V 40 32. (d) 33. (d) : During connection of switch with point a for long time, capacitor gets fully charged.
35. (b) : Suppose there is a hole in the cylinder at depth y below water level.
Velocity of efflux v = 2 gy Time taken by water to reach on the plane PQ will be
\
h 2 + (h − y ) 2 3h − 2 y t= = g g Horizontal distance x travelled by the liquid is 3h − 2 y x = vt = 2 gy g x = 2 y(3h − 2 y )
Charge on capacitor, q = CE
q2 2C When switch is thrown to position b, there is an LC oscillating circuit. Suppose amplitude of current in LC circuit is I0. Using energy conservation principle, Maximum electrical energy = Maximum magnetic energy Energy stored in capacitor, U =
q2 1 2 = LI0 2C 2
C2 E2 C = LI02 ; \ I0 = E C L 34. (c) : For a certain gas, AT − BT 2 pressure, p = V 2 or, V = AT − BT ... (i) p Work done by the gas at constant pressure, W = pDV = p(V2 – V1) AT − BT 2 AT − BT 2 2 1 = p 2 − 1 [Using eqn (i)] p p = A(T2 – T1) – B(T22 – T12)
For x to be maximum, 1 2 2 y(3h − 2 y )
dx =0 dy
× 2(3h − 4 y ) = 0
3h 4 3h h Hence, x will be maximum at y = = h − 4 4 where hole number 2 is present. 36. (a,b,d) : de-Broglie wavelength of a particle is given by, h h h l= = = ; K = kinetic energy p mv 2 mK or, 3h – 4y = 0 \ y =
(a) If both the particles are allowed to fall through same height, their speed will be same. v = 2 gh 1 m So heavier block will have smaller de-Broglie wavelength. i.e., option (a) is correct. (b) For same kinetic energy (K), 1 l∝ m Again heavier block will have smaller de-Broglie wavelength, i.e., option (b) is correct. \ l∝
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(c) For same momentum, both particles have same de Broglie wavelength. So option (c) is incorrect. (d) For same speed, 1 l∝ m Again, heavier particle will have smaller de-Broglie wavelength, i.e., option (d) is correct. 37. (a, c, d) : In case of pure rolling, v = wR So, velocity at the point of contact, vg = v – wR = 0 Velocity at the top, vt = v + wR = 2 v Velocity at a point on the rim of disc,vr = v is possible if v and wR are at 120°. Velocity at a point on the rim of disc, vr = –v is not possible because –1 ≤ cosq ≤ 1.
38. (b, d) : An emf will be induced in the loop if there is change in magnetic flux through it. Flux changes through the loop only in options (b) and (d). 39. (b, c) : In Fraunhoffer diffraction, the incident rays on the slit are parallel. This can be achieved by keeping : (a) point source at the focus of a converging lens (b) point source at infinite distance 40. (b, c, d) :
Net force on the dipole = +qE – qE = 0 So, option (b) is correct. Torque on the dipole, t = pEsinq ≠ 0 Here, p, E and q are independent of choice of coordinates. So, options (c) and (d) are correct.
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By : Prof. Rajinder Singh Randhawa*
1. A boy standing in an elevator observes a screw falls from the ceiling. The ceiling is 3 m above the floor. (a) If the elevator is moving upward with a speed of 2.2 m s–1, how long does it take for the screw to hit the floor? (b) How long is the screw in air if the elevator starts from rest when the screw falls, and moves upwards with a constant acceleration of a = 4 m s–2? 2. An enemy fighter jet is flying at a constant height of 250 m with a velocity of 500 m s–1. The fighter jet passes over an anti-aircraft gun that can fire at any time and in any direction with a speed of 100 m s–1. Determine the time interval during which the fighter jet is in danger of being hit by the gun bullets. 3. A particle is to be projected from a point P so that it may strike the incline perpendicularly. Determine the required velocity of projection if it is horizontal initially.
4. A steel ball which can slide on a smooth rod of length L is attracted by an electromagnet. The force of attraction imparts an acceleration a = K/(L – x)2, where K is constant and x is distance travelled. If the ball is released from rest at x = 0, determine the velocity v with which it strikes the pole face.
5. The current velocity of a river grows in proportion to the distance from its bank and reaches the maximum value v0 in the middle. Near the banks the velocity is zero. A boat is moving along the river in such a manner that it is always perpendicular to current and the speed of the boat in still water is u. Find the distance through which the boat crossing the river will be carried away by the current if the width of river is c. Also trace the trajectory of boat. 6. A gas-filled spherical balloon is expanding. The radius of sphere at time t is r. Find the radius when the rates of increase of the surface area and the radius are numerically equal. 7. A projectile is fired with velocity u from a gun adjusted for a maximum range. It passes through two points P and Q whose heights above the horizontal are h each. Show that the separation u 2 between the two is u − 4 gh . g 8. A particle is moving in a circle of radius r in such a way that at any instant, the total acceleration makes an angle of 45° with radius. Initial speed of particle is u. Find the time taken to complete the first revolution.
*Randhawa Institute of Physics, S.C.O. 208, First Fl., Sector-36D & S.C.O. 38, Second Fl., Sector-20C, Chandigarh, Ph. 09814527699
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SOLUTIONS
1. (a) Consider the elevator floor to be origin. It moves with a constant velocity. Equation for floor yf = vf t = 2.2t ...(i) The screw falls with acceleration due to gravity 1 ...(ii) ys = h + 2.2 t − gt 2 2 At time t, ys = yf 1 i.e., 2.2 t = h + 2.2t − gt 2 2 2×3 2h or t = = = 0.78 s g 9. 8 1 (b) Equation for floor is yf = af t 2 2 1 2 Equation for screw is ys = h − gt 2 When the screw meets the floor, ys = yf 1 1 \ a f t 2 = 3 − gt 2 2 2 2×3 or t = = 0.66 s g + af 2. The equation of trajectory of bullet is 1 2 gx y = x tan q − 2 2 sec2 q u 1 2 gx = x tan q − 2 2 (1 + tan2 q) ...(i) u For a given value of x, maximum y can be determined from 1 2 gx u2 dy = x − 2 2 (2 tan q) = 0 or tan q = gx d(tan q) u ...(ii) Putting (ii) in (i), we get, 1 2 u2 2 gx ymax = − 2 2g u The bullet can hit an area defined by 1 2 u2 2 gx y≤ − 2 2g u On substituting given values, we get x2 ≤ 250 or − 500 2 ≤ x ≤ + 500 2 2000
The fighter jet can travel 1000 2 m while it can be hit. So the plane is in danger for a period of 1000 2 =2 2s 500 3. Let v be the velocity when the particle strikes the incline. The velocity components are vx = v0 and vy = gt v v v \ tan q = x = 0 ⇒ t = 0 cot q g v y gt Since, x = v0t, we have v2 x = 0 cot q g The equation of trajectory of particle is
1 gx 2 2 v02 The incline is a straight line, its equation is y = xtanq On solving (i), (ii) and (iii) we get y =h−
v0 =
...(ii) ...(iii)
2 gh
2 + cot 2 q
K
4. Acceleration of ball, a =
(L − x)2 dv dv dx vdv Since, a = = × = dt dx dt dx vdv K \ = dx (L − x)2 v dv =
...(i)
(given)
Kdx
(L − x)2 On integrating, v
L − D /2
0
0
∫ v dv =
∫
K dx (L − x)2 L − D /2
v2 1 =K 2 L − x 0
or
v2 2 1 =K − D L 2
2 1 \ v = 2K − D L physics for you | june ‘15
77
5. Current velocity, v = ky c v = v0 when y = , 2 2v k \ v0 = c or k = 0 c 2 2v0 ut y = ut \ v = c v 2v u \ Rate of change of velocity = = 0 t c \ Acceleration of the boat along the water 2v u current, a = 0 x c 1 2v0u 2 \ x =0×t + t 2 c v u y2 v x = 0 2 = 0 y2 ( y = ut ) uc c u uc \ y2 = x v0 It represents the equation of a parabola. 1 cv0 c When y = , x = 2 4 u cv0 Hence, drift = 2 x = 2u 6. Surface area of sphere is S = 4pr2, ds dr So, = 8pr dt dt ds dr (given) As = dt dt 1 Hence 8pr = 1 or r = p 8
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7. The trajectory of projectile is given by gx 2
(1 + tan2 q) 2u2 Gun is adjusted for maximum range, i.e., q = 45° y = x tan q −
\ y=x−
gx 2
u2 For y = h, we have, u2 u2 g h = x − 2 x 2 or x 2 − x + h = 0 g g u
If x1 and x2 are roots of the above equation, x1 + x2 =
u2 u2 and x1x2 = h g g
2
u2 u2 \ (x1 − x2 ) = (x1 + x2 ) − 4x1x2 = − 4 h g g u x1 − x2 = u2 − 4 gh g 2
2
8. Since, tangential acceleration (at) = radial acceleration (ar) ( q = 45°) dv v 2 = ⇒ dt r
v
2pr
T
dv
t
dt ∫ v2 = ∫ r u 0 1 1 t ur ⇒ − = or v = \ u v r r − ut ⇒
dx ur = dt r − ut
ur dt ⇒ 2pr = − r ln[(r − ut]T0 r − ut 0 0 r which gives T = (1 − e −2p ) u
∫
dx = ∫
nn
1. A particle in uniform circular motion can possess (a) radial acceleration only (b) tangential acceleration only (c) both radial and tangential acceleration (d) neither radial nor tangential acceleration 2. Which of the following physical quantities is not dimensionless ? (a) Relative density (b) Relative velocity (c) Relative refractive index (d) Relative permittivity 3. The numerical ratio of displacement to the distance covered is always (a) less than one (b) equal to one (c) equal to or less than one (d) equal to or greater than one 4. A student unable to answer a question on Newton’s laws of motion attempts to pull himself up by tugging on his hair. He will not succeed because (a) the force exerted is small. (b) the frictional force while gripping is small. (c) Newton’s law of inertia is not applicable to living beings. (d) the force applied is internal to the system. 5. If mass-energy equivalence is taken into account when water is cooled to form ice, the mass of water should (a) increase (b) remain unchanged (c) decrease (d) first increase then decrease 6. Which one of the following waves does not involve oscillations of particles of a medium? (a) Waves in a hanging spring (b) Ripples on a water surface (c) A light wave (d) Ultrasonic waves 80
Physics for you | june ‘15
7. If a body is projected with an angle q to the horizontal, then (a) its velocity is always perpendicular to its acceleration. (b) its velocity becomes zero at its maximum height. (c) its velocity makes zero angle with the horizontal at its maximum height. (d) the body just before hitting the ground, the direction of velocity coincides with the acceleration. 8. The internal energy of an ideal gas depends on which of the following factors? (a) Pressure only (b) Volume only (c) Temperature only (d) Pressure, volume and temperature 9. Bernoulli’s principle is not involved in the working/ explanation of (a) movement of spinning ball (b) blades of a kitchen mixer (c) heart attack (d) dynamic lift of an aeroplane 10. The temperature of the system decreases in the process of (a) isothermal expansion (b) adiabatic expansion (c) isothermal compression (d) adiabatic compression 11. A wire is stretched under a force. If the wire suddenly snaps, the temperature of the wire, (a) remains the same (b) decreases (c) increases (d) first decreases then increases 12. For a constant volume gas thermometer, one should fill the gas at (a) low temperature and low pressure (b) low temperature and high pressure (c) high temperature and low pressure (d) high temperature and high pressure
13. Which of the following forces is a contact force? (a) Gravitational force (b) Electrostatic force (c) Magnetic force (d) Buoyant force 14. Which of the following characteristics does not change due to the damping of simple harmonic motion ? (a) Angular frequency (b) Time period (c) Initial phase (d) Amplitude 15. A cricket bat is cut at the location of its centre of mass as shown. Then (a) (b) (c) (d)
the two pieces will have the same mass. the bottom piece will have larger mass. the handle piece will have larger mass. mass of handle piece is double the mass of bottom piece. 16. Which of the following thermodynamic variables is an extensive variable ? (a) Temperature (b) Pressure (c) Internal energy (d) Density 17. Angle between wave velocity and particle velocity of a longitudinal wave is (a) 90° (b) 60° (c) 0° (d) 120° 18. If there were no gravity, which of the following will not be there for a fluid ? (a) Viscosity (b) Surface tension (c) Pressure (d) Archimedes’ upward thrust 19. A lead ball falling freely from a height strikes the ground, as a result its temperature rises. This is due to (a) friction of air (b) conversion of chemical energy into heat (c) conversion of mass into heat (d) conversion of mechanical energy into heat 20. Which of the following quantities is zero on an average for the molecules of an ideal gas in equilibrium ? (a) Kinetic energy (b) Momentum (c) Density (d) Speed 21. In a simple harmonic oscillator, at the mean position (a) kinetic energy is minimum, potential energy is maximum. (b) both kinetic and potential energies are maximum. (c) kinetic energy is maximum, potential energy is minimum. (d) both kinetic and potential energies are minimum. 22. In planetary motion, the angular momentum conservation leads to the law of (a) orbits (b) areas
23.
24.
25.
1.
(c) periods (d) conservation of linear momentum Energy is not transferred by (a) transverse progressive wave (b) longitudinal progressive wave (c) stationary wave (d) electromagnetic wave Which of the following is not an illustration of Newton’s third law ? (a) Flight of a jet plane (b) A cricket player lowers his hands while catching a cricket ball (c) Walking on the floor (d) Rebounding of a rubber ball Hydrostatic paradox states that the pressure exerted by a liquid (a) depends on the shape of the containing vessel and independent of height of liquid column. (b) depends on both shape of the containing vessel and height of liquid column. (c) independent of both shape of the containing vessel and height of liquid column. (d) depends on height only and independent of the shape of the containing vessel. soLuTioNs (a) : A particle in uniform circular motion can possess radial acceleration only.
2. (b) : Relative velocity has the dimensions of velocity. 3. (c) : Since displacement is always less than or equal to distance covered, but never greater than distance covered. Hence the numerical ratio of displacement to the distance covered is always equal to or less than one. 4. (d) : According to Newton’s first law of motion, the state of rest or motion of body in a straight line with uniform velocity remains the same unless an external force is applied on the body. Internal forces are the forces which are exchanged by the particles in the system. 5. (c) : When water is cooled to form ice, its thermal energy decreases. By mass-energy equivalence, the mass of water should decrease. 6. (c) : A light wave is an electromagnetic wave. It does not require a medium for propagation. A man may imagine things that are false, but he can only understand things that are true, for if the things be false, the apprehension of them is not understanding -Sir Isaac Newton
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7. (c) : In angular projection, the body at the highest point has velocity =ucosq in the horizontal direction which makes zero angle with the horizontal direction. 8. (c) : The internal energy of an ideal gas depends only on temperature. 9. (b) : Movement of spinning ball, heart attack and dynamic lift of an aeroplane, all are based on the Bernoulli’s principle. 10. (b) : In an adiabatic expansion the temperature of the system decreases while in an adiabatic compression it increases. In an isothermal expansion or isothermal compression the temperature of the system remains constant. 11. (c) : The work done on the wire to produce a strain in it will be stored as energy which is converted into heat when the wire snaps suddenly. Due to it, the temperature of the wire increases. 12. (c) 13. (d) : Among the given forces buoyant force is a contact force whereas all others are non-contact forces. 14. (c) 15. (b): The centre of mass is closer to massive part of the body therefore the bottom piece of bat has larger mass. 16. (c) : Among the given thermodynamics variables internal energy is an extensive variable whereas all others are intensive variables. 17. (c) : In a longitudinal wave the particles of the medium oscillate along the direction of wave motion. Thus the angle between wave velocity and particle velocity is 0°. 18. (d) : For a fluid, Archimedes’ upward thrust will not be present if there were no gravity. 19. (d) 20. (b) 21. (c) : At the mean position, the kinetic energy is maximum and potential energy is minimum. 22. (b) : Angular momentum conservation leads to the law of areas. 23. (c) : Energy is not transferred by stationary wave but it is transferred by all others. 24. (b) : A cricket player lowers his hands while catching a cricket ball is an illustration of Newton’s second law, whereas all others are illustrations of Newton’s third law. 25. (d) nn
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Y U ASK
WE ANSWER Do you have a question that you just can’t get answered? Use the vast expertise of our mtg team to get to the bottom of the question. From the serious to the silly, the controversial to the trivial, the team will tackle the questions, easy and tough. The best questions and their solutions will be printed in this column each month.
The fluttering has often been attributed to vortex formation by the flag. Indeed, whether the wind merely extends the flag or causes it to flutter, the free end of the flag sheds vortexes, that is, alternating on the left and right side of the flag, form and then move downstream. The vortexes are larger if the flag flutters, but they are a product of, not the cause of, the fluttering and they can be present even if the flag is not fluttering.
Q1. How does cooking happen in a microwave oven?
– Abhishek Yadav (U.P.)
Ans. Microwave oven heats and cooks food by exposing it to electromagnetic radiation in the microwave spectrum. Passage of microwaves through food results in increased agitation of molecules. Microwave induces polarized molecules in the food to rotate and produce in a process known as dielectric heating. At microwave frequencies, the alternating electric field of the radiation interacts with the electric dipole moment of water molecules, making them vibrate faster. The absorption length of the microwave energy is long, that means not all energy falling on a chunk of cooking food gets absorbed while traversing through it. It also mean that the heating occurs all through the body of the food. That is the reason you do not get a crisp surface on the outside, as you would in normal cooking where the heat has to travel inwards from outside. When you cook a big chunk of raw food in a pan on open fire, the surface might become crisp, while the inside might remain relatively raw. Also remember that a microwave oven is an enclosure whose walls have excellent reflecting properties. Therefore, the fact that little microwave energy is absorbed upon initial traversal through the chunk of food does not mean the rest of the energy is lost, the reflective metallic walls of the oven bounce the microwaves back and forth to ensure repeated interaction with water contained in the food. Q2. What causes a flag on a flagpole to flap even in a moderate breeze? – Anoop Kumar (Bihar)
Ans. Imagine that the plane of the flag is at an angle to the direction of the passing air so that the air pushes against one side of the flag. That push can simply straighten out the flag, making it extend in the direction of the airflow. Instead, the push can bend the flag. If the airspeed is above a certain critical value, this bending can become unstable and the flag will then flutter. 84
physics for you | June ‘15
Q3. What causes water waves? How are they generated? – Akshi Sharma (Delhi)
Ans. These two questions are not yet clearly answered. However, a simple explanation is as follows. A breeze or some disturbance in the air or water creates ripples. The ripples can then grow into larger waves as wind moves across them. In particular, the wind pushes on the windward side of a crest, comes over the crest, and then breaks up into vortexes, and so the pressure difference between the windward and leeward sides of the crest can push the crest downwind and also make it taller. In other words, the wind can feed energy into the crest. If the wind becomes stronger, the waves grow larger and their wavelengths also change. Q4. Electric field for a point charge is proportional to 1/r2 but electric field for a dipole is proportional to 1/r3. Why? – Anubhab Banerjee (W.B.) Ans. Experimentally, electric field due to a point charge is
1 q 1 ;E∝ 2. 2 4πε0 r r Now electric dipole is the distribution of some charges which give net charge on dipole zero and 1 p 1 ,V ∝ potential due to it, V = 2 4πε0 r r2 (For simplicity we take dipole as two oppositely charges of same magnitude separated by a very small distance.) dV Also, E = − dr 1 So electric field due to dipole, E ∝ r3 1 Note : For infinitely line charge, E ∝ . So electric r field due to any charge system depends on its distribution.
given by E =
Readers can send their answer with complete address before 15th of every month to win exciting prizes. Winners' name with their valuable feedback will be published in next issue.
3.
5. 11. 13.
16.
19. 20.
21. 22. 23. 24. 25. 26.
1. 2. 4.
across The ratio of the total absorbed radiation to the total incident radiation. (11) The fraction of solar energy reflected from the earth back into space. (6) Short form of quasi-stellar radio source. (6) Black body radiation that is predicted to be released by black holes, due to quantum effects near the event horizon. (7, 9) A collective excitation in a periodic, elastic arrangement of atoms or molecules in condensed matter, like solids and some liquids. (6) Carrier of strong nuclear force. (5) A hypothetical set of multiple possible universe which exist in parallel with each other. (10) A unit of atmospheric pressure equivalent to 1000 dyne per cm2. (8) A branch of physics applied to technology that makes use of gas or pressurized air. (10) A mechanism that converts rotational motion to linear motion and a torque to a linear force. (5) CGS unit of magnetic field strength. (7) The division between two regions of differing physical properties. (8) The phenomenon that consists of a given system being driven by another vibrating system or by external forces to oscillate with greater amplitude at some preferential frequencies. (9)
DoWN Single disturbance moving through a medium or fluid. (4, 5) The dark lines in the spectrum of sun. (10) The temperature of steam over pure boiling water under 1 atm pressure. (5, 5)
6. 7. 8. 9. 10. 12. 14.
15. 17. 18. 20.
A hypothetical type of nuclear reaction that would occur at or near room temperature. (4, 6) A group of lines in the ultraviolet region in the spectrum of hydrogen. (5, 6) The study of magnets and magnetism. (11) Subatomic particle composed of three quarks. (6) The lowest possible energy of molecular vibration. (4, 5, 6) The branch of mechanics that deals with matter and forces in equilibrium. (7) A hypothetical tunnel through space time that connects widely distant regions thus providing a shortcut through space time. (8) The melting point of ice under 1 atm pressure. (3, 5) An electrical device that converts thermal energy into electrical energy. (10) The theory that nature contains no vacuums. (7) A magnesium alloy used to enclose uranium fuel element in certain types of nuclear reactors. (6)
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