September 2015
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edit Vol. XXIII
No. 9
A New star is Born in heaven
September 2015
Corporate Office :
rial
T
he name of the new star is Abdul Kalam. Born in the temple town
Plot 99, Sector 44 Institutional area, Gurgaon -122 003 (HR). Tel : 0124-4951200 e-mail : [email protected] website : www.mtg.in
of Rameswaram, he had his higher education in St. Joseph’s College
Regd. Office
Trichy (Tiruchirapalli). He had taken Physics Hons. as his choice. He, like
406, Taj Apartment, Near Safdarjung Hospital, Ring Road, New Delhi - 110029.
the other top scientists of India developed the rockets which could carry
Managing Editor Editor
: :
atomic missiles, fulfilling the dreams of Dr. Bhabha, Vikram Sarabhai and
Mahabir Singh Anil Ahlawat (BE, MBA)
the great scientists like Raja Ramanna and his team. The next step was sending rockets to the moon. Now India is sending
contents Physics Musing (Problem Set-26) AIPMT (Re-Exam)
simultaneously five satellites for other countries to study various planets
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simultaneously. Yes, our respected ex-president was more than a great
10
scientist. After his term as president, he had a new dream- igniting the
Solved Paper 2015 Ace Your Way CBSE XI
minds of the young to teach them to excel themselves. No film hero had
25
Series 1
such a following as the star of science. Our advice to the young is this. Try to study his books when you are grown
JEE Accelerated Learning Series
31
up. We cannot classify him as an engineer, physicist or astronomer. He was
Brain Map
46
just a great scientist without barriers, inspiring students from the age of
Thought Provoking Problems
58
five to eighty and more. We pray for the great man who is no more with
JEE Workouts
61
Ace Your Way CBSE XII
65
us physically. May god inspire us to do great things together.
AA lg oh;Za djokogsAA
Series 4 Core Concept
73
Physics Musing (Solution Set-25)
78
Exam Prep
80
You Ask We Answer
84
Crossword
85
Owned, Printed and Published by Mahabir Singh from 406, Taj Apartment, New Delhi - 29 and printed by Personal Graphics and Advertisers (P) Ltd., Okhla Industrial Area, Phase-II, New Delhi. Readers are adviced to make appropriate thorough enquiries before acting upon any advertisements published in this magazine. Focus/Infocus features are marketing incentives MTG does not vouch or subscribe to the claims and representations made by advertisers. All disputes are subject to Delhi jurisdiction only. Editor : Anil Ahlawat Copyright© MTG Learning Media (P) Ltd. All rights reserved. Reproduction in any form is prohibited.
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Physics for you | SEPTEMBER ‘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.
Set 26 single option correct type 1. A wall consists of alternate blocks with a length d and coefficients of thermal conductivity l1 and l2. The area of cross-section of the blocks is same.The effective thermal conductivity of the wall is ll (a) l = 1 2 (b) l = 2(l1 + l2)
l1 + l2 l1 + l2 l + l2 (c) l = 1 (d) l = 4 2 2. A copper plate is soldered between two steel plates. All the plates have the same cross-section area A and length l. The coefficients of thermal expansion are ac and as and their Young’s moduli are Yc and Ys Steel respectively. What force will arise Copper in the plates if the temperature is increased by T°C?(Assume that Steel the plates suffer the same net expansion) 2 AYcYs (ac − a s )T AYcYs (ac − a s )T (a) (b) 2Ys + Yc 2Ys + Yc 2AYcYs (ac − a s )T AYcYs (ac − a s )T (c) (d) Ys + Yc Ys + Yc 3. A disc of radius R has a mass 9m. A hole of radius R is cut from it as shown in figure. 3 R/ 3 O The moment of inertia of the R remaining part about an axis passing through centre O of the disc and perpendicular to the plane of disc is (a) 8mR2 (b) 4mR2 40 37 (c) mR2 (d) mR2 9 9 4. The surface density of a circular disc of radius a depends on the distance from the center as r(r) = A+ Br.
The moment of inertia about the line perpendicular to the plane of the disc through its centre is
Aa5
(a) 2 π
5 Aa 4
(c) 2 π
4
+
Ba6 6
(b) 2 π
Aa3
+
Ba5 5
(d) π
3
Aa3 3
+
physics for you | september ‘15
Ba 4 4
Ba 4 4
5. The maximum length of an open organ pipe that produces a fundamental note just audible to a person of normal hearing is (Take velocity of sound in air = 340 m s–1) (a) 4.25 m (b) 8.5 m (c) 12.75 m (d) 1 m 6. A string B has twice the length, twice the diameter of another string A. Both strings have same density. Which of the following alternatives express the relation between the frequency of A and B ? (a) uB = 4uA (b) uA = 4uB (c) uA = 2uB (d) 2uB = uA 7. How many octaves does the audible range for normal human hearing cover approximately ? (a) 3 (b) 5 (c) 10 (d) 20 8. The maximum number of overtones emitted by an open organ pipe of length 15 cm that can be heard by a person with normal hearing is
(Take, velocity of sound = 330 m s–1) (a) 16 (b) 17 (c) 18 (d) 19 9. A battery of 10 V is connected to a 20 W resistance through a variable resistance R. The amount of charge which has passed in the circuit in 4 minutes, if the variable resistance R is increased at the rate of 5 W min–1 is (a) 120 C (b) 120 ln 2 C (c) 240 ln 2 C (d) 60 ln 2 C 10. In a series grouping of N cells, current in the external circuit is I. Number of cells to be reversed in polarity such I that current becomes is 3 N N N 2N (a) (b) (c) (d) 3 2 4 3 nn
By Akhil Tewari, Author Foundation of physics for Jee main & Advanced, senior professor physics, rAO IIt ACADemY, mumbai.
8
+
AIPMT
Re-Ex a held m t 25 h J on uly
SOLVED PAPER 2015 1. A photoelectric surface is illuminated successively by monochromatic light of wavelength l and l . 2 If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface of the material is (h = Planck’s constant, c = speed of light) 2hc hc hc hc (a) (b) (c) (d) l 2l l 3l 2. The input signal given to a CE amplifier having p a voltage gain of 150 is Vi = 2 cos 15t + . 3 The corresponding output signal will be 5p (a) 2 cos 15t + 6
4p (b) 300 cos 15t + 3
2p (c) 300 cos 15t + p (d) 75 cos 15t + 3 3 3. A series R-C circuit is connected to an alternating voltage source. Consider two situations : (a) When capacitor is air filled. (b) When capacitor is mica filled. Current through resistor is i and voltage across capacitor is V then (a) ia > ib (b) Va = Vb (c) Va < Vb (d) Va > Vb 4. Point masses m1 and m2 are placed at the opposite ends of a rigid rod of length L, and negligible mass. The rod is to be set rotating about an axis perpendicular to it. The position of point P on this rod through which the axis should pass so that the work required to set the rod rotating with angular velocity w0 is minimum, is given by m2 L m (a) x = 2 L (b) x = m1 + m2 m1 10
Physics for you | September ‘15
(c) x =
m1L m1 + m2
(d) x =
m1 L m2
5. A parallel plate air capacitor has capacity C, distance of separation between plates is d and potential difference V is applied between the plates. Force of attraction between the plates of the parallel plate air capacitor is CV 2 C 2V 2 (a) (b) d 2d 2 C 2V 2 CV 2 (d) 2d 2d 6. An ideal gas is compressed to half its initial volume by means of several processes. Which of the process results in the maximum work done on the gas? (a) Isochoric (b) Isothermal (c) Adiabatic (d) Isobaric (c)
7. A beam of light consisting of red, green and blue colours is incident on a right angled prism. The refractive index of the material of the prism for the above red, green and blue wavelengths are 1.39, 1.44 and 1.47 respectively.
The prism will (a) not separate the three colours at all (b) separate the red colour part from the green and blue colours (c) separate the blue colour part from the red and green colours (d) separate all the three colours from one another 8. Two vessels separately contain two ideal gases A and B at the same temperature, the pressure of A being twice that of B. Under such conditions, the density of A is found to be 1.5 times the density of B. The ratio of molecular weight of A and B is
3 1 2 (c) (d) 4 2 3 9. A remote-sensing satellite of earth revolves in a circular orbit at a height of 0.25 × 106 m above the surface of earth. If earth’s radius is 6.38 × 106 m and g = 9.8 ms–2, then the orbital speed of the satellite is (a) 9.13 km s–1 (b) 6.67 km s–1 –1 (c) 7.76 km s (d) 8.56 km s–1 (a) 2
(b)
10. The energy of the em waves is of the order of 15 keV. To which part of the spectrum does it belong? (a) Ultraviolet rays (b) g-rays (c) X-rays (d) Infra-red rays 11. A proton and an alpha particle both enter a region of uniform magnetic field B, moving at right angles to the field B. If the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton is 1 MeV, the energy acquired by the alpha particle will be (a) 1.5 MeV (b) 1 MeV (c) 4 MeV (d) 0.5 MeV 12. If vectors A = cos wt i + sin wt j and wt wt B = cos i + sin j are functions of time, 2 2 then the value of t at which they are orthogonal to each other is p p p (a) t = (b) t = 0 (c) t = (d) t = w 2w 4w
13. A rectangular coil of length 0.12 m and width 0.1 m having 50 turns of wire is suspended vertically in a uniform magnetic field of strength 0.2 Weber/m2. The coil carries a current of 2 A. If the plane of the coil is inclined at an angle of 30° with the direction of the field, the torque required to keep the coil in stable equilibrium will be (a) 0.24 Nm (b) 0.12 Nm (c) 0.15 Nm (d) 0.20 Nm 14. An automobile moves on a road with a speed of 54 km h–1. The radius of its wheels is 0.45 m and the moment of inertia of the wheel about its axis of rotation is 3 kg m2. If the vehicle is brought to rest in 15 s, the magnitude of average torque transmitted by its brakes to the wheel is (a) 10.86 kg m2 s–2 (b) 2.86 kg m2 s–2 2 –2 (c) 6.66 kg m s (d) 8.58 kg m2 s–2 15. Two metal wires of identical dimensions are connected in series. If s1 and s2 are the conductivities of the metal wires respectively, the effective conductivity of the combination is 12
Physics for you | September ‘15
s1 + s 2 s1s 2 2s1s 2 (c) s1 + s 2 (a)
s1s 2 s1 + s 2 s1 + s 2 (d) 2s1s 2 (b)
16. If potential (in volts) in a region is expressed as V(x, y, z) = 6xy – y + 2yz, the electric field (in N/C) at point (1, 1, 0) is (a) −(2i + 3j + k ) (b) −(6i + 9 j + k ) (c) −(3i + 5j + 3k ) (d) −(6i + 5j + 2k ) 17. Two particles A and B, move with constant velocities v1 and v2 . At the initial moment their position vectors are r1 and r2 respectively. The condition for particles A and B for their collision is (a) r1 × v1 = r2 × v2 (b) r1 − r2 = v1 − v2 r1 − r2 v2 − v1 (c) = (d) r1 ⋅ v1 = r2 ⋅ v2 r1 − r2 v2 − v1
18. 4.0 g of a gas occupies 22.4 litres at NTP. The specific heat capacity of the gas at constant volume is 5.0 JK–1 mol–1. If the speed of sound in this gas at NTP is 952 ms–1, then the heat capacity at constant pressure is (Take gas constant R = 8.3 JK–1 mol–1) (a) 7.0 JK–1 mol–1 (b) 8.5 JK–1 mol–1 –1 –1 (c) 8.0 JK mol (d) 7.5 JK–1 mol–1 19. A force F = a i + 3j + 6k is acting at a point r = 2i − 6 j − 12k . The value of a for which angular momentum about origin is conserved is (a) zero (b) 1 (c) –1 (d) 2
20. At the first minimum adjacent to the central maximum of a single-slit diffraction pattern, the phase difference between the Huygen’s wavelet from the edge of the slit and the wavelet from the midpoint of the slit is p radian (a) p radian (b) 8 p p radian radian (c) (d) 4 2 21. The heart of a man pumps 5 litres of blood through the arteries per minute at a pressure of 150 mm of mercury. If the density of mercury be 13.6 × 103 kg/m3 and g = 10 m/s2 then the power of heart in watt is (a) 3.0 (b) 1.50 (c) 1.70 (d) 2.35 22. A ball is thrown vertically downwards from a height of 20 m with an initial velocity v0. It collides
with the ground, loses 50 percent of its energy in collision and rebounds to the same height. The initial velocity v0 is (Take g = 10 ms–2) (a) 28 ms–1 (b) 10 ms–1 –1 (c) 14 ms (d) 20 ms–1 23. The cylindrical tube of a spray pump has radius R, one end of which has n fine holes, each of radius r. If the speed of the liquid in the tube is V, the speed of the ejection of the liquid through the holes is VR2
V 2R VR2 VR2 (c) (d) nr n3r 2 n2r 2 nr 2 24. A string is stretched between fixed points separated by 75.0 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. The lowest resonant frequency for this string is (a) 10.5 Hz (b) 105 Hz (c) 155 Hz (d) 205 Hz (a)
(b)
25. If dimensions of critical velocity vc of a liquid flowing through a tube are expressed as [hxryrz] where h, r and r are the coefficient of viscosity of liquid, density of liquid and radius of the tube respectively, then the values of x, y and z are given by (a) –1, –1, –1 (b) 1, 1, 1 (c) 1, –1, –1 (d) –1, –1, 1 26. A nucleus of uranium decays at rest into nuclei of thorium and helium. Then (a) The helium nucleus has more momentum than the thorium nucleus. (b) The helium nucleus has less kinetic energy than the thorium nucleus. (c) The helium nucleus has more kinetic energy than the thorium nucleus. (d) The helium nucleus has less momentum than the thorium nucleus. 27. An electron moves on a straight line path XY as shown. The abcd is a coil adjacent to the path of electron. What will be the direction of current, if any, induced in the coil?
(a) The current will reverse its direction as the electron goes past the coil
(b) No current induced (c) abcd (d) adcb 28. Water rises to a height h in capillary tube. If the length of capillary tube above the surface of water is made less than h, then (a) water rises upto a point a little below the top and stays there. (b) water does not rise at all. (c) water rises upto the tip of capillary tube and then starts overflowing like a fountain. (d) water rises upto the top of capillary tube and stays there without overflowing. 29. In an astronomical telescope in normal adjustment a straight black line of length L is drawn on inside part of objective lens. The eye-piece forms a real image of this line. The length of this image is I. The magnification of the telescope is L L+I (a) (b) I L−I L L (c) (d) −1 +1 I I 30. A circuit contains an ammeter, a battery of 30 V and a resistance 40.8 ohm all connected in series. If the ammeter has a coil of resistance 480 ohm and a shunt of 20 ohm, the reading in the ammeter will be (a) 2 A (b) 1 A (c) 0.5 A (d) 0.25 A 31. On a frictionless surface, a block of mass M moving at speed v collides elastically with another block of same mass M which is initially at rest. After collision the first block moves at an angle q to its v initial direction and has a speed . The second 3 block’s speed after the collision is 3 3 v (a) (b) v 2 2 3 (c) 2 2 v (d) v 4 3 32. A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small compared to the mass of the earth. Then, (a) the linear momentum of S remains constant in magnitude. (b) the acceleration of S is always directed towards the centre of the earth. (c) the angular momentum of S about the centre of the earth changes in direction, but its magnitude remains constant. (d) the total mechanical energy of S varies periodically with time. Physics for you | September ‘15
13
33. In the given figure, a diode D is connected to an external resistance R = 100 W and an e.m.f. of 3.5 V. If the barrier potential developed across the diode is 0.5 V, the current in the circuit will be (a) 20 mA (b) 35 mA (c) 30 mA (d) 40 mA 34. A potentiometer wire of length L and a resistance r are connected in series with a battery of e.m.f. E0 and a resistance r1. An unknown e.m.f. E is balanced at a length l of the potentiometer wire. The e.m.f. E will be given by LE0r E l (a) 0 (b) (r + r1 )l L (c)
LE0r lr1
(d)
E0r l . (r + r1 ) L
35. Two stones of masses m and 2m are whirled in r horizontal circles, the heavier one in a radius 2 and the lighter one in radius r. The tangential speed of lighter stone is n times that of the value of heavier stone when they experience same centripetal forces. The value of n is (a) 4 (b) 1 (c) 2 (d) 3 36. Two slits in Youngs experiment have widths in the ratio 1 : 25. The ratio of intensity at the maxima and I minima in the interference pattern, max is I min 49 9 121 4 (a) (b) (c) (d) 121 4 49 9 37. The Young’s modulus of steel is twice that of brass. Two wires of same length and of same area of cross section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weights added to the steel and brass wires must be in the ratio of (a) 4 : 1 (b) 1 : 1 (c) 1 : 2 (d) 2 : 1 38. The coefficient of performance of a refrigerator is 5. If the temperature inside freezer is –20°C, the temperature of the surroundings to which it rejects heat is (a) 11°C (b) 21°C (c) 31°C (d) 41°C 39. Light of wavelength 500 nm is incident on a metal with work function 2.28 eV. The de Broglie 14
Physics for you | September ‘15
wavelength of the emitted electron is (a) ≥ 2.8 × 10–9 m (b) ≤ 2.8 × 10–12 m –10 (c) < 2.8 × 10 m (d) < 2.8 × 10–9 m 40. A source of sound S emitting waves of frequency 100 Hz and an observer O are located at some distance from each other. The source is moving with a speed of 19.4 ms–1 at an angle of 60° with the source observer line as shown in the figure. The observer is at rest. The apparent frequency observed by the observer (velocity of sound in air 330 ms–1), is (a) 106 Hz (b) 97 Hz (c) 100 Hz (d) 103 Hz 41. The value of coefficient of volume expansion of glycerin is 5 × 10–4 K–1. The fractional change in the density of glycerin for a rise of 40°C in its temperature, is (a) 0.025 (b) 0.010 (c) 0.015 (d) 0.020 42. The position vector of a particle R as a function of time is given by R = 4 sin(2 pt )i + 4 cos(2 pt )j Where R is in meters, t is in seconds and i and j
denote unit vectors along x-and y-directions, respectively. Which one of the following statements is wrong for the motion of particle? (a) Magnitude of the velocity of particle is 8 meter/second. (b) Path of the particle is a circle of radius 4 meter. (c) Acceleration vector is along − R. v2 (d) Magnitude of acceleration vector is , where R v is the velocity of particle.
43. A plank with a box on it at one end is gradually raised about the other end. As the angle of inclination with the horizontal reaches 30°, the box starts to slip and slides 4.0 m down the plank in 4.0 s. The coefficients of static and kinetic friction between the box and the plank will be, respectively (a) 0.5 and 0.6 (b) 0.4 and 0.3 (c) 0.6 and 0.6 (d) 0.6 and 0.5 44. In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is 27 5 9 4 (a) (b) (c) (d) 5 27 4 9
45. A particle is executing a simple harmonic motion. Its maximum acceleration is a and maximum velocity is b. Then, its time period of vibration will be (a)
b2 a
(b)
b2 2pb (c) 2 a a
(d)
a b
solutions 1. (c) : Let f0 be the work function of the surface of
the material. Then, According to Einstein’s photoelectric equation, the maximum kinetic energy of the emitted photoelectrons in the first case is hc K max = − f0 1 l and that in the second case is hc 2hc K max = − f0 = − f0 2 l l 2 But Kmax = 3Kmax (given) 2
\
1
2hc hc − f0 = 3 − f0 l l 2hc 3hc − f0 = − 3f0 l l 3hc 2hc 3f0 − f0 = − l l hc hc 2f 0 = or f0 = 2l l
2. (b) : Here,
Input signal, Vi = 2 cos 15t + p 3
and voltage gain, Av = 150 V As Av = o Vi \ Output signal, Vo = AvVi Since CE amplifier gives a phase difference of p(=180°) between input and output signals, p \ Vo = 150 2 cos 15t + + p 3 4p = 300 cos 15t + 3 3. (d) : Current through resistor, i
= Current in the circuit V0 V0 = = R2 + XC2 R2 + (1/wC )2
Voltage across capacitor, V = iXC V0 1 = × R2 + (1/wC )2 wC =
V0
R 2 w 2C 2 + 1 As Ca < Cb \ ia < ib and Va > Vb 4. (b) :
Moment of inertia of the system about the axis of rotation (through point P) is I = m1x2 + m2(L – x)2 By work energy theorem, Work done to set the rod rotating with angular velocity w0 = Increase in rotational kinetic energy 1 1 W = I w20 = [m1x 2 + m2 (L − x )2 ]w20 2 2 dW For W to be minimum, =0 dx 1 i.e. [2m1x + 2m2 (L − x )(−1)]w20 = 0 2 or m1x – m2(L – x) = 0 (... w0 ≠ 0) m2 L or (m1 + m2)x = m2L or x = m1 + m2 5. (d) : Force of attraction between the plates of the
parallel plate air capacitor is
Q2 2e0 A where Q is the charge on the capacitor, e0 is the permittivity of free space and A is the area of each plate. But Q = CV e A and C = 0 or e0 A = Cd d C 2V 2 CV 2 \ F= = 2Cd 2d 6. (c) : The P-V diagram of an ideal gas compressed V from its initial volume V0 to 0 by several processes 2 is shown in the figure. F=
Physics for you | September ‘15
15
Work done on the gas = Area under P–V curve As area under the P–V curve is maximum for adiabatic process, so work done on the gas is maximum for adiabatic process.
density of the gas respectively and R is the universal gas constant. \ The molecular weight of A is r RT MA = A A PA and that of B is r RT MB = B B PB Hence, their corresponding ratio is M A r A TA PB = M B rB TB PA Here,
7. (b) : As beam of light is incident normally on
the face AB of the right angled prism ABC, so no refraction occurs at face AB and it passes straight and strikes the face AC at an angle of incidence i = 45°. For total reflection to take place at face AC, i > ic or sini > sinic where ic is the critical angle. 1 But as here i = 45° and sin ic = m 1 1 1 \ sin 45° > or > m 2 m or m > 2 = 1.414 As mred (= 1.39) < m(= 1.414) while mgreen( = 1.44) and mblue(= 1.47) > m (= 1.414), so only red colour will be transmitted through face AC while green and blue colours will suffer total internal reflection. So the prism will separate red colour from the green and blue colours as shown in the following figure.
8. (d) : According to an ideal gas equation, the
molecular weight of an ideal gas is rRT rRT M= as P = M P where P, T and r are the pressure, temperature and
16
Physics for you | September ‘15
\
rA 3 T = 1.5 = , A = 1 and rB 2 TB
PA =2 PB
MA 3 1 3 = (1) = MB 2 2 4
9. (c) : The orbital speed of the satellite is
vo = R
g ( R + h)
where R is the earth’s radius, g is the acceleration due to gravity on earth’s surface and h is the height above the surface of earth. Here, R = 6.38 × 106m, g = 9.8 m s–2 and h = 0.25 × 106 m \ vo = (6.38 × 106 m)
(9.8 m s −2 ) (6.38 × 106 m + 0.25 × 106 m)
= 7.76 × 103 m s–1 = 7.76 km s–1 (... 1 km = 103 m) hc E where the symbols have their usual meanings. Here, E = 15 keV = 15 × 103 V and hc = 1240 eV nm 1240 eV nm \ l= = 0.083 nm 15 × 103 eV
10. (c) : As l =
As the wavelength range of X-rays is from 1 nm to 10–3 nm, so this wavelength belongs to X-rays. 11. (b) : The kinetic energy acquired by a charged particle in a uniform magnetic field B is 2mK mv q2 B2 R2 K= as R = qB = qB 2m where q and m are the charge and mass of the particle and R is the radius of circular orbit. \ The kinetic energy acquired by proton is
Kp =
q2p B2 R2p 2m p
and that by the alpha particle is q2 B2 Ra2 Ka = a 2ma 2
q m p Ra Thus, K a = a Kp q p ma R p 2
2
2
qa m p Ra or K a = K p q p ma R p mp 1 q = Here, K p = 1 MeV, a = 2, qp ma 4 Ra =1 and Rp 1 \ K a = (1 MeV)(2)2 (1)2 = 1 MeV 4 12. (a) : Two vectors A and B are orthogonal to each other, if their scalar product is zero i.e. A ⋅ B = 0. Here, A = cos wt i + sin wt j wt wt i + sin j and B = cos 2 2 wt wt \ A ⋅ B = (cos wti + sin wt j) ⋅ cos i + sin j 2 2 wt wt + sin wt sin 2 2 ( i ⋅ i = j ⋅ j = 1 and i ⋅ j = j ⋅ i = 0)
= cos wt cos
wt = cos wt − 2
(... cos(A – B) = cosAcosB + sinAsinB) But A ⋅ B = 0 (as A and B are orthogonal to each other) \ cos wt − wt = 0 2 wt p p wt = or wt − cos wt − = cos 2 2 2 2 p wt p or t = = w 2 2 13. (d) : The required torque is
t = NIABsinq where N is the number of turns in the coil, I is the current through the coil, B is the uniform magnetic field, A is the area of the coil and q is the angle
between the direction of the magnetic field and normal to the plane of the coil. Here, N = 50, I = 2 A, A = 0.12 m × 0.1 m = 0.012 m2 B = 0.2 Wb/m2 and q = 90° – 30° = 60° \ t = (50)(2 A)(0.012 m2)(0.2 Wb/m2) sin60° = 0.20 Nm 14. (c) : Here, Speed of the automobile, 5 v = 54 km h −1 = 54 × m s −1 = 15 m s −1 18 Radius of the wheel of the automobile, R = 0.45 m Moment of inertia of the wheel about its axis of rotation, I = 3 kg m2 Time in which the vehicle brought to rest, t = 15 s The initial angular speed of the wheel is −1
1500 100 v 15 m s rad s −1 = rad s −1 = = 45 3 R 0.45 m and its final angular speed is wf = 0 (as the vehicle comes to rest) \ The angular retardation of the wheel is 100 w f − wi 0 − 3 100 rad s −2 a= = =− t 15 s 45 The magnitude of required torque is 100 t = I | a | = (3 kg m2 ) rad s −2 45 20 = kg m2s −2 = 6.66 kg m2s −2 3 15. (c) : As both metal wires are of identical dimensions, so their length and area of cross-section will be same. Let them be l and A respectively. Then The resistance of the first wire is l ... (i) R1 = s1 A wi =
and that of the second wire is l R2 = s2 A
... (ii)
As they are connected in series, so their effective resistance is Rs = R1 + R2 l l = + s1 A s 2 A (using (i) and (ii)) =
1 l 1 + A s1 s 2 Physics for you | September ‘15
... (iii) 17
If seff is the effective conductivity of the combination, then 2l ... (iv) Rs = s eff A Equating eqns. (iii) and (iv), we get 1 2l l 1 = + s eff A A s1 s 2 s + s1 2 = 2 s eff s1s 2 2s1s 2 s eff = s1 + s 2 16. (d) : The electric field E and potential V in a region are related as ∂V ∂V ∂V i+ j+ k E = − ∂y ∂z ∂x Here, V(x, y, z) = 6xy – y + 2yz ∂ ∂ \ E = − (6 xy − y + 2 yz )i + (6 xy − y + 2 yz )j ∂y ∂x ∂ + (6 xy − y + 2 yz )k ∂z = −[(6 y )i + (6 x − 1 + 2z )k + (2 y )k] At point (1, 1, 0), E = −[(6(1))i + (6(1) − 1 + 2(0))j + (2(1))k ] = −(6i + 5j + 2k ) 17. (c) : Let the particles A and B collide at time t. For
their collision, the position vectors of both particles should be same at time t, i.e. r1 + v1t = r2 + v2t r1 − r2 = v2t − v1t = (v2 − v1 ) t ... (i) Also, | r1 − r2 | = | v2 − v1 | t |r − r | or t = 1 2 | v2 − v1 |
gRT M where g is the ratio of two specific heats, R is the universal gas constant and T is the temperature of the gas. v=
Mv 2 RT Here, M = 4.0 g mol–1= 4.0 × 10–3 kg mol–1, v = 952 ms–1, R = 8.3 JK–1 mol–1 and T = 273 K (at NTP) \ g=
\ g=
(4.0 × 10−3 kg mol −1 )(952 ms −1 )2
= 1.6 (8.3 JK −1 mol −1 )(273 K) By definition, Cp g= or Cp = gCv Cv But g = 1.6 and Cv = 5.0 JK–1 mol–1 \ Cp = (1.6)(5.0 JK–1 mol–1) = 8.0 JK–1 mol–1 19. (c) : For the conservation of angular momentum about origin, the torque t acting on the particle will be zero. By definition, t = r × F Here, r = 2i − 6 j − 12 k and F = ai + 3j + 6 k i j k \ t = 2 −6 −12 a 3 6 = i(−36 + 36) − j(12 + 12a) + k (6 + 6a) = − j(12 + 12a) + k (6 + 6a) But t = 0 \ 12 + 12a = 0 or a = – 1 and 6 + 6a = 0 or a = – 1 20. (a) : The situation is shown in the figure.
Substituting this value of t in eqn. (i), we get | r1 − r2 | r1 − r2 = (v2 − v1 ) | v2 − v1 | (v − v ) r −r or 1 2 = 2 1 | r1 − r2 | | v2 − v1 |
18. (c) : Since 4.0 g of a gas occupies 22.4 litres at NTP,
so the molecular mass of the gas is M = 4.0 g mol–1 As the speed of the sound in the gas is
18
Physics for you | September ‘15
In figure A and B represent the edges of the slit AB of width a and C represents the midpoint of the slit. For the first minimum at P,
asinq = l ... (i) where l is the wavelength of light. The path difference between the wavelets from A to C is a 1 Dx = sin q = (a sin q) 2 2
23. (d) : Let the speed of the ejection of the liquid
l (using (i)) 2 The corresponding phase difference Df is 2p 2p l Df = Dx = × =p l l 2 21. (c) : Here, Volume of blood pumped by man’s heart, V = 5 litres = 5 × 10–3 m3 (... 1 litre = 10–3 m3) Time in which this volume of blood pumps, t = 1 min = 60 s Pressure at which the blood pumps, P = 150 mm of Hg = 0.15 m of Hg = (0.15 m)(13.6 × 103 kg/m3)(10 m/s2) (... P = hrg) 3 2 = 20.4 × 10 N/m PV \ Power of the heart = t
(n + 1)v nv v − = 2L 2L 2L which is also the lowest resonant frequency (n = 1). Thus the lowest resonant frequency for the given string = 420 Hz –315 Hz = 105 Hz
=
=
(20.4 × 103 N/m2 )(5 × 10−3 m3 ) 60 s
= 1.70 W
22. (d) : The situation is shown in the figure.
Let v be the velocity of the ball with which it collides with ground. Then according to the law of conservation of energy, Gain in kinetic energy = loss in potential energy 1 1 i.e. mv 2 − mv02 = mgh 2 2 (where m is the mass of the ball)
or
v 2 − v02 = 2 gh
... (i)
Now, when the ball collides with the ground, 50% of its energy is lost and it rebounds to the same height h. 50 1 2 \ mv = mgh 100 2 1 2 v = gh or v2 = 4gh 4 Substituting this value of v2 in eqn. (i), we get 4 gh − v02 = 2 gh
or
v02 = 4 gh − 2 gh = 2 gh
or v0 = 2 gh Here, g = 10 ms–2 and h = 20 m \ v0 = 2(10 ms −2 )(20 m) = 20 ms −1 through the holes be v. Then according to the equation of continuity, pR2V VR2 pR2V = npr2v or v = = npr 2 nr 2 24. (b) : For a string fixed at both ends, the resonant frequencies are nv un = where n = 1, 2, 3, ..... 2L The difference between two consecutive resonant frequencies is Dun = un+1 – un =
25. (c) : [vc] = [hxry rz] (given)
... (i) Writing the dimensions of various quantities in eqn. (i), we get [M0LT–1] = [ML–1T–1]x[ML–3T0]y[M0LT0]z = [Mx + y L–x –3y + z T–x] Applying the principle of homogeneity of dimensions, we get x + y = 0; –x – 3y + z = 1; – x = –1 On solving, we get x = 1, y = –1, z = –1 26. (c) : If pTh and pHe are the momenta of thorium and helium nuclei respectively, then according to law of conservation of linear momentum 0 = pTh + pHe or pTh = − pHe –ve sign shows that both are moving in opposite directions. But in magnitude pTh = pHe If mTh and mHe are the masses of thorium and helium nuclei respectively, then p2 Kinetic energy of thorium nucleus is K Th = Th 2mTh and that of helium nucleus is K He =
2 pHe 2mHe
Physics for you | September ‘15
19
2
K Th pTh mHe \ = K He pHe mTh But pTh = pHe and mHe < mTh \ KTh < KHe or KHe > KTh Thus the helium nucleus has more kinetic energy than the thorium nucleus. 27. (a) :
When the electron moves from X to Y, the flux linked with the coil abcd (which is into the page) will first increase and then decrease as the electron passes by. So the induced current in the coil will be first anticlockwise and will reverse its direction (i.e. will become clockwise) as the electron goes past the coil. 28. (d) : Water will not overflow but will change its radius of curvature. 29. (b) : The situation is shown in the figure.
Let fo and fe be the focal lengths of the objective and eyepiece respectively. For normal adjustment distance of the objective from the eyepiece (tube length) = fo+ fe. Treating the line on the objective as the object and eyepiece as the lens. \ u = –(fo + fe) and f = fe 1 1 1 As − = v u f 1 1 1 \ − = v −( fo + fe ) fe f +f −f fo 1 1 1 = − = o e e = v fe fo + fe fe ( fo + fe ) fe ( fo + fe ) f (f + f ) or v = e o e fo fe ( fo + fe ) fo f I v = e Thus, = u = fo ( fo + fe ) L 20
Physics for you | September ‘15
or
fo L = fe I
... (i)
\ The magnification of the telescope in normal adjustment is f L (using (i)) m= o = fe I 30. (c) : The circuit is shown in the figure.
Resistance of the ammeter is (480 W)(20 W) RA = = 19.2 W (480 W + 20 W) (As 480 W and 20 W are in parallel) As ammeter is in series with 40.8 W, \ Total resistance of the circuit is R = 40.8 W + RA = 40.8 W + 19.2 W = 60 W By Ohm’s law, Current in the circuit is V 30 V 1 I= = = A = 0.5 A R 60 W 2 Thus the reading in the ammeter will be 0.5 A. 31. (c) : The situation is shown in the figure.
Let v′ be speed of second block after the collision. As the collision is elastic, so kinetic energy is conserved. According to conservation of kinetic energy, 1 1 v 2 1 Mv 2 + 0 = M + Mv ′2 2 2 3 2 v2 v 2 = + v ′2 9 or
v 2 9v 2 − v 2 8 2 = = v 9 9 9 8 2 8 2 2 v′ = v = v= v 9 3 3 v ′2 = v 2 −
Physics for you | September ‘15
21
32. (b) : The gravitational force on the satellite S acts
towards the centre of the earth, so the acceleration of the satellite S is always directed towards the centre of the earth. 33. (c) :
The potential difference across the resistance R is V = 3.5 V – 0.5 V = 3 V By Ohm’s law, The current in the circuit is 3V V I= = R 100 W = 3 × 10–2 A = 30 × 10–3 A = 30 mA 34. (d) :
2 2 n2 mv = 4 mv r r n2 = 4 or n = 2 36. (c) : As, intensity I ∝ width of slit W Also, intensity I ∝ square of amplitude A I W A2 \ 1= 1= 1 I2 W2 A22 But \
\
W1 1 = W2 25
A12
A22
=
1 25
2
I max ( A1 + A2 )2 = = 2 I min ( A1 − A2 )2 A 1 − 1 A 2
37. (d) :
1 + 1 5 1 −1 5
2 2
=
6 5
2
4 − 5
2
=
36 9 = 16 4
(r / 2) But (Fc)lighter = (Fc)heavier (given)
Let L and A be length and area of cross section of each wire. In order to have the lower ends of the wires to be at the same level (i.e. same elongation is produced in both wires), let weights Ws and Wb are added to steel and brass wires respectively. Then By definition of Young’s modulus, the elongation produced in the steel wire is WL W/ A DLs = s as Y = Ys A DL / L and that in the brass wire is WL DLb = b Yb A But DLs = DLb (given) Ws L Wb L Ws Ys \ = or = Ys A Yb A Wb Yb Y (given) As s = 2 Yb
\
\
35. (c) : Let v be tangential speed of heavier stone. Then,
Centripetal force experienced by lighter stone is m(nv )2 r and that of heavier stone is (Fc )lighter =
(Fc )heavier =
22
A1 1 1 = = A2 25 5
or
A1 A + 1 2
=
The current through the potentiometer wire is E0 I= (r + r1 ) and the potential difference across the wire is E r V = Ir = 0 (r + r1 ) The potential gradient along the potentiometer wire is E0r V k= = L (r + r1 )L As the unknown e.m.f. E is balanced against length l of the potentiometer wire, E0r l \ E = kl = (r + r1 ) L
(given)
2 mv 2
m(nv )2 2mv 2 = r (r / 2)
Physics for you | September ‘15
Ws 2 = Wb 1
38. (c) : The coefficient of performance of a refrigerator
is
a=
T2 T1 − T2
where T1 and T2 are the temperatures of hot and cold reservoirs (in kelvin) respectively. Here, a = 5, T2 = –20°C = –20 + 273 K = 253 K T1 = ? 253 K \ 5= T1 − 253 K 5T1 – 5(253 K) = 253 K 5T1 = 253 K + 5(253 K) = 6(253 K) 6 T1 = (253 K) = 303.6 K = 303.6 – 273 5 = 30.6°C ≈ 31°C 39. (a) : According to Einstein’s photoelectric equation, the maximum kinetic energy of the emitted electron is hc K max = − f0 l where l is the wavelength of incident light and f0 is the work function. Here, l = 500 nm, hc = 1240 eV nm and f0 = 2.28 eV 1240 eV nm \ K max = − 2.28 eV 500 nm = 2.48 eV – 2.28 eV = 0.2 eV The de Broglie wavelength of the emitted electron is h l min = 2 mK max
As the velocity of source along the source observer line is vscos60° and the observer is at rest, so the apparent frequency observed by the observer is v u = u0 v − vs cos 60° 330 ms −1 = (100 Hz) 330 ms −1 − (19.4 ms −1 ) 1 2 330 ms −1 = (100 Hz) 330 ms −1 − 9.7 ms −1
330 ms −1 = (100 Hz) = 103 Hz 320.3 ms −1 41. (d) : Let r0 and rT be densities of glycerin at 0°C and T°C respectively. Then, rT = r0(1 – gDT) where g is the coefficient of volume expansion of glycerine and DT is rise in temperature. r rT = 1 − gDT or gDT = 1 − T r0 r0 r −r Thus, 0 T = gDT r0
Here, g = 5 × 10–4 K–1 and DT = 40°C = 40 K \ The fractional change in the density of glycerin r −r = 0 T = gDT = (5 × 10−4 K −1 )(40 K) = 0.020 r0 42. (a) : Here, R = 4 sin(2 pt )i + 4 cos(2 pt )j
where h is the Planck’s constant and m is the mass of the electron. As h = 6.6 × 10–34 J s, m = 9 × 10–31 kg and Kmax = 0.2 eV = 0.2 × 1.6 × 10–19 J
The velocity of the particle is dR d v= = [4 sin(2 pt )i + 4 cos(2 pt )j] dt dt = 8 p cos(2 pt )i − 8 p sin(2 pt )j
\
Its magnitude is | v | = (8 p cos(2 pt ))2 + (−8 p sin(2 pt ))2
l min =
6.6 × 10−34 J s
2(9 × 10−31 kg )(0.2 × 1.6 × 10−19 J)
6. 6 × 10−9 m = 2.8 × 10−9 m 2. 4 So, l ≥ 2.8 × 10–9 m 40. (d) : Here, Frequency of source, u0 = 100 Hz Velocity of source, vs = 19.4 ms–1 Velocity of sound in air, v = 330 ms–1 =
= 64 p2 cos2 (2 pt ) + 64 p2 sin2 (2 pt ) = 64 p2[cos2 (2 pt ) + sin2 (2 pt )] = 64 p2
(as sin2q + cos2q = 1)
= 8p m/s 43. (d) : Let ms and mk be the coefficients of static and kinetic friction between the box and the plank respectively. When the angle of inclination q reaches 30°, the block just slides, Physics for you | September ‘15
23
\
m s = tan q = tan 30° =
1 3
= 0. 6
If a is the acceleration produced in the block, then ma = mgsinq – f k (where f k is force of kinetic friction) = mgsinq – mkN (as f k = mkN) = mgsinq – mkmgcosq (as N = mgcosq) a = g(sinq – mkcosq) As g = 10 ms–2 and q = 30° \ a = (10 ms–2)(sin30° – mkcos30°) ...(i) If s is the distance travelled by the block in time t, then 1 s = at 2 (as u = 0) 2 2s or a = t2 But s = 4.0 m and t = 4.0 s (given) 2(4.0 m) 1 \ a= = ms −2 2 2 ( 4. 0 s ) Substituting this value of a in eqn. (i), we get 1 ms −2 = (10 ms −2 ) 1 − mk 3 2 2 2 1 = 1 − 3 mk 10 1 9 or 3 m k = 1 − = = 0. 9 10 10 0.9 mk = = 0. 5 3 44. (b) : The wavelength of a spectral line in the Lyman series is 1 1 1 = R 2 − 2 , n = 2, 3, 4, ....... lL 1 n and that in the Balmer series is 1 1 1 = R 2 − 2 , n = 3, 4, 5, ....... lB 2 n For the longest wavelength in the Lyman series, n=2 1 1 1 1 1 4 − 1 3R = \ = R 2 − 2 = R − = R lL 1 4 4 4 1 2 4 or l L = 3R
24
Physics for you | September ‘15
For the longest wavelength in the Balmer series, n=3 1 1 1 1 1 \ = R 2 − 2 = R − 4 9 2 lB 3 9 − 4 5R = R = 36 36 36 or l B = 5R 4 l L 3R 4 5R 5 Thus , = = × = l B 36 3R 36 27 5R 45. (b) : If A and w be amplitude and angular frequency of vibration, then a = w2A ...(i) and b = wA ...(ii) Dividing eqn. (i) by eqn. (ii), we get
\
a w2 A = =w b wA Time period of vibration is 2p 2p 2 pb T= = = w (a / b) a
nn
Haryana teen tops AiPMt Vipul Garg, a 17-year-old from Haryana’s Jind district, topped the All India Pre-Medical Test (AIPMT) entrance examination 2015, the results for which were announced on 17th August 2015. Vipul, who is the first in his family to go to medical school and has mostly relied on scholarships so far to get ahead, scored 695 marks out of 720. “My family worked very hard and faced lots of hardships to meet my expenses. I scored cent per cent in Class X and was given a fee waiver by the school for the remaining years. A private coaching institute agreed to waive off the fee for me while I was preparing for my medical entrance.” He was disappointed when the earlier AIPMT was cancelled. “I had done well and was sad. But then I realised it was a good opportunity to work on my weak points and things I knew I had difficulty doing in the first test. I had no idea I would top the entrance,” he said, adding he plans to become a cardiologist. Occupying the second slot is 17-year old Khushi Tiwari from Rajasthan, who was sure she would ace the examination. Khushi, who always wanted to be a doctor like her parents, scored 688 out of 720. “Although, I have not decided on my specialisation, I know I studied 14 hours a day to be able to get into Maulana Azad Medical College,” she said. The Central Board of Secondary Education had re-conducted the test on July 25 on direction from the Supreme Court after allegations of irregularities surfaced in the first test held on May 3. Courtesy : The Hindu
Series 1 Chapterwise Unit test : Units and Measurement | Kinematics GENERAL INSTRUCTIONS (i) (ii) (iii) (iv) (v) (vi) (vii)
All questions are compulsory. Q. no. 1 to 5 are very short answer questions and carry 1 mark each. Q. no. 6 to 10 are short answer questions and carry 2 marks each. Q. no. 11 to 17 are also short answer questions and carry 3 marks each. Q. no. 18 is a value based question and carries 4 marks. Q. no. 19 and 20 are long answer questions and carry 5 marks each. Use log tables if necessary, use of calculators is not allowed.
1. Can a quantity have units but still be dimensionless? 2. A stone tied at the end of string is whirled in a circle.
If the string breaks, the stone flies away tangentially. Why?
3. Express (3.0 × 10–4 – 1.7 × 10–6) with proper
significant figures.
4. Is the acceleration of a car greater when the
accelerator is pushed to the floor or when brake pedal is pushed hard ?
5. Two straight lines drawn on the same x-t curve
make angles 30° and 60° with time axis. Which line represents greater velocity? What is the ratio of the two velocities?
6. If x = a + bt + ct2 where x is in metres and t in
seconds, find the units of b.
7. A ball is thrown vertically upwards. Draw its height-
time and velocity-time graph.
8. At what angle the two forces A + B and A – B act so
that their resultant is
3 A2 + B 2 ?
9. The displacement of a particle is proportional to the
cube of time elapsed. How does the acceleration of the body depend on time elapsed?
OR Two balls of different masses (one lighter and other heavier) are thrown vertical upwards with the same speed. Which one will pass through the point of projection in their downward direction with the greater speed? 10. The lengths of two cylinders are measured to be
l1 = (5.62 ± 0.01) cm and l2 = (4.34 ± 0.02) cm. Calculate difference in lengths with error limits.
11. The vernier scale of a travelling microscope
has 50 divisions which coincide with 49 main scale divisions. If each main scale division is 0.5 mm, calculate the minimum inaccuracy in the measurement of distance.
12. A man runs across the roof-top of a tall building and
jumps horizontally with the hope of landing on the roof of the next building which is of a lower height than the first. If his speed is 9 m s–1, the (horizontal) distance between the two buildings is 10 m and the height difference is 9 m, will he be able to land on the next building? (Take g = 10 m s–2)
13. A boy throws a ball in air at 60° to the horizontal
along a road with a speed of 10 m s–1 (36 km h–1). Another boy sitting in a passing by car observes the ball. Sketch the motion of the ball as observed by the boy in the car, if car has a speed of 18 km h–1. Give explanation to support your diagram. Physics for you | SEPTEMBER ‘15
25
OR A particle falling vertically from a height hits a plane surface inclined to horizontal at an angle q with speed vo and rebounds elastically (see figure). Find the distance along the plane where it will hit second time.
14. A car moving along a straight highway with speed of
126 km h–1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?
15. There are two angles of projection for which the
horizontal range is the same. Prove that the sum of the maximum heights for these two angles does not depend upon the angle of projection.
16. An object is thrown vertically upward with some
speed. It crosses 2 points p, q which are separated by h metre. If tp is the time between p and highest point and coming back and tq is the time between q and highest point and coming back, relate acceleration due to gravity, tp , tq and h. OR Two ends of a train moving with a constant acceleration passes a certain point with velocities u and v. Show that the velocity with which the middle point of the train passes the same point is
20. An artificial satellite is revolving around a planet
of mass M and radius R, in a circular orbit of radius r. From Kepler’s third law about the period of a satellite around a common central body, square of the period of revolution T is proportional to the cube of the radius of the orbit r. Show using
k r3 , where k is R g a dimensionless constant and g is acceleration due to gravity. OR Figure gives a speed-time graph of a particle in one dimensional motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest? Choosing the positive direction as the constant direction of motion, give the signs of v and a in the three intervals. What are the accelerations at the points A, B, C and D? dimensional analysis, that T =
(u2 + v 2 ) . 2 17. The speed of sound, v through a medium may be
assumed to depend upon : (i) the density of the medium, d and (ii) its modulus of elasticity, E. Modulus of elasticity is a ratio of stress to strain and stress is the force per unit area. Deduce by the method of dimensions, the formula for the speed of sound.
18. Read the given passage and answer the following
questions.
Shyam goes to college with his sister Shreya in their own car. The college is about 10 km from their home. They drive on alternate days. Shreya is a very careful driver, but Shyam is a rasher. He takes 3 minutes lesser than Shreya in reaching the college. Shreya advises Shyam to drive safely, but he hardly listens. (i) What values are displayed by Shreya ? Do you agree with her? (ii) What is the difference between average speeds of Shyam and Shreya if latter takes 15 minutes to drive to the college ? 19. A projectile is fired at a certain angle with the
horizontal. Derive the equation of trajectory of the projectile. Also write expression for : (i) the maximum height attained (ii) the time of its flight and (iii) the horizontal range.
26
Physics for you | SEPTEMBER ‘15
D
Speed
B C
A
1
2
3
Time
solutions 1.
Yes, an angle is measured in radian but it has no dimensions.
2.
When a stone is going around a circular path, the instantaneous velocity of stone is acting tangentially to the circle. When the string breaks, the centripetal force stops to act. Due to inertia, the stone continue to move along the tangent to circular path. That is why, the stone flies off tangentially to the circular path.
3.
3.0 × 10–4 – 1.7 × 10–6 = (3.0 – 0.017) × 10–4 = (3.0 – 0.0) × 10–4 = 3.0 × 10–4
4.
Acceleration of a car is greater when brake pedal is pushed hard, because car suddenly comes to rest, i.e., the rate of change of velocity of car is large.
5.
The line which has greater slope has greater velocity. Thus, the line which makes an angle of 60° with time axis has greater velocity.
tan 30° 1 / 3 1 = = tan 60° 3 3 6. As L.H.S. represents distance, every term on R.H.S. must represent distance. [x ] [L] \ [bt ] = [ x ] or [b ] = = = [LT−1 ] [t ] [ T ] Therefore, b represents velocity and hence its unit is m s–1. Ratio of two velocities =
7.
8.
The h-t graph for the motion is a parabola and is shown in figure (i). The v-t graph for the motion is a straight line as shown in figure (ii) because acceleration is constant during the entire motion of the body.
Here, P = A + B ; Q = A – B and R = 3A2 + B2 R2 = (A + B)2 + (A – B)2 + 2(A + B) (A – B) cos q or 3A2 + B2 = 2(A2 + B2) + 2(A2 – B2) cos q 1 or cos q = 2 \ q = 60°
9.
Let x be the displacement at time t of an object in motion. Given, x = k t3, where k is a constant of proportionality. dx Velocity of object, v = = 3 kt 2 and acceleration dt dv of object, a = = 3k × 2t = 6 kt dt i.e., a ∝ t. It means acceleration ∝ time. OR
Let u be the initial velocity of projection of body and v be the velocity of the same body while passing downwards through point of projection. The displacement of body s = 0. Using the relation v2 = u2 + 2as, and u = u, v = ? ; a = – g, s = 0, we have v2 = u2 + 2 (– g) × 0 = u2 or v = u It means that the final speed is independent of mass of the body. Hence, both the bodies will acquire the same speed while passing through point of projection. 10. Here, l1 = (5.62 ± 0.01) cm
l2 = (4.34 ± 0.02) cm l′ = l1 – l2 = 5.62 – 4.34 = 1.28 cm. Dl′ = ± (Dl1 + Dl2) = ± (0.01 + 0.02) = ± 0.03 ± 0. 03 Percentage error = × 100 = ± 2. 34 % 1. 28 Hence, difference in lengths = (1.28 ± 0.03) cm = 1.28 cm ± 2.34 %
11. Given, 50 VSD = 49 MSD ⇒
1 VSD =
49 MSD 50
1 MSD = 0.5 mm In vernier callipers, Minimum inaccuracy in the measurement of distance by vernier callipers = vernier constant = 1 MSD – 1 VSD 49 1 = 1 MSD − MSD = MSD 50 50 1 = × 0.5 mm = 0.01 mm 50 12. Suppose man is at building A and wants to land on building B. Horizontal speed of man, vx = 9 m s–1
Vertical speed of man, vy = 0 Distance between buildings, x = 10 m Difference between height of the buildings, h = 9 m Suppose t is the time taken by the man to fall vertically downward by a height h, Physics for you | SEPTEMBER ‘15
27
h = v yt + ⇒
1 2 gt 2
9=0×t +
1 × 10 × t 2 2
9 = 1.34 s 5 If distance covered by the man along x-axis during this time t is x′, then x′ = vxt = 9 × 1.34 = 12.06 m Here, x′ > x So, man will land successfully from building A to building B. ⇒
5t 2 = 9 ⇒ t =
–1
= 36 km h–1 The ball is thrown at angle of 60° with the horizontal. Horizontal speed of the ball, ux = ucos60° 1 = 36 × = 18 km h −1 2 Vertical speed of the ball, uy = usin60°
13. Initial speed of the ball, u = 10 m s
3 = 18 3 km h −1 2 Speed of the car, v = 18 km h–1
an angle q with the horizontal direction, then Horizontal range, R =
uy u 60°
Case (i) : If q = a, let R = R1 and H = H1, then u2 sin 2a …(i) R1 = g u2 sin2 a 2g Case (ii) : If q = (90° – a), let R = R2 and H = H2, then and H1 =
R2 =
Since, ux = v, so boy sitting in the car will observe only the vertical motion of the ball. Sketch of motion is shown in the figure. 14. Here,
u = 126 km h −1 =
126 × 1000 60 × 60
m s−1 = 35 m s−1 ;
v = 0, s = 200 m, a = ? and t = ?
We know, v2 = u2 + 2as \ 0 = (35)2 + 2 × a × 200 2
− ( 35 ) −49 = = − 3.06 m s−2 a= 2 × 200 16 As, v = u + at − 49 \ 0 = 35 + t 16 or
28
Physics for you | SEPTEMBER ‘15
…(ii)
u2 sin 2 (90° − a) g
u2 u2 sin(180° − 2a) = sin 2a g g
u2 u2 sin2 (90° − a) = cos2 a 2g 2g From (i) and (iii), R1 = R2 From (ii) and (iv); H2 =
H1 + H 2 =
ux= 5 m s–1
u2 sin 2q g
2 2 and maximum height, H = u sin q 2g
=
uy
=
15. If a projectile is projected with velocity u, making
= 36 ×
uy
35 × 16
80 = 11.43 s 49 7 Negative sign shows that acceleration is negative, which is called retardation i.e. car is uniformly retarded at a = 3.06 m s–2. or t =
…(iii) …(iv)
u2 u2 (sin2 a + cos2 a) = 2g 2g
16. Let u′ be the velocity of the object while crossing
point p and v′ be its velocity while crossing point q as shown in figure. A is the highest point of vertical motion of object. As per question, the time taken by tp and the time the object in going from p to A = 2 tq taken by the object in going from q to A = . 2 Taking vertical upward motion of object from p to A, we have tp u = u′, v = 0, a = − g , t = 2 As, v = u + at tp gt \ 0 = u′ + (− g ) or u′ = p …(i) 2 2
Taking vertical upward motion of object from q to A, we have, t u = v′, v = 0, a = –g, t = q 2 As, v = u + at tq tq \ 0 = v ′ + (− g ) or v ′ = g 2 2 Taking vertical upward motion of object from p to q, we have, u = u′, v = v′, a = – g, s = h As, v2 = u2 + 2as \ v′2 = u′2 + 2(– g)h g 2t 2p g 2tq2 2 2 or 2 gh = u′ − v ′ = − 4 4 8h or g = (t 2p − tq2 ) OR Let x be the total length of the train, V be the velocity of the middle point of the train while passing a certain point and a be the uniform acceleration of the train. Taking the motion of the train when middle point is passing from the given point, we have x u = u, v = V, s = ; a = a 2 Using, v2 = u2 + 2 as, we have 2ax …(i) V 2 = u2 + = u2 + ax 2 Taking the motion of train when the last end of train is passing from the given point, then u = u, v = v, a = a, s = x Now, we have, v2 = u2 + 2ax
v 2 − u2 or ax = 2 Putting this value in (i), we get V 2 = u2 +
v 2 − u2 u2 + v 2 = 2 2
or V =
(u2 + v 2 ) 2
17. We are given that,
E= =
Stress Strain
Force/ Unit area
Change in configuration/ Original configuration
or E =
MLT−2 / L2 = [ML−1 T−2 ] 1
Let v ∝ da Eb or v = k daEb …(i) where a, b are constants. Writing the dimensional formula of the various quantities on both the sides, [M0L1T–1] = [ML–3]a [ML–1T–2]b, we get [M0L1T–1] = Ma + b L–3a – b T–2b] Applying the principle of homogeneity of dimensions, we get a + b = 0, –3a – b = 1 and –2b = –1 1 Clearly, b = 2 1 Also, a = − b = − 2 From eqn (i), E v = kd −1/ 2 E1/ 2 = k d 18. (i) Shreya displays safety concerns for her brother. Rash driving can lead to any unfortunate incident. We agree with Shreya that driving must be careful and safe. (ii) Average speed of Shreya, 10 km Distance v1 = = = 40 km h −1 Time taken 15 / 60 hr \
Time taken by Shyam to reach the school = 15 – 3 = 12 minute. Average speed of Shyam,
\
10 km Distance = = 50 km h −1 12 / 60 hr Time v2 – v1 = 50 – 40 = 10 km h–1 v2 =
19. Suppose a projectile is fired with velocity u at an
angle q with the horizontal. Let it reach the point (x, y) after time t. Then Components of initial velocity, ux = u cos q, uy = u sin q Components of acceleration at any instant, ax = 0, ay = – g Position after time t, x x = (u cos q) t ⇒ t = u cos q 1 y = (u sin q) t − gt 2 …(i) 2 Putting value of t in (i) 2 x 1 x y = usinq – g u cos q 2 u cos q y = x tan q −
2
g
2u cos2 q
x2
Physics for you | SEPTEMBER ‘15
29
Maximum height, H = Time of flight, T =
2u y g
Horizontal range, R =
u2y 2g =
=
u2 sin2 q 2g
2u sin q
2ux u y
g =
u2 sin 2q
g g Maximum horizontal range is attained at q = 45° and its value is u2 g Velocity after time t, vx = u cos q, vy = u sin q – gt Rmax =
OR
Since the collision of particle with inclined plane is elastic, so its rebound speed is same but direction of velocity is changed. After rebounding, motion of the particle is projectile on inclined plane. Let T = time taken by the particle to hit the plane second time.
2
2v 1 2v 4v 2 = v0 sin q 0 + ( g sin q) 0 = 0 sin q g g 2 g 20. According to Kepler’s third law,
T2 ∝ r3 ⇒ T ∝ r3/2 Also, T depends on g and R. Let, T ∝ r3/2 gx Ry where x and y are exponents of g and R respectively. ⇒ T = kr3/2 gx Ry where k is dimensionless constant of proportionality. Writing dimensions of the physical quantities on both sides [M0L0T1] = [L]3/2 [LT–2]x [L]y ⇒ [M0 L0 T1] = [M0L(3/2 + x + y)T–2x] Using the principle of homogeneity of dimensions 3 x+y+ =0 … (i) 2 –2x = 1 …(ii) 1 From eqns. (i) and (ii), x = − , y = −1 2 k r3 T = kr 3/ 2 g −1/ 2 R −1 ⇒ T = R g which is required quantity.
\
(i)
Consider motion along perpendicular to inclined plane, y = 0, uy = v0 cosq, ay = –g cosq, t = T 1 y = u yt + a yt 2 2 1 ⇒ 0 = (v 0 cos q)T + (− g cos q)T 2 2 ⇒ T=
2v0 g
Consider motion along inclined plane, 2v x = L, ux = v0sinq, ax = gsinq, t = T = 0 g 1 \ x = ux t + a x t 2 2 30
Physics for you | SEPTEMBER ‘15
(ii) (iii) (iv)
(v)
OR The magnitude of the average acceleration is given by Change in speed = Time interval i.e. average acceleration in a small interval of time is equal to the slope of (v-t) graph in that time interval. As the slope of (v-t) graph is maximum in the interval 2 as compared to intervals 1 and 3, hence the magnitude of average acceleration is greatest in interval 2. The average speed is greatest in the interval 3 as peak D is at maximum on speed axis. v > 0 i.e. positive in all the three intervals. The slope is positive in intervals 1 and 3, so a i.e. acceleration is positive in these intervals while the slope is negative in interval 2, so acceleration is negative in it. So, a > 0 i.e. positive in intervals 1 and 3 and a < 0 i.e. negative in interval 2. As slope is zero at points A, B, C and D, so the acceleration is zero at all the four points. nn
Class XI
ACCELERATED LEARNING SERIES
Unit
3
Rotational Motion | Gravitation | Properties of Solids and Liquids ROtatIOnal MOtIOn
Centre of Mass of n-particles System
Centre of Mass
The centre of mass of a system of particles is the point that moves as though • all the mass of the system is concentrated there and • all external forces are applied there. Centre of Mass of a two Particle System
In figure (a), two particles of masses m1 and m2 are separated by distance d and the origin coincide with the particle of mass m1. The position of the centre of mass (com) of this two particles is given by m2 y xcom = d m1 + m2
y
O
xcom m1
com
m2
x
d (a)
For a system of n-particles of masses m1, m2, m3, ...., mn having the position vectors r1, r2 , ....., rn respectively with respect to coordinate system, the position of the centre of mass is given by a position vector n mi ri ∑ m r + m2r2 + .....mnrn i =1 Rcom = 1 1 = m1 + m2 + ..... + mn M where, M is total mass of the system. The coordinates of centre of mass is given by n
n
∑ mi xi
xcom =
i =1
M
n
∑ mi yi
i =1 , ycom = M
∑ mi zi
xcom
In figure (b), in which m1 m2 x the coordinate system O com x1 d has been shifted leftward. x The position of the centre 2 (b) of mass is m x + m2 x2 xcom = 1 1 m1 + m2 Similarly for y and z-axis m y + m2 y2 m z + m2 z2 ycom = 1 1 and z com = 1 1 m1 + m2 m1 + m2
and z com = i =1 M Centre of Mass of Rigid Body (or continuous distribution of mass Solid bodies contain so many particles (atoms) that we can treat them as a continuous distribution of matter. The ‘particles’ then become differential mass elements dm, the sum become integrals, and the coordinates of the centre of mass are defined as 1 1 xcom = ∫ xdm , ycom = ∫ ydm M M 1 and z com = ∫ zdm M where M is the mass of the body. PHYSICS FOR YOU | September ‘15
31
KEY POINT • The position of the centre of mass of a system is independent of the choice of coordinate system. • The position of the centre of mass depends on the shape of the body and the distribution of its mass. Hence it may lie within or outside the material of the body.
SELF CHECK
1. Distance of the centre of mass of a solid uniform cone from its vertex is z0. If the radius of its base is R and its height is h then z0 is equal to 5h 3h 2 h2 3h (a) (b) (c) (d) 4R 8 8R 4 (JEE Main 2015) Velocity of the Centre of Mass of a System of n-particles
Velocity of centre of mass is given by m v + m2v2 + ... + mnvn vcom = 1 1 m1 + m2 + ... + mn n n ∑ mi vi ∑ mi vi = i =1
n
∑ mi
= i =1 M
i =1
acceleration of the Centre of Mass of a System of n-particles
Acceleration of centre of mass is given by m a + m2a2 + ... + mnan acom = 1 1 m1 + m2 + ... + mn n n m a ∑ i i ∑ mi ai = i =1
n
∑ mi
= i =1 M
i =1
If total external force acting on the system is zero, then the total linear momentum of the system is conserved. Also, when the total external force acting on the system is zero, the velocity of centre of mass remains constant. torque
Torque is the turning or twisting action on a body, about the axis of rotation due to a force. If a force F is acting at any point in the body whose position vector relative to any arbitrary point on the axis of rotation is r then torque of the force on the body about the axis of rotation is given by t=r ×F 32
PHYSICS FOR YOU | September ‘15
Also, | t | = rF sin q where q is the angle between r and F . Torque is a vector quantity. Its SI unit is N m. Torque has the same dimensions as that of work i.e. [ML2T–2]. But both are different. Work is a scalar quantity whereas torque is a vector quantity. By sign convention, anticlockwise torque is taken as positive and clockwise torque is taken as negative. Work done by torque
Work done by torque t, W = t ⋅ q , where q = angular displacement. For variable torque, dW = t ⋅ dq angular Momentum Angular momentum of a particle about a given point is defined as L=r × p where, r = position vector of the particle with respect to the given point, p = linear momentum of the particle. Direction of L can be determined by using the rule for vector product. Angular momentum is a vector quantity. Its S I unit is kg m2 s– 1. Its dimensional formula is [ML2T–1].
SELF CHECK
2. A bob of mass m attached to an inextensible string of length l is suspended from a vertical support. The bob rotates in a horizontal circle with an angular speed w rad s–1 about the vertical. About the point of suspension (a) angular momentum changes both in direction and magnitude. (b) angular momentum is conserved. (c) angular momentum changes in magnitude but not in direction. (d) angular momentum changes in direction but not in magnitude. (JEE Main 2014) Conservation of angular Momentum
When there is no net external torque acting on a particle, dL then we can put = 0 ⇒ L = constant dt Therefore, the angular momentum of the particle remains unchanged in the absence of an external torque. Relationship Momentum
between
torque
and
angular
Rate of change of angular momentum of a body is equal to the external torque acting upon the body. dL i.e., text = dt
SELF CHECK
3. A hoop of radius r and mass m rotating with an angular velocity w0 is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. What will be the velocity of the centre of the hoop when it ceases to slip? rw 0 rw 0 rw 0 (a) rw0 (b) (c) (d) 4 2 3 (JEE Main 2013) Equilibrium of Rigid Bodies
A rigid body is in mechanical equilibrium, if • it is in translational equilibrium i.e. the total external force on it is zero. i.e. SFi = 0. • it is in rotational equilibrium, i.e., the total external torque on it is zero, i.e. Sti = 0. Moment of Inertia
The property of a body by virtue of which it opposes any change in its state of rest or of rotation is defined as its moment of inertia. The moment of inertia of a particle is given by I = mr2 where m is the mass of the particle and r is the distance of the particle from axis of rotation. Moment of inertia of a system of particles depends on • Axis of rotation • Mass of the system • Distribution of mass in the body About a particular axis of rotation, moment of inertia of a rigid body is constant. Moment of inertia plays same role in rotational motion as mass plays in translational motion. KEY POINT • Moment of inertia is a scalar quantity. Its SI unit is kg m2 and its dimensional formula is [ML2T0].
Radius of gyration of a body about an axis of rotation may also be defined as the root mean square distance of the particles from the axis of rotation r12 + r22 + ... + rN2 N theorems of Moment of Inertia Theorem of perpendicular axes: The moment of inertia of a planar lamina about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendiculzar axes concurrent with perpendicular axis and lying in the plane of the body. Iz = Ix + Iy where x and y are two perpendicular axes in the plane and z axis is perpendicular to its plane. Theorem of parallel axes : The moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes. I = ICM + Md2 where ICM is the moment of inertia of the body about an axis passing through the centre of mass and d is the perpendicular distance between two parallel axes. Moment of inertia of some regular bodies about specific axis is given in the table : i.e., k =
S. No.
1.
Radius of Gyration
It is defined as the distance from the axis of rotation at which, if whole mass of the body were supposed to be concentrated, the moment of inertia would be same as with the actual distribution of the mass of body into small particles. I M The moment of inertia of a body about a given axis is equal to the product of mass of the body and square of its radius of gyration about that axis. i.e., I = Mk2. The SI unit of radius of gyration is metre and its dimensional formula is [M0LT0]. Radius of gyration, k =
2.
Body
Axis of rotation
Moment of inertia (I)
about an axis passing through its centre and MR2 perpendicular to its plane
Uniform circular ring about a diameter of mass M and radius R about a tangent in its own plane about a tangent perpendicular to its plane about an axis passing through its centre and perpendicular to its plane Uniform circular disc about a diameter of mass M and radius R about a tangent in its own plane about a tangent perpendicular to its own plane
1 MR2 2
3 MR2 2
2MR2 1 MR2 2
1 MR2 4 5 MR2 4 3 MR2 2
PHYSICS FOR YOU | September ‘15
33
3.
4.
5.
6.
2 2 Solid sphere about its diameter 5 MR of radius R 7 and mass M about a tangential MR2 axis 5 Hollow about its diameter sphere of radius R and about a tangential mass M axis
2 MR2 3 5 MR2 3
about its own axis
1 MR2 2
Solid cylinder of length l, radius R and mass M
Hollow cylinder of mass M, length l and radius R
about an axis l 2 R2 passing through M + its centre and 12 4 perpendicular to its own axis about the diameter l 2 R2 of one of the faces M + 4 3 of cylinder about its own axis MR2
about an axis passing through R2 l 2 its centre and M + 2 12 perpendicular to its own axis
7.
about an axis passing through 2 its centre and ML perpendicular to 12 Thin rod of the rod length L about an axis passing through ML2 one end and 3 perpendicular to the rod
8.
about an axis Rectangular passing through 2 2 lamina of its centre and M l + b length l and 12 perpendicular to breadth b its plane
SELF CHECK
4. Consider a uniform square plate of side a and mass m. The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is 34
PHYSICS FOR YOU | September ‘15
(a) (c)
2 ma 2 3 1 ma 2 12
5 ma 2 6 7 (d) ma 2 12
(b)
(AIEEE 2008)
Kinematic Equations of Rotational Motion
The equations of motion with uniform acceleration have rotational counterparts which are, by comparison, ...(i) w = w0 + at 1 2 ...(ii) q = w 0t + at 2
w02
2
...(iii) w = + 2aq where w0 is the initial angular velocity and w is the final angular velocity (both in rad s–1) after the body has rotated through angular displacement q (in rad) with constant angular acceleration a (in rad s–2) in a time interval t (in s). Analogy between translational motion and rotational motion 1.
Translational motion Displacement s
2.
Velocity v =
3.
4. 5.
Acceleration dv a= dt Mass M Force F = Ma
6.
Work dW = Fds
ds dt
Rotational motion Angular displacement q Angular velocity dq w= dt Angular acceleration dw a= dt Moment of inertia I Torque t = Ia Work dW = tdq
Kinetic energy of a Kinetic energy of a rotational motion translational motion 2 Mv I w2 KT = KR = 2 2 8. Power P = Fv Power P = tw 9. Linear momentum Angular momentum p = Mv L = Iw Equations of 10. Equations of rotational motion translational motion (i) w = w0 + at (i) v = u + at 1 2 1 2 (ii) s = ut + at (ii) q = w0t + at 2 2 2 2 2 2 (iii) v – u = 2as (iii) w – w0 = 2aq a (iv) (iv) snth = u + (2n − 1) a 2 qnth = w0 + (2n − 1) 2
7.
a Rolling Rigid Body
1 Kinetic energy of rotational motion, K R = I w2 . 2 Kinetic energy of a rolling body = translational kinetic energy (KT) + rotational kinetic energy (KR) K2 1 1 1 = Mv 2 + I w2 = Mv 2 1 + 2 2 2 2 R When a body rolls down an inclined plane of inclination q without slipping its velocity at the bottom of incline 2 gh where h is the height of the is given by v = K2 1+ R2 incline. Its acceleration down the inclined plane is given g sin q . by a = K2 1+ R2 Time taken by the body to reach the bottom is given by t=
K2 2l 1 + R2 where l is the length of the inclined plane. g sin q
KEY POINT • The motion of a rigid body, which is not pointed or fixed in some way, is either purely translational or a combination of translational and rotational motions. • Angular displacement in the anticlockwise direction is positive and one in the clockwise direction is negative. GRaVItatIOn
•
•
•
•
It holds good for long distances like interplanetary distances and also for short distances like interatomic distances. The interaction means that both the particles experience forces of equal magnitude in opposite directions. If F1 , F2 are the forces exerted on particle 1 by particle 2 and on particle 2 by particle 1, respectively then F1 = − F2 . Since the forces F1 and F2 are exerted on different bodies, they are known as actionreaction pair. It is a conservative force. Therefore, the work done by the gravitational force on a particle is independent of the path described by the particle. It depends upon the initial and final positions of the particle. Therefore, no work is done by the gravity if a particle moves in a closed path. If a particle A is acted upon by n particles say the net force exerted on it must be equal to the vector sum of the forces due to all the surrounding particles. i =n FA = ∑ Fi i =1
where Fi is force acted on the particle A by the ith particle.
SELF CHECK
5. Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is (a)
Universal law of Gravitation
Every particle in this universe attracts every other particle with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them. The gravitational force F between two particles of masses m1 and m2, distance r apart is given by mm F =G 1 2 r2 where G is a constant, called the universal gravitational constant, and assumed to have the same value everywhere for all matter. Characteristics of the Gravitational Force • Gravitational force is always attractive and directed along the line joining the particles. • It is independent of the nature of the medium surrounding the particles.
(c)
1 GM (1 + 2 2 ) (b) 2 R 2 2
GM R
(d)
GM R GM (1 + 2 2 ) R (JEE Main 2014)
Kepler’s laws of Planetary Motion
The sun has eight planets and the earth is one of them. The planets closer to the sun moves in nearly circular orbits whereas the farther planets move in elliptical orbits. some planets have satellites orbiting around them. For example the moon is the earth’s satellite. It is a natural satellite. The earth has many artificial satellite orbiting it. The orbital motion of planets and satellites obey certain laws known as Kepler’s laws of planetary motion. First law (law of orbits)
The orbits of the planets are elliptical with sun at one of its foci. PHYSICS FOR YOU | September ‘15
35
Second law (law of areas)
KEY POINT
The radial line connecting the planet and the sun sweeps out equal areas in equal time intervals. dA remains constant for a planet. Areal velocity = dt Angular momentum ( L ) of a planet is related with areal dA dA velocity dt by the relation L = 2m dt The area covered by the radius vector in dt seconds =
• The value of acceleration due to gravity is
independent of the shape, size, mass etc. of the body but depends upon mass and radius of the earth or planet due to which there is a gravity pull. • The value of the acceleration due to gravity on the moon is about one sixth of that on the earth and on the sun is about 27 times that on the earth. • The value of acceleration due to gravity is minimum at planet Mercury and maximum at planet Jupiter.
1 2 r dq. 2
\ The areal velocity = 1 r 2 dq = 1 r 2w = 1 rv. 2
dt
2
2
KEY POINT • Kepler’s second law follows from the law of conservation of angular momentum. • According to Kepler’s second law, the speed of the planet is maximum, when it is closest to the sun and is minimum when the planet is farthest from the sun.
Variation of acceleration due to gravity (g) •
third law (law of periods)
The square of the period of revolution is proportional to the cube of the semi-major axis of the orbit. i.e., T2 ∝ r3 \ For two planets with time period, T1, T2 and semimajor axis r1 and r2
r3 = 1 T22 r23 acceleration Due to Gravity It is defined as the acceleration of a particle caused by the gravitation force, at the point under consideration. According to Newton’s second law, F = ma If the force on body is due to gravity of earth, then acceleration in the body is called acceleration due to gravity. It is denoted by g, i.e. a = g, \ F = mg Acceleration due to gravity is a vector quantity. It is directed towards the centre of earth. Its SI unit is m s–2. Its dimensional formula is [M0LT–2]. The value of g on the surface of earth is taken to be 9.8 m s–2 and it varies with altitude, depth, shape and the rotation of earth.
•
T12
Relationship between g and G
4 G pRe3r 4 g= = 3 = pGRe r 2 2 3 Re Re where Me is the mass of the earth, Re is the radius of the earth and r is the uniform density of the material of the earth. GMe
36
PHYSICS FOR YOU | September ‘15
•
•
•
Due to altitude (h) : The acceleration due to gravity at height h above the surface of earth is given by −2 GMe GMe h gh = = g 1 + g = Re (Re + h)2 Re2
2h For h Bgas For gases, bulk modulus are of two types: • Isothermal bulk modulus Biso = P (pressure exerted by the gas) • Adiabatic bulk modulus Bad = gP where g = CP/CV. Bad \ = g > 1 ; Bad > Biso Biso Therefore adiabatic bulk modulus is greater than isothermal bulk modulus. h is the characteristic of solid materials only as the liquids and gases do not have fixed shape. h for liquid is zero.
Breaking force = Breaking stress × Area of cross section of the wire. Work done in a stretched wire, 1 W = × stress × strain × volume 2 1 F DL 1 = × × AL = F × DL 2A L 2 1 = × load × elongation 2 This work done is stored in the wire as its elastic potential energy or strain energy. Elastic potential energy stored per unit volume of a stretched wire, 1 1 u = × stress × strain = × Y × (strain)2 2 2 Pressure It is defined as the thrust acting per unit area on the surface in contact with liquid. thrust (F ) F i.e. P = = area (A) A It is a scalar quantity. Its dimensional formula is [ML–1T–2]. Its SI unit is N m–2. It has been named as pascal (Pa) in the honour of French scientist Blaise Pascal. • For a point at a depth h below the surface of a liquid of density r, hydrostatic pressure P is given by P = P0 + hrg where P0 represents the atmospheric pressure.
KEY POINT • Young’s modulus is defined only for solids, but not for liquids and gases.
lateral strain −Dr / r = longitudinal strain DL / L where L is the original length of the wire, DL the increase in length, r is the original radius and Dr is the change in radius. s has no units and dimensions. 1 Theoretically, s lies between –1 and + . 2 1 + . Practically s lies between zero and 2 Relations among elastic constants (Y, B, h and s) • Y = 3B(1 – 2s) • Y = 2h(1 + s) 3B − 2h • s= 2h + 6 B Poisson’s ratio (s) :
•
9 1 3 = + Y B h
The pressure P0 + hrg is also known as absolute pressure. • Pressure depends on the depth of the point below the surface (h), nature of liquid (r) and acceleration due to gravity (g) while it is independent of the amount of liquid, shape of the container or crosssectional area considered. So if a given liquid is filled in vessels of different shapes to same height, the pressure at the base in each vessel will be the same, though the volume or weight of the liquid in different vessels will be different.
PA = PB = PC but WA < WB < WC PHYSICS FOR YOU | September ‘15
41
Pascal’s law
Stoke’s law
It states that pressure in a fluid at rest is same at all points which are at the same height. A change in pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and the walls of the containing vessel. • Hydraulic lift : It is used to lift heavy loads such as car at the service station. It is based on the Pascal’s law. It consists of two pistons which are separated by the space filled with a liquid as shown in the figure.
When a sphere of radius r moves with a velocity v through a fluid of viscosity h, the viscous force opposing the motion of the sphere is given by F = 6phrv Importance of Stoke’s law • • •
•
This law is used in the determination of electronic charge with the help of milikan’s experiment. This law accounts the formation of clouds. This law accounts why the speed of rain drops is less than that of a body falling freely with a constant velocity from the height of clouds. This law helps a man coming down with the help of parachute.
terminal Velocity
•
A piston of small cross-section A1 is used to exert a force F1 directly on the liquid. The pressure P = F1/A1 is transmitted throughout the liquid to the larger cylinder attached with larger piston of area A2 which results in an upward force of P × A2. FA F2 = PA2 = 1 2 A1 Hydraulic brakes in automobiles is based on the Pascal’s law.
Viscosity
When a fluid flows such that a velocity gradient is set up within it, force act within the fluid so as to prevent the velocity gradient from existing. This force is due to property called viscosity. dv Thus, F = −h A dz (The negative sign is put to account for the fact that the viscous force is opposite to the direction of motion.) where h is a constant called coefficient of viscosity, its CGS unit is poise. Dimensions is [ML–1T–1]. The SI unit of viscosity is poiseuille (Pl). 1 Pl = 10 poise Effect on viscosity • •
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Effect of temperature : On increasing temperature viscosity of a liquid decreases. Effect of pressure : On increasing pressure viscosity of liquid increases but viscosity of water decreases. PHYSICS FOR YOU | September ‘15
If for a sphere, viscous force become equal to the net weight acting downward, the velocity of the body become constant and is known as terminal velocity and is given by, 2 r−s vc = r 2 g 9 h Streamline Flow
When a liquid (fluid) flows, such that each particle of the liquid passing a point moves along the same path and has the velocity as its predecessor then the flow is called streamline flow. It is also called laminar flow. turbulent Flow
When the velocity at a point in the liquid changes with time, the flow is called unsteady flow. The unsteady flow is called turbulent when there are bends in the path of a fast moving liquid. Critical Velocity
The critical velocity is that velocity of liquid flow, upto which its flow is streamlined and above which its flow becomes turbulent. kh vc = rr Reynold number : Reynold number is a pure number which determines the nature of flow of liquid through a pipe. According to Reynold, the critical velocity vc of a liquid flowing through a tube of diameter D is given by N h rDvc vc = R or NR = rD h where h is the coefficient of viscosity of the liquid, r is the density of liquid and NR is a constant called Reynold number.
If the value of Reynold number lies between 0 to 2000, the flow of liquid is streamline or laminar. For values of NR above 3000, the flow of liquid is turbulent and for values of NR between 2000 to 3000, the flow of liquid is unstable changing from streamline to turbulent flow.
SELF CHECK
9. If it takes 5 minutes to fill a 15 litre bucket from a 2 water tap of diameter cm then the Reynolds p number for the flow is (density of water = 103 kg/m3 and viscosity of water = 10–3 Pa.s) close to (a) 5500 (b) 11,000 (c) 550 (d) 1100 (JEE Main 2015) Bernoulli’s theorem
It states that in a streamline flow of incompressible and non-viscous fluid through a tube of non-uniform cross-section, the sum of the pressure energy per unit volume, the potential energy per unit volume and the kinetic energy per unit volume is same at every point in the tube, 1 i.e., P + rgh + rv 2 = constant 2 This equation is known as Bernoulli’s equation. It represents conservation of mechanical energy in case of moving fluids. If the liquid is flowing through a horizontal tube, then h is constant, then Bernoulli’s theorem states that 1 P + rv 2 = a constant 2 applications of Bernoulli’s theorem • Bunsen’s burner • Atomiser or sprayer • Aerofoil or lift on aircraft wing • Blowing off the roof during storm • Curved motion of a spinning ball (Magnus effect) • Venturimeter Surface tension
The free surface of a liquid contracts so that its exposed surface area becomes minimum i.e., it behaves as if it were under tension, some what like a stretched elastic membrane. This property is known as surface tension. The surface tension of a liquid varies with temperature as well as dissolved impurities, etc. When soap is mixed with water, the surface tension of water decreases. Surface tension of a liquid is measured by the normal force acting per unit length. F i.e., S = . L
Surface tension is a scalar quantity. SI unit of surface tension is N m–1. Dimensions of surface tension is [ML0T–2]. Surface Energy
It is defined as the amount of work done against the force of surface tension in increasing the liquid surface of a given area at a constant temperature. i.e., surface energy = work done = surface tension × increase in surface area of the liquid The SI unit of surface energy is joule and CGS unit is erg. When a bigger drop splits into smaller drops, energy is required to break it but when smaller drops coalesce to form a bigger drop energy is released. Work done in forming a liquid drop of radius r, surface tension S is, W = 4pr2S. Work done in forming a soap bubble of radius r, surface tension S is, W = 2 × 4pr2S = 8pr2S Work done in increasing the radius of a liquid drop from r1 to r2 is W = 4 pS (r22 − r12 ) Work done in increasing the radius of a soap bubble from r1 to r2 is W = 8 pS (r22 − r12 ) When n number of smaller drops of a liquid, each of radius r, surface tension S are combined to form a bigger drop of radius R, then R = n1/3r The surface area of bigger drop = 4pR2 = 4pn2/3r2. It is less than the area of n smaller drops. Work done in breaking a liquid drop of radius R into n
equal small drops W = 4pR2 (n1/3 – 1)S where S is the surface tension.
SELF CHECK
10. A thin liquid film formed between a U-shaped wire and a light slider supports a weight FILM of 1.5 × 10–2 N (see figure). The length of the slider is 30 cm and its weight negligible. The surface tension of the liquid w film is (a) 0.1 Nm–1 (b) 0.05 Nm–1 (c) 0.025 Nm–1 (d) 0.0125 Nm–1 (AIEEE 2012) 11. Work done in increasing the size of a soap bubble from a radius of 3 cm to 5 cm is nearly (Surface tension of soap solution = 0.03 N m–1) (a) 4p mJ (b) 0.2p mJ (c) 2p mJ (d) 0.4p mJ (AIEEE 2011) PHYSICS FOR YOU | September ‘15
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Excess Pressure
angle of Contact
The pressure on the concave side of the liquid surface is always greater than the pressure on the convex side. The difference of pressure is known as excess pressure. Excess pressure inside a liquid drop is given by 2S P= r Excess pressure inside a soap bubble is given by 4S P= . r Excess pressure inside an air bubble in a liquid is given by 2S P= r When an air bubble of radius r is at depth h below the free surface of liquid of density r and surface tension S, then the excess pressure inside the bubble, 2S P = + hrg r If r1 and r2 are the radii of curved liquid surface, then excess pressure inside the liquid surface is given by
It is defined as the angle between the tangents to solid and liquid surfaces at a point of contact inside the liquid. It depends on the nature of solid and liquid. The angle of contact is different for different pairs of solids and liquids. Angle of contact does not depend upon the inclination of the solid surface to the liquid surface. Angle of contact increases with increase in temperature of liquid. Angle of contact decreases on adding soluble impurity to a liquid. The value of angle of contact lies between 0° and 180°. The value of angle of contact for pure water and glass (without grease) is zero. For all those liquids which wet the solid surface and which rise up in a capillary tube, the angle of contact is an acute angle (q < 90°), e.g. water and glass. For all those liquids which do not wet a solid surface and which depress in a capillary tube, the angle of contact is an obtuse angle (q > 90°). e.g. glass and mercury. For all those liquids which neither rise nor get depressed in a capillary tube, the angle of contact is right angle (q = 90°). e.g. silver, and water.
1 1 P =S + r1 r2 When two bubbles of different sizes are in communication with each other, air passes from smaller one to larger one and larger one grows at the expense of smaller one. This happens due to pressure inside the smaller bubble being higher than that inside the larger bubble. When two soap bubbles of radii r1 and r2 coalesce to form a new soap bubble of radius r, under isothermal conditions then r = r12 + r22 . When two soap bubbles of radii r1 and r2 are in contact with each other and r is the radius of the interface, then rr r= 12 . r2 − r1 The total pressure inside an air bubble of radius r at a depth h below the surface of liquid of density r is 2S P = P0 + hrg + r where P0 is the atmospheric pressure and S is the surface tension of liquid. KEY POINT • If a small drop of water is squeezed between two parallel glass plates so that a very thin layer of large area is formed then the pressure inside the water layer is less than the pressure on the plates by (2S/d) (where d is the distance between the plates). 44
PHYSICS FOR YOU | September ‘15
Capillarity
The phenomenon of rise or fall of liquid in a capillary tube is known as capillarity. The rise or fall in a capillary tube is given by 2S cos q 2S r h= = cos q = rrg Rrg R where S is the surface tension of the liquid, θ is the angle of contact, r is the density of liquid, r is the radius of capillary tube, R is the radius of the meniscus and g is the acceleration due to gravity. • If q > 90°, i.e., meniscus is convex, h will be negative, i.e., the liquid will fall in a capillary tube. • If q = 90°, i.e., meniscus is plane, h = 0, so no phenomenon of capillarity. • If q < 90°, i.e., meniscus is concave, h will be positive, i.e., the liquid will rise in the capillary. If a capillary tube is of insufficient length as compared to height to which liquid can rise in the capillary tube, then the liquid rises upto the full length of capillary tube but there is no overflowing of the liquid in the form of fountain. It is so because the liquid meniscus adjusts its radius of curvature so that hR = a constant i.e. hR = h′R′. Heat
It is a form of energy that flows from one body to another by virtue of temperature difference between them. The SI unit of heat is joule.
The practical unit of heat is calorie. 1 kcal = 103 cal 1 calorie is the amount of heat required to raise the temperature of 1 g of water from 14.5°C to 15.5°C. Joule’s mechanical equivalent of heat : Joule found that when mechanical work is converted into heat, the ratio of work done (W) to heat produced (Q) always remains the same and constant. W i.e. = constant = J or W = JQ Q The constant J is called Joule’s mechanical equivalent of heat. J is not a physical quantity but a conversion factor involved when work is converted into heat or vice-versa. The value of J = 4.186 J cal–1 1 calorie = 4.186 joule. temperature
It is a measure of degree of hotness or coldness of the body. Relationship between Different Temperature Scales If TC, TF, TR and TK are the temperatures of a body on Celsius, Fahrenheit, Reaumur and Kelvin scales respectively, then TC − 0 TF − 32 TR − 0 T − 273.15 = = = K 100 − 0 212 − 32 80 − 0 373.15 − 273.15 T T − 32 TR TK − 273.15 = = or C = F 5 9 4 5 KEY POINT • Celsius and Fahrenheit scales show the same reading at –40° i.e. –40°C = –40°F thermometer
It is an instrument used to measure the temperature of a body. Different types of thermometers are as follows : • Liquid thermometers • Gas thermometers • Resistance thermometers • Thermoelectric thermometers • Pyrometers • Vapour pressure thermometers • Bimetallic thermometers • Magnetic thermometers thermal Expansion
The increase in the dimensions of a body due to the increase in its temperature is called thermal expansion. Thermal expansion is present in solids, liquids and gases. In the case of solids the increase will be in length, area and volume. In liquids and gases only expansion in volume is
possible as they do not possess any fixed shape. In the case of gases the state of a gas at any instant is dependent on its volume, pressure and temperature. Hence a gas can be heated at constant volume or at constant pressure. The property of thermal expansion of substance is different for different substances and it also depends on the state of the substance viz, solid, liquid or gas. thermal Expansion in Solids
Thermal expansion in solids is of three types : • Linear expansion : The increase in length is called linear expansion. • Area expansion or superficial expansion : The increase in area is called area expansion or superficial expansion. • Volume expansion or cubical expansion : The increase in the volume is called volume expansion or cubical expansion. Coefficient of linear expansion : It is defined as the increase in length per unit original length per degree rise in temperature. Increase in length a= Original length × Rise in temperature L − L0 a= T or LT = L0 (1 + aDT ) L0 × DT Coefficient of area expansion : It is defined as the increase in surface area per unit original surface area per degree rise in temperature. Increase in area b= Original area × Rise in temperature A − A0 b= T or AT = A0 (1 + bDT ) A0 × DT Coefficient of volume expansion : It is defined as the increase in volume per unit original volume per degree rise in temperature. Increase in volume g= Original volume × Rise in temperature V − V0 g= T or VT = V0 (1 + gDT ) V0 × DT Relation between a, b and g The three coefficients of thermal expansion are related as b g a= = . 2 3 The units of a, b and g are the same. They are measured in °C–1 or K–1. The value of a, b and g depend on the nature of the material of the solid. When temperature increases (during summer), the length of the pendulum increases due to which the time period increases and the clock will lose time and run PHYSICS FOR YOU | September ‘15
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slow. On the other hand, when temperature decreases (during winter season), the length of the pendulum decreases due to which the time period decreases and the clock will gain time and run fast. 1 The loss or gain in time Dt = a(DT )t 2 where a is the coefficient of linear expansion. DT = change in temperature and t = time interval in which we have to find loss or gain of time. Bimetallic strip : A bimetallic strip consists of two strips of equal length but of different metals, riveted together keeping one over the other. When such a bimetallic strip is heated, it bends with metal of greater a on outer side, i.e., convex side. Expansion in Liquids
When a liquid is heated its volume changes. When a liquid is heated, the containing vessel also expands and hence the measured increase in volume of the liquid is the apparent increase in volume. The real increase in volume of the liquid is equal to the sum of the apparent increase in the volume of the liquid and the increase in volume of the containing vessel. A liquid has two coefficients of expansion : • Coefficient of real expansion • Coefficient of apparent expansion Coefficient of real expansion Real increase in volume gr = Original volume × Rise in temperaature Coefficient of apparent expansion Apparent increase in volume ga = mperature Original volume × Rise in tem gr = ga + gg, where gg is coefficient of volume expansion of the container. Anomalous Expansion of Water
Generally matter expands on heating and contracts on cooling. In case of water, it expands on heating if its temperature is greater than 4°C. In the range 0°C to 4°C, water contracts on heating and expands on cooling, (i.e., its coefficient of volume expansion in this range is negative). This behaviour of water in the range from 0°C to 4°C is called anomalous expansion. Water has a maximum density a 4°C. Specific Heat or Specific Heat Capacity
It is defined as the amount of heat required to raise the temperature of unit mass of the substance through 1°C. 1 DQ The specific heat of a substance is given by s = m DT where m is the mass of the substance and DQ is the heat required to change its temperature by DT. 48
PHYSICS FOR YOU | September ‘15
The SI unit of specific heat is J kg–1 K–1. The practical unit of specific heat is cal g–1 °C–1. Specific heat depends on the nature of the substance and its temperature. Specific heat for hydrogen is maximum (= 3.5 cal g–1 °C–1) and for water, it is 1 cal g–1 °C–1. For all other substances, the specific heat is less than 1 cal g–1 °C–1. Gas has two types of Molar Specific Heats • •
Molar specific heat at constant volume (CV) Molar specific heat at constant pressure (CP)
Change of State
Matter normally exists in three states, i.e., solid, liquid and gas. A transition from one of these states to another is known as change of state. Melting : The change of state from solid to liquid is known as melting. Fusion : The change of state from liquid to solid is known as fusion. Melting point : The temperature at which the solid and liquid states of the substance are in thermal equilibrium with each other is called its melting point. It is the characteristic of the substance. It also depends on pressure. The melting point of a substance at standard atmospheric pressure is called its normal melting point. Vaporisation : The change of state from liquid to vapour is known as vaporisation. Boiling point : The temperature at which the liquid and the vapour states of the substance co-exist is called its boiling point. It is the characteristic of the substance. It also depends on pressure. The boiling point of a substance at standard atmospheric pressure is called its normal melting point. Sublimation : The change from solid state to vapour state without passing through the liquid state is known as sublimation. Latent Heat
It is defined as the amount of heat required to change the state of a unit mass of the substance at a constant temperature. Q The latent heat of a substance is given by L = m where m is the mass of a substance. The SI unit of latent heat is J kg–1 while practical unit is cal g–1. Latent heat of fusion : It is the amount of heat required to change unit mass of the solid into liquid at constant temperature e.g. latent heat of fusion of ice = 80 cal g–1.
Latent heat of vaporisation : It is the amount of heat required to change unit mass of the liquid into vapour at constant temperature. e.g. latent heat of vaporisation of water = 540 cal g–1. Principle of Calorimetry
When two bodies at different temperatures are placed in contact with each other then heat will pass from the body at higher temperature to the body at lower temperature until both reach a common temperature. i.e. heat lost by one body = heat gained by the other. Principle of calorimetry obeys law of conservation of energy. Conduction
If a metal spoon is put in a cup of hot tea, after some time its other end also becomes warm. Heat is transferred from one end of the spoon to the other end due to molecular vibrations within the spoon. This process is called thermal conduction. In this process there is no movement of mass. In the beginning of conduction process, a part of heat is absorbed at every cross-section and the remaining part is transported to the next section. But after time a steady-state is reached and the temperature of all the parts of the conductor becomes constant. Convection
Convection is that method of heat transfer in which the heat is carried from one place to another by actual movement of heated matter. This process can occur only in fluids. If the heated matter is forced to move by some agent (like a fan) then the process is known as forced convection. If the matter flows on its own due to difference of pressure it is known as natural or free convection. Radiation
Radiation is that method of heat transfer which does not require any material medium. In this process heat energy is transferred in the form of electro-magnetic waves. The properties of radiation are similar to those of light. • It travels with the velocity of light (3 × 108 m s–1 in vacuum). • It shows all the properties of light, i.e. reflection, interference, polarization etc. • It obeys inverse square law. Wien’s Displacement Law
The wavelength corresponding to highest intensity lm is inversely proportional to the absolute temperature.
Thus
b T where b (= 2.89 × 10–3 meter kelvin) is known as the Wien’s constant. lm =
Stefan’s law
The energy emitted per second per unit area of a black body is proportional to the fourth power of corresponding absolute temperature. Thus, the energy of thermal radiation emitted per unit time by a black body of surface area A is given by Q = sAT4 where s = 5.67 × 10–8 W m–2 K–4 is a universal constant, known as Stefan’s constant. For a non-black body Q = esAT4, where e is the emissivity of that body. For a body kept in surroundings of temperature T0, the net loss of thermal energy per unit time DQ = esA(T4 – T04) as the body absorbs esAT04 energy during the same period. Newton’s Law of Cooling
It states that the rate of cooling of a body is proportional to the excess temperature of the body over the surroundings dQ = − K (T − T0 ) dt where T0 is the temperature of the surrounding medium and T is the temperature of the body.
SELF CHECK
12. Three rods of Copper, Brass and Steel are welded together to form a Y-shaped structure. Area of crosssection of each rod = 4 cm2. End of copper rod is maintained at 100°C where as ends of brass and steel are kept at 0°C. Lengths of the copper, brass and steel rods are 46, 13 and 12 cms respectively. The rods are thermally insulated from surroundings except at ends. Thermal conductivities of copper, brass and steel are 0.92, 0.26 and 0.12 CGS units respectively. Rate of heat flow through copper rod is (a) 6.0 cal/s (b) 1.2 cal/s (c) 2.4 cal/s (d) 4.8 cal/s (JEE Main 2014) ANSWER kEYS (SELF CHECk) 1. (d) 2. (d) 3. (d) 4. (a) 5. (a) 6. (d) 7. (b) 8. (b) 9. (a) 10. (c) 11. (d) 12. (d) PHYSICS FOR YOU | September ‘15
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1. A tank with a square base of area 2.0 m2 is divided into two compartments by a vertical partition in the middle. There is a small hinged door of face area 20 cm2 at the bottom of the partition. Water is filled in one compartment and an acid of relative density 1.5 in the other, both to a height of 4 m. The force necessary to keep the door closed is (Take g = 10 m s–2) (a) 10 N (b) 20 N (c) 40 N (d) 80 N 2. Two bodies, each of mass M, are kept fixed with a separation 2L. A particle of mass m is projected from the midpoint of the line joining their centres, perpendicular to the line. The gravitational constant is G. The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is GM GM (a) 4 (b) 2 . L L 2GM (c) (d) None of these L 3. A particle undergoes uniform circular motion. About which point on the plane of the circle, will the angular momentum of the particle remain conserved? (a) Centre of the circle (b) On the circumference of the circle (c) Inside the circle (d) Outside the circle 4. Two thin uniform circular rings each of radius 10 cm and mass 0.1 kg are arranged such that they have common centre and their planes are perpendicular to each other. The moment of inertia of this system about an axis passing through common centre and perpendicular to the plane of either of the rings in kg m2 is (a) 15 × 10–3 (b) 5 × 10 –3 –4 (c) 15 × 10 (d) 18 × 10–4 5. A metal cylinder of length L is subjected to a uniform compressive force F as shown in the figure. The material of the cylinder has Young’s modulus Y and Poisson’s ratio s. The change in volume of the cylinder is 50
Physics for you | September ‘15
F
(1− s)FL sFL (b) Y Y (1 − 2s)FL (1 + 2s)FL (c) (d) Y Y 6. The cylindrical tube of spray pump has a crosssection of 8 cm2, one end of which has 40 fine holes each of area 10–8 m2. If the liquid flows inside the tube with a speed of 0.15 m min–1, the speed with which the liquid is ejected through the holes is (a) 50 m s–1 (b) 5 m s–1 –1 (c) 0.05 m s (d) 0.5 m s–1 (a)
7. A launching vehicle carrying an artificial satellite of mass m is set for launch on the surface of the earth of mass M and radius R. If the satellite is intended to move in a circular orbit of radius 7R, the minimum energy required to be spent by the launching vehicle on the satellite is (Gravitational constant = G) 13 GMm GMm (a) (b) 14 R R GMm GMm (c) (d) 7R 14 R 8. A uniform disc of radius R lies in xy plane with its centre at origin. Its moment of inertia about the axis x = 2R and y = 0 is equal to the moment of inertia about the axis y = d and z = 0. Find the value of d interms of R. 4 15 R (d) R 3 2 2 9. A body of area 1 cm is heated to a temperature 1000 K. The amount of energy radiated by the body in 1 s is (Stefan’s constant s = 5.67 × 10–8 W m–2 K–4) (a) 5.67 J (b) 0.567 J (c) 56.7 J (d) 567 J (a)
17 R (b) 2
13 R (c)
10. Two spheres each of mass M and radius R are separated by a distance of r. The gravitational potential at the midpoint of the line joining the centres of the spheres is GM 2GM (a) − (b) − r r GM 4GM (c) − (d) − 2r r
11. A child is standing with his two arms outstretched at the centre of a turntable that is rotating about its central axis with an angular speed w0. Now, the child folds his hands back so that moment of inertia becomes 3 times the initial value. The new angular speed is w w (a) 3w0 (b) 0 (c) 6w0 (d) 0 6 3 12. Two bodies have their moments of inertia I and 2I respectively about their axis of rotation. If their kinetic energies of rotation are equal, their angular momenta will be in the ratio (a) 1 : 2
(b)
2 : 1 (c) 1 : 2 (d) 2 : 1
13. A square gate of size 4 m × 4 m is hinged at topmost point. A fluid of density ρ fills Hinge the space left of it. The force which acting 1 m from 1m lowest point can hold the gate stationary is 256 256 128 128 ρg (b) ρg (c) ρg (a) ρg (d) 3 9 3 9 14. A coin of mass m and radius r, having moment of inertia I about the axis passing through its centre and perpendicular to its plane, is beaten uniformly to form a disc of radius 2r. What will be the moment of inertia of the disc about the same axis? (a) I (b) 2I (c) 4I (d) 16 I 15. A spherical ball is F P dropped in a long column Q of viscous liquid. Which of the following graphs R represent respectively the variation of (i) gravitational force with time (ii) viscous force with time (iii) net force acting on the ball with time? (a) Q, R, P (b) R, Q, P (c) P, Q, R (d) P, R, Q
t
16. A planet revolves around the sun in an elliptical orbit of eccentricity e. If T is the time period of the planet, then the time spent by the planet between the ends of the minor axis and major axis close to the sun is Tp 2e − 1 (a) (b) T 2e p (c)
Te 2p
1 e (d) T − 4 2p
17. A rigid bar of mass M is supported symmetrically by three wires each of length L. Those at each end are of copper and the middle one is of iron. The ratio of their diameters, if each is to have the same tension, is equal to Ycopper Yiron (b) (a) Ycopper Yiron (c)
2 Yiron
2 Ycopper
(d)
Yiron Ycopper
18. Two strips of metal are riveted together at their ends by four rivets, each of diameter 6 mm. Assume that each rivet is to carry one quarter of the load. If the shearing stress on the rivet is not to exceed 6.9 × 107 Pa, the maximum tension that can be exerted by the riveted strip is (a) 2 × 103 N (b) 3.9 × 103 N 3 (c) 7.8 × 10 N (d) 15.6 × 103 N 19. A solid cylinder of radius R made of a material of thermal conductivity K1 is surrounded by a cylindrical shell of inner radius R and outer radius 2R made of a material of thermal conductivity K2. The two ends of the combined system are maintained at two different temperatures. Then there is no loss of heat across the cylindrical surface and the system is in steady state. The effective thermal conductivity of the system is K1 K 2 (a) K1 + K2 (b) K1 + K 2 K + 3K 2 3K1 + K 2 (b) (d) 1 4 4 20. Three uniform spheres of mass M and radius R each are kept in such a way that each touches the other two. The magnitude of gravitational force on any of the spheres due to other two is (a) (c)
3 GM 2 2 R2 3GM 2
(b) (d)
3 GM 2 2 R2
3 GM 2 4 R2
R2 21. A circular platform is mounted on a frictionless vertical axle. Its radius R = 2 m and its moment of inertia about the axle is 200 kg m2. It is initially at rest. A 50 kg man stands on the edge of the platform and begins to walk along the edge at the speed of 1 m s–1 relative to the ground. Time taken by the man to complete one revolution is p 3p s s (c) 2p s (d) (a) p s (b) 2 2 Physics for you | September ‘15
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22. Three particles of masses 1 kg, 2 kg and 3 kg are situated in the xy plane at the corners of an equilateral triangle of side b with mass 1 kg at the origin and 2 kg on the x-axis The coordinates of the centre of mass are 3 3b 7b 7b 3 3b , , 0 (a) 0, , (b) 12 12 12 12 7b 3 3b , 0 (c) , 12 12
7b 3 3b (d) , 0, 12 12
y 23. Four holes of radius R each are cut from a thin square plate of side 4R and mass M, as shown in x figure. The moment of inertia of the remaining portion about z-axis is 4 p p (a) (b) − MR2 MR2 3 4 12 8 5p 4 p 2 (c) − MR2 (d) − MR 3 6 3 8 24. A point P lies on the axis of a a ring of mass M and radius a, P at a distance a from its centre O a O. A small particle starts from P and reaches O under gravitational attraction only. Its speed at O will be
(a)
2GM a
(b)
2GM 1 1− a 2
2GM ( 2 − 1) (d) zero (c) a 25. What will be the total pressure (in N m–2) inside a spherical air bubble of radius 0.2 mm at a depth of 2 m below the surface of a liquid of density 1.1 g cm–3 and surface tension 50 dyne cm–1? Atmospheric pressure is 1.01 × 105 N m–2. (Take g = 10 m s–2) (a) 1.235 × 105 (b) 1.112 × 105 5 (c) 1.215 × 10 (d) 1.122 × 105 26. How many grams of ice at – 14°C is needed to cool 200 gram of water from 25°C to 10°C? (Take specific heat of ice = 0.5 cal g–1°C–1 and latent heat of ice = 80 cal g–1.) (a) 14 g (b) 31 g (c) 50 g (d) 80 g 27. A mass M is divided into two parts xm and (1 – x)m. For a given separation, the value of x for which the gravitational attraction between the two pieces becomes maximum is 52
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(a)
1 2
(b)
3 5
(c) 1
(d) 2
28. A rod of length L is composed of a uniform length 1 L of wood whose mass is mw and a uniform 2 1 length L of brass whose mass is mb. The moment 2 of inertia I of the rod about an axis perpendicular to the rod and through its centre is equal to (a) (mw + mb)
L2 12
(c) (mw + mb)
L2 3
L2 6 L2 (d) (mw + mb) 2 (b) (mw + mb)
29. A block of ice at –8°C is slowly heated and converted to steam at 100°C. Which of the following curves represents the phenomena qualitatively? (a)
(b)
(c)
(d)
30. Match the Column I with Column II. Column I
Column II
(A) The shape of rubber (p) Young’s modulus of heel changes under elasticity is involved stress (B) In a suspended bridge, (q) B u l k m o d u l u s o f there is a strain in the elasticity is involved ropes by the load of the bridge (C) In an automobi l e (r) Modulus of rigidity tyre, when air is is involved compressed, the shape of tyre changes (D) A s o l i d b o d y i s (s) A l l t h e m o d u l i subjected to a of elasticity are deforming force involved (a) (b) (c) (d)
A – q, B – r, C – s, D – p A – p, B – q, C – r, D – s A – r, B – q, C – p, D – s A – r, B – p, C – q, D – s
SolutionS 1. (c) : The situation is as shown in the figure.
4 m Water
Acid
Door
For compartment containing water, h = 4 m, ρw = 103 kg m–3 Pressure exerted by the water on the door at the bottom is Pw = ρwhg = 103 kg m–3 × 4 m × 10 m s–2 = 4 × 104 N m–2 For compartment containing acid, ρa = 1.5 × 103 kg m–3, h = 4 m Pressure exerted by the acid on the door at the bottom is Pa = ρahg = 1.5 × 103 kg m–3 × 4 m × 10 m s–2 = 6 × 104 N m–2 \ Net pressure on the door = Pa – Pw = (6 × 104 – 4 × 104) N m–2 = 2 × 104 N m–2 Area of the door= 20 cm2 = 20 × 10–4 m2 \ Force on the door = 2 × 104 N m–2 × 20 × 10–4 m2 = 40 N Thus, to keep the door closed the force of 40 N must be applied horizontally from the water side. 2. (b) : The situation is as shown in the figure. v
M
•
L
• m 2L
M
•
L
Applying the law of conservation of mechanical energy, we get GMm GMm 1 2 − − + mv = 0 + 0 L L 2 1 2 2GMm 4GM GM mv = =2 ⇒ v= L 2 L L 3. (a) : Here, L = mvr The direction of L (about the centre) is perpendicular to the plane containing the circular path. Both magnitude L and direction of the angular momentum of p= mv r the particle moving in O a circular path about its centre O is constant.
4. (c) : Because both the rings have common centre and their planes are mutually perpendicular, hence, an axis which is passing through the centre of one of the rings and perpendicular to the plane of its plane, will be along the diameter of other ring. Hence, moment of inertia of the system about the given axis is 2 1 2 3 2 I = ICM + Idiameter = MR + MR = MR 2 2 3 = (0.1 kg )(0.1 m)2 = 15 × 10−4 kg m2 2 5. (d) : Volume of the cylinder, V = pr2L DV D(pr 2 L) = = Volumetric strain V pr 2 L 2 DV pr DL + 2 pr L Dr DL 2Dr = + ...(i) = L r V pr 2 L Dr sDL (Dr / r ) or =− Poisson’s ratio, s = − r L (DL / L) Dr On substituting this value of in eq. (i), we get r DV DL ...(ii) = (1 − 2s) V L DL F (F / pr 2 ) or = Young’s modulus, Y = L (DL / L) pr 2Y DL On substituting this value of in eq. (ii), we L get DV F (1 − 2s) = V pr 2Y DV F = (1 − 2s) 2 pr L pr 2Y FL DV = (1 − 2s) Y 6. (b) : According to equation of continuity, a1v1 = a2v2 0.15 \ (40 × 10−8 ) × v1 = 8 × 10−4 × 60 8 × 10−4 × 0.15 or v1 = = 5 m s −1 −8 40 × 10 × 60 7. (b) : The energy of the satellite on the surface of the earth is GMm GMm Es = KE + PE = 0 + − = − R R The energy of the satellite in an orbit of radius r is 1 GMm Eo = mvo2 + − 2 r 1 GM GMm GM = m − As vo = 2 r r r GMm =− 2r Physics for you | September ‘15
53
The minimum energy required to be spent by the vehicle on the satellite is DE = Eo – Es GMm GMm ( r = 7R) =− − − 2(7 R) R GMm GMm 13 GMm =− + = 14 R R 14 R 8. (a) :
y
y=d z=0
Initial moment of inertia = Ii Final moment of inertia If = 3Ii According to the law of conservation of angular momentum, we get Li = Lf Ii wi = If wf I I Iw w f = i i = i wi = i wi ( If = 3Ii) If 3Ii If w w = i = 0 ( wi = w0) 3 3 12. (c) : As K R = K R 1
O
x = 2R y=0
x
The axis x = 2R, y = 0 is in z direction as shown in figure. Using theorem of parallel axes, 1 9 I1 = m R2 + m(2 R)2 = m R2 2 2 The axis y = d, z = 0 is shown as dotted line in figure. Again, using the theorem of parallel axes, 1 I2 = m R2 + md 2 4 As I2 = I1 (Given) 1 2 2 9 2 \ m R + md = m R 4 2 17 2 9 1 md = − m R2 = m R2 2 4 4 17 d= R 2 9. (a) : Here, A = 1 cm2 = 10–4 m2 T = 1000 K, t = 1 s s = 5.67 × 10–8 W m–2 K–4 According to Stefan-Boltzmann law, energy radiated by a body is E = sAT4t = 5.67 × 10–8 × 10–4 × (1000)4 × 1 = 5.67 J 10. (d) :
M R
M
P
r/2
R
r/2 r
Let P is the midpoint of the line joining the centres of the spheres. The gravitational potential at point P is 4 GM GM GM 2GM 2GM VP = − − =− − =− r /2 r /2 r r r 11. (b) : Here, Initial angular speed, wi = w0 54
Physics for you | September ‘15
2
1 1 \ I w2 = I w2 2 1 1 2 2 2 w I or 1 = 2 w2 I1 L1 I1w1 I1 = = L2 I2w2 I2 L1 1 1 = = 2 L2 2
…(i) I2 I = 1 I1 I2
(Using (i))
13. (b) : Side of square gate, a = 4 m Consider an elementary strip of thickness dy of fluid at distance y from the point O. Area of the strip = ady Force acting on the elementary strip at a distance y from O O dF = (ρgy) (ady) y Torque due to dF about O dy dt = y (dF) F dt = (ρgy2a)dy Net torque due to fluid a=4
a
y3 t = ∫ (ρgay )dy = ρga = 256ρg clockwise 3 3 0 0 Torque due to applied force t = F × 3 anticlockwise 256 256ρg ρg For equilibrium, F × 3 = ⇒ F= 9 3 14. (c) : Moment of inertia of a coin of mass m and radius r about the axis passing through the centre of mr 2 mass and perpendicular to its plane is I = . 2 Moment of inertia of disc of mass m and radius 2r about the axis passing through the centre of mass and perpendicular to its plane is 2
I′ =
m(2r )2 = 4I 2
Physics for you | September ‘15
55
15. (c) : Gravitational force remains constant on the falling spherical ball. It is represented by straight line P. The viscous force (F = 6phrv) increases as the velocity increases with time. Hence, it is represented by curve Q. Net force = gravitational force – viscous force. As viscous force increases, net force decreases and finally becomes zero. Then the body falls with a constant terminal velocity. It is thus represented by curve R. A 16. (d) : B
S
O
As areal velocity of a planet around the sun is constant. Therefore, the desired time is area ABS t AB = × time period area of ellipse If a = semi-major axis and b = semi-minor axis of ellipse, then area of ellipse = pab 1 Area ABS = (area of ellipse) 4 – Area of triangle ASO 1 1 = × p ab − (ea) × (b) 4 2 p (ab) 1 4 − 2 eab × T = T 1 − e \ t AB = 4 2p p ab 17. (b) : The situation is as shown in the figure. Let T be tension in each wire. As the bar is supported symmetrically by the three wires, therefore extension in each wire is same. F/A As Y = Copper Copper Iron DL/L T T T If D is the diameter of the wire, then M 4F L F / p(D /2)2 Y = = DL/L p D 2 DL As per the conditions of the problem, F(tension), length L, and extension DL is same for each wire. 1 1 \ Y ∝ 2 or D ∝ Y D Dcopper Yiron \ = Diron Ycopper 18. (c) : Radius of a rivet, 6 r = mm = 3 mm = 3 × 10−3 m 2 56
Physics for you | September ‘15
Maximum stress = 6.9 × 107 Pa Maximum load on a rivet = Maximum stress × Area of cross-section = 6.9 × 107 × p × (3 × 10–3)2 = 1950 N .. . Maximum tension that can be exerted by rivet strip = 4 × 1950 N = 7.8 × 103 N 19. (d) :
T2
T1 K2
R
K1
2R
L
Area of cross-section of inner cylinder = pR2 Area of cross-section of outer shell = p(2R)2 – pR2 = 3pR2 Rate of heat flow in inner cylinder K pR2 (T1 − T2 ) H1 = 1 L Rate of heat flow in outer shell K 3pR2 (T1 − T2 ) H2 = 2 L Rate of heat flow in the combined system K 4 pR2 (T1 − T2 ) H= L At steady state, H = H1 + H2 \
K 4 pR2 (T1 − T2 ) K1pR2 (T1 − T2 ) K 2 3pR2 (T1 − T2 ) = + L L L
K + 3K 2 4K = K1 + 3K2 or K = 1 4
20. (d) : The situation is as shown in the figure. A 2R B
2R FCA
60°
2R FCB C
Gravitational force on sphere C due to sphere A is GM × M GM 2 FCA = = along CA (2R)2 4 R2 Gravitational force on sphere C due to sphere B is GM × M GM 2 FCB = = along CB (2R)2 4 R2 These two forces are equal in magnitude and inclined at an angle 60°. \ The total gravitational force on sphere C due to other two spheres is
mR2 M (16R2 + 16R2 ) − 4 + m( 2R)2 2 12 M 2 2 8 2 10 p = × 32R − 10mR = MR − MR2 12 3 16 8 5p 2 I = − MR 3 8
2 2 Ftotal = FCA + FCB + 2FCA FCB cos 60° 2
2
GM 2 GM 2 GM 2 = + +2 4 R2 4 R2 4 R2 =
=
GM 2 1 4 R2 2
3GM 2
4 R2 21. (c) : As the system is initially at rest, therefore, initial angular momentum Li = 0. According to the principle of conservation of angular momentum, final angular momentum, Lf = 0. \ Angular momentum of man = Angular momentum of platform in opposite direction i.e., mvR = Iw mvR 50 × 1 × 2 1 or w = = = rad s −1 I 200 2 Angular velocity of man relative to platform is v 1 1 wr = w + = + = 1 rad s −1 R 2 2 Time taken by the man to complete one revolution is 2p 2p T= = = 2p s wr 1 22. (c) : The coordinates of points A, B and C are (0, 0, 0), (b, 0, 0) and b b 3 2, 2 ,
Y C b
3 kg b
0 respectively. B A X b 2 kg 1 kg Now as the triangle is in XY plane, i.e., Z coordinate of all the masses is zero, so ZCM = 0. 1 × 0 + 2 × b + 3(b / 2) 7b = Now, XCM = 1+ 2 + 3 12 1 × 0 + 2 × 0 + 3 3 (b / 2) 3 3b YCM = = 1+ 2 + 3 12 So, the coordinates of centre of mass are 7b 3 3b 12 , 12 , 0
23. (d) : If M is mass of the square plate before cutting the holes, then mass of each hole, M p m= × pR2 = M 2 16 16R \ Moment of inertia of remaining portion about the given axis is I = Isquare – 4Ihole
24. (b) : Gravitational potential at P, GM GM VP = − =− 2a a2 + a2 GM Gravitational potential at O, VO = − a Let m be mass of the particle and v be the velocity of it while reaching O. According to law of conservation of mechancial energy, we get (PE)at P = (KE)at O + (PE)at O GM m 1 2 GM m \ − = mv + − a 2a 2 or or
2GM 2 GM 2GM 1 − = 1− a a 2 a 2 2GM 1 v= 1− a 2 v2 =
25. (a) : Here, r = 0.2 mm = 0.2 × 10 –3 m = 2 × 10– 4 m, h = 2 m, ρ = 1.1 g cm–3 = 1.1 × 103 kg m–3, S = 50 dyne cm–1 = 50 × 10–3 N m–1, Pa = 1.01 × 105 N m–2 Total pressure inside the bubble = atmospheric pressure + pressure due to column of liquid + excess pressure 2S = Pa + hρg + r 2 × (50 × 10−3 ) = 1.01 × 105 + 2 × 1.1 × 103 × 10 + 2 × 10−4 5 4 = 1.01 × 10 + 2.2 × 10 + 500 = 1.235 × 105 N m–2 26. (b) : Heat lost by water in cooling from 25°C to 10°C, Q1 = msDT = 200 × 1 × (25 – 10) = 3000 cal Heat gained by m g of ice at –14°C to change into water at 10°C, Q2 = (msDT)ice + mL + (msDT)water = m × 0.5 × 14 + m × 80 + m × 1 × 10 = 97m cal By principle of calorimetry, Q1 = Q2 or 97m = 3000 3000 or m = g = 31 g 97 27. (a) 28. (a) 29. (a) 30. (d) : A – r, B – p, C – q, D – s Physics for you | September ‘15
57
smooth surface
1. A uniform ladder rests in equilibrium with its lower end on a rough horizontal plane and its upper end against a smooth vertical wall as shown in figure. Find the maximum inclination q of the ladder to the vertical.
2. What is the magnitude of the maximum couple C A which may be applied to R O the cylinder if it is not to B spin. The cylinder has a weight W, radius R and the coefficient of friction m is W same at A and B as shown in figure.
ladder
3. A short semicircular right cylinder of radius r and weight W rests on a horizontal surface and is pulled at right angle to its geometric axis by a horizontal force F applied at B of the front edge. Find the angle q that the flat face will make with the horizontal plane just before sliding begins if the coefficient of friction at the line of contact A is m as shown in figure.
B
F
A
By : Prof. Rajinder Singh Randhawa*
FRICTION
4. One end of a heavy Ring D C uniform rod AB can slide A along a rough horizontal rod CD to which it is attached by a ring, B and B C are joined by a string. If ABC is a right angle and a is the angle between AB and vertical when the rod is on the point of sliding, find the coefficient of friction between ring and horizontal rod CD as shown in figure. 5. Figure shows a small block of mass m kept at the left end of a larger block of mass M and length L. The system can slide on a horizontal surface. The system is started moving towards right with an initial velocity v. The coefficient of friction between the bigger block and floor is m and that between two blocks is m/2. L Find the time m elapsed before v M the smaller block separates from the bigger block. 6. Figure shows that two blocks in contact are sliding down an inclined surface of inclination q = 30°. The friction coefficient between the block of mass m = 2 kg and the incline is m1 = 0.20 and that between the block of mass M = 4 kg and m M the incline is m2 = 0.30. Find the acceleration of 2 kg block.
*Randhawa Institute of Physics, S.C.O. 208, First Fl., Sector-36D & S.C.O. 38, Second Fl., Sector-20C, Chandigarh, Ph. 09814527699
58
physics for you | SEPTEMBER ‘15
SOLUTIONS
1. Free body diagram of ladder is shown in figure. NB
G
W NA
A
⇒ F(r − r sin q) = W ×
NA
Under the equilibrium condition,
SFV = 0 = W – NA ⇒ NA = W SFH = 0 = NB – mNA ⇒ NB = mNA = mW (using (i)) StA = 0 = NB × AB cosq – W × AG sinq \ W × AGsinq = NB × ABcosq
⇒ W×
...(i)
AB sin q = mW × AB cos q 2
sin q = 2m or tan q = 2m or q = tan −1(2m) cos q 2. Free body diagram of cylinder is shown in figure. Under the equilibrium NA condition of cylinder, SFV = mNA + NB = W ...(i) C O SFH = 0 = NA – mNB NA A W ⇒ NA = mNB ...(ii) Putting (ii) in (i), we get, NB B (m2 + 1)NB = W NB W ⇒ NB = 2 m +1 ⇒
\
NA =
mW
1+ m Taking moment about O, StO = 0 ⇒ mNA × R + mNB × R = C ...(iii) mR[NA + NB] = C Putting NA and NB, we get m(1 + m)WR W mW + = C or C = mR 2 2 1 + m2 1 + m 1 + m 3. Free body diagram is shown in figure. N
N
\ sin q =
4r sin q 3π
F 4W F+ 3π
...(iv)
Putting (iii) in (iv), we get sin q =
⇒ sin q =
mW 4W mW + 3π 3mπ 3mπ ⇒ q = sin −1 3mπ + 4 3mπ + 4
4. Let W be the weight of rod AB acting at its centre of gravity G. Let N′ be the resultant of N and mN, then N′ is inclined to N at an angle l given by m = tanl. Three forces, tension T, weight W of rod and resultant reaction N′ meet at O. Then ∠AOG = l, ∠BOG = 90° – a and ∠BGO = a. Applying trigonometrical theorem (m + n)cotq = mcota – ncotb. N
(using (ii))
2
O G
N D
...(i) ...(ii) ...(iii)
Taking moment about point A, F × (AD) = W × (GD) F(OA – OD) = W(GD)
B
⇒ F = mN and SFV = 0 = N – W ⇒ N = W Putting (ii) in (i), F = mW
B
F
A W
Let N′ is the resultant of normal force N and friction force mN and l is the friction angle, i.e. m = tan l. Under equilibrium, SFH = 0 = F – mN
T
G
B 90
°–
A
N N
W
O
From DAOB, we have (BG + AG)cota = AG cotl – BG cot (90° – a) 2 cota = cotl – tana (Q AG = BG) cot l = 2 cot a + tan a = \ cot l = ⇒ m=
2 + tan2 a tan a tan a
2 + tan2 a
2 + tan a tan a
⇒ tan l =
tan a
2 + tan2 a ( m = tanl)
physics for you | SEPTEMBER ‘15
59
5.
m /2N1
/2N1
a
M
mA N2
mg N1 FBD of m
Mg FBD of M
MA N2
From free body diagram of upper block N1 = mg m N = ma 2 1 From free body diagram of lower block N2 = N1 + Mg mA −
...(i)
m N1 m mg mg +a= +a= +a 2 m 2 m 2
Putting the value of N1, N2 and A in (iv), we get, m m mg + M g + a 2 2
a N
...(iii)
m ...(iv) N + MA 2 1 From (i) and (iii), N2 = mg + Mg = (M + m)g
m(m + M )g =
6. From free body diagram of mass m, N1 = mgcosq ma = mgsinq – m1N1 – N From free body diagram of mass M, N2 = Mgcosq Ma = Mgsinq + N – m2N2
...(ii)
mN 2 =
From (ii), A =
m M + m g 2 M 1 2L 4ML As L = at 2 ⇒ t = = m(M + m)g 2 a
⇒ a=
N1
m
1N1
mg N1
a
Mg
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physics for you | SEPTEMBER ‘15
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M
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...(iii) ...(iv)
N2
Putting values of N1 and N2 and adding equations (ii) and (iv), we get (M + m)a = (M + m)g sinq – (m1m + m2M)gcosq (M + m) g sin q − (m1m + m2 M )g cos q ⇒ a= M +m Here, M = 4 kg, m = 2 kg, q = 30°, m1 = 0.20, m2 = 0.30, g = 10 m s–2 \ a = 2.7 m s–2 nn
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....(i) ...(ii)
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integer tyPe Questions class-Xi 1. The loss of pressure when a fluid flows through a pipe is given by P = kral vbdch, where d and l are diameter and length of the pipe respectively, r and h are the mass density and coefficient of viscosity of the fluid, v is the mean velocity of flow through the pipe and k is a numerical constant. The value of a + b – c is 2. The acceleration of a marble in a certain fluid is proportional to the speed of the marble squared −v 2 for v > 0. If and is given (in SI units) by a = 3 the marble enters this fluid with a speed of 1.50 m s–1, how long will it take before the marble’s speed is reduced to half of its initial value (in s)? 3. A 2 kg block is A gently pushed from R = 1.5 m B rest at A and it θ R slides down along the fixed smooth circular surfaces as shown in figure. If the attached spring has a force constant k = 20 N m–1, what is unstretched length of spring (in m) so that it does not allow the block to leave the surface until angle with the vertical is q = 60°? 4. A disc of mass m is k connected to an ideal spring of force constant k. If disc is released from rest, what is maximum friction force θ on disc during subsequent motion. Assuming friction is sufficient to put it in rolling. (Take, mass of disc = 1 kg, radius of disc = 10 cm, angle of incline = 37°.) 5. A cubical container with side 2 m has a small hole with a cap at point C as shown. The water level is
upto point D. (Here, BC = 0.5 m and BD = 1.5 m). If container is given an A acceleration of 8 m s–2 D and the hole is opened simultaneously. The C amount of water that will spill out of the container is B 200 a litre. Find the value 2m of a. 6. A cylindrical block of length 0.4 m and area of cross-section 0.04 m2 is placed coaxially on a thin metal disc of mass 0.4 kg and of the same crosssection. The upper face of the cylinder is maintained at a constant temperature of 400 K and the initial temperature of the disc is 300 K. If the thermal conductivity of the material of the cylinder is 10 W m–1 K–1 and the specific heat of the material of the disc is 600 J kg–1 K–1, it takes 30 n ln 2 s for the temperature of the disc to increase to 350 K? Assume, for the purpose of calculation, the thermal conductivity of the disc to be very high and the system to be thermally insulated except for the upper face of the cylinder. 7. A solid copper cube and sphere, both of same mass and emissivity are heated to same initial temperature and kept under identical conditions. The ratio of 1
3 their initial rate of the fall of temperature is p . π
Find the value of p.
8. n moles of an ideal gas at 27°C are expanded isothermally to five times its volume and then heated at this constant volume to raise its pressure equal to initial pressure (before expansion). If heat given to the system in the process is 83.14 kJ then C find the value of n. The γ = p of the gas is 1.42. Cv [Take ln5 = 1.61] Physics for you | SEPTEMBER ‘15
61
9. Two identical wires are stretched by the same tension of 100 N, and each emits a note of frequency 200 Hz. The tension in one wire is increased by 1 N. Calculate the number of beats heard per second when the wires are plucked.
–4 MJ. Find the additional energy (in MJ) that should be given to the satellite so that it escapes from the gravitational field of earth. Assume earth’s gravitational force to be the only gravitational force on the satellite and no atmospheric resistance.
10. The potential energy function for the force between two atoms in a diatomic molecule is approximately a b given by U (x ) = − 6 12 x x where a and b are constants and x is the distance between the atoms. If the dissociation energy of the b2 , then molecule is D = [U (x = ∝) – Uequilibrium] = pa find the value of p.
15. The rate at which ice would melt in a wooden box 2.0 cm thick and of inside measurements 200 cm × 120 cm × 120 cm assuming that the external temperature is 30°C and coefficient of thermal conductivity of wood is 0.0004 cal s–1 cm–1°C–1, is approximately b g s–1. Find the value of b.
11. Three aircrafts make a turn 60 0 in the horizontal plane at 60 0 uniform speed, moving 60 0 along concentric circular trajectories that are shown 3 2 1 in figure. The aircrafts move such that they are at constant distance of 600 m from each other at any time. The aircraft closest to the centre moves in a circle of radius R = 600 m. The aircraft 2 is moving at a speed of v2 = 720 km h–1. The acceleration of third aircraft is 10a m s–2. Find the value of a. 12. A massless rod of length L is A C suspended by two identical strings AB and CD of equal lengths. A block of mass m B D O L is suspended from point m O such that BO is equal to L . Further it is observed that the frequency of 1st n harmonic in AB is equal to 2nd harmonic frequency in CD. Find the value of n. 13. An open organ pipe containing air resonates in fundamental mode due to a tuning fork. The measured values of length l(in cm) of the pipe and radius r(in cm) of the pipe are l = 94 ± 0.1, r = 5 ± 0.05. The velocity of the sound in air is accurately known. The maximum percentage error in the measurement of the frequency of that tuning fork by this experiment is given by a2%. Find the value of 10 a. 14. The gravitational potential energy of a satellite revolving around the earth in circular orbit is 62
Physics for you | SEPTEMBER ‘15
solutions
1. (3) : Given, P = kral vbdch Considering dimensions on both sides [P] = [kral vbdch] or [ML–1 T–2] = [ML–3]a [L] [LT–1]b [L]c [ML–1 T–1] Equating the dimensions on both sides, we get a + 1 = 1 ; –3a + 1 + b + c – 1 = –1 ; –b – 1 = – 2 On solving these equations, we get a = 0, b = 1, c = –2 ∴ a + b – c = 0 + 1 – (– 2) = 3 2. (2) : Given a =
−v 2 3
dv dt dv −v 2 = or 2 = − 3 v 3 dt On integrating both sides, 0.75
∫
1.5
t
−1 = ∫ dt 2 3 0 v
dv
0.75
t 1 1 1 t = ⇒ − = v 3 0 . 75 1 .5 3 1. 5 1× 3 or t = = 2s 1. 5 3. (1) : When the block has fallen by 60°, 1 2 1 mv = mgR(1 − cos q) ⇒ v = 2 gR × = gR 2 2 Applying Newton’s law of motion along the radial direction
mv 2 ⇒ x = 0. 5 m R ∴ Extension in spring = 0.5 m ⇒ Natural length = (R – x) = 1 m 4. (2) : Let maximum extension of spring be xm. From work-energy theorem mg cos q + kx =
kxm
1 2 mgxm sinq = kxm 2 2mg sinq k At lower extreme position kxm – mg sinq – f = ma Also, a = Ra or xm =
N
mgcos f
mR2 a 2 mg sin q 1 × 10 × 3 / 5 Thus, f = = =2 N 3 3 5. (6) : P and fR =
2m Q
θ
mgsin
...(i) ...(ii) ...(iii)
dT σeA 4 =− (T − T04 ) . Here e is same, m is dt ms same, both are of copper and both are heated upto same temperature and kept under same conditions, so (T 4 – T 40 ) is same initially. But (T 4 – T40 ) will not remain same for both every time. dT ∴ ∝ −A dt dT dt cube = Acube Asphere dT dt sphere
7. (6) :
C R 0.5 m
3M 4 For sphere, r × πR3 = M ; R = 3 4 πr
B
∴
Here, tanq = a/g = 8/10 = 4/5 ⇒ RB = PQ – QB tanq = 2 – 2 × 4/5 = 0.4 m PQ + BR 2 Volume of water contained, Vf = × 4 m 2 2
Initial volume of water, Vi = BD × 4 m Amount of water spilled = Vi – Vf = (4BD – 2PQ – 2BR) m3 = 2(2BD – PQ – BR) × 1000 litre = 2(3 – 2 – 0.4) × 1000 = 1200 litre = 200a litre ∴ a = 6 6. (8) : If in any time dt, the temperature of disc is increased by dT. Then amount of heat supply will be (ms dT) in time dt, so rate of heat supply will dT be ms and it will be equal to the rate of heat dt dT KA(T0 − T ) = supply by conduction. ∴ ms dt L where T is the temperature of disc at any time t, T0 = 400 K 350 dT t KA ∴ ∫ =∫ dt 300 T − T 0 msL 0 350 KAt = − ln(T0 − T ) 300 msL KAt = – [ln(400 – 350) – ln(400 – 300)] msL 1 KAt msL 50 KAt = −ln or t = = −ln ln 2, or msL 2 msL KA 100 0.4 × 600 × 0.4 t= ln 2 = 240 ln 2 s = 8 × 30 ln 2 s 10 × 0.04 ∴ n=8
1/3
,
2 /3
3M Asphere = 4 πR2 = 4 π 4 πr M rL3 = M; L = r
For cube,
M Acube = 6L2 = 6 r
1/3
,
2/3
2/3
M dT 6× 1/3 dt Acube 6 cube r = = = 2/3 Asphere dT π 3M dt 4π sphere 4 πr
∴ p=6 8. (3) : As DQ = DU + DW For isothermal process, DU1 = 0 DW1 = nRT0 ln(Vf /Vi), where T0 = 300 K ∴ DQ1 = nRT0 ln 5 J (Q Vf = 5Vi) For isochoric process, DW2 = 0 DU2 = nCvDT ⇒ DQ2 = nCvDT, DT = T – T0, PV P × 5V = ⇒ T = 1500 K T0 T
⇒ DQ2 = n
R × 1200 J γ −1
Now, DQ1 + DQ2 = 83.14 × 103 J On solving, we get n = 3 9. (1) : The frequency of the fundamental note emitted by each wire before the tension change occurs is u= ∴
1 T 2L µ
1/2
...(i)
−1/2 du 1 1 T 1 = dT 2L 2 µ µ
Physics for you | SEPTEMBER ‘15
63
du 1 T = dT 4 L µ
−1/2
12. (5) : Frequency of first harmonic in AB = Frequency of second harmonic in CD
1 µ 1/2
2 or du = 1 T µ 2
dT
4 LT T µ du u ∴ = dT 2T u DT Hence, Du ≈ 2 T
1 T = 4 LT µ
1/2
(Using (i))
where Du is the frequency difference induced in the string as a result of change in tension DT. In other words, Du is the number of beats observed if the string’s tension is changed by an amount DT. Using the given data, 200 1 Du = = 1 Hz 2 100 a b − x12 x 6 dU 12a 6b ∴ F=− = − dx x13 x 7
10. (4) : U =
−
2
b b2 =− 2a 4a b
60
0
or
5 = 1 or n = 5 n
∴ a = 0.4. Hence 10a = 4 14. (2) : Total energy of the satellite
60
0
3
2
60
0
1
1000 3600 = 1 rads −1 1200 6
720 ×
2
1 ∴ a3 = ω R3 = × (1800) = 50 m s–2 6
As 10a = 50 ∴ a=5 64
4 1 =1− n n
0.1 + 1.2 × 0.05 Du = × 100 = 0.16% = a2 % u × 100 . 94 + 1 2 × 5 max
11. (5) : Angular velocity of all aircrafts will be same.
2
L L = T2(L – ) n n 4mg mg 1 = 1− or 5n 5 n
T1
For maximum % error
b2 b2 = 0−− = 4a 4a
v2 = R2
For rotational equilibrium about O, Torque due to T1 about O = Torque due to T2 about O
v Du Dl + 1.2Dr × 100 = − × 100 u l + 1.2r
∴ D = U(x = ∞) – Uequilibrium
ω=
mg 4mg and T2 = 5 5
Here Dv = 0 (given)
1/6
2a b
... (ii)
v where e = end correction = 0.6r 2(l + 2e) v v ∴ u= = 2(l + 2 × 0.6r ) 2(l + 1.2r ) Du Dv D(l + 1.2r ) Dv Dl + 1.2 Dr = − ∴ = − u v l + 1.2r v l + 1.2r
12a 6b − =0 x13 x 7 2a ∴ x6 = b
a
T1 =
...(i)
13. (4) : u =
F=
U equilibrium =
1 T1 1 T2 = 2l µ l µ
or T1 = 4T2 For translational equilibrium, T1 + T2 = mg From (i) and (ii), we get
or
At equilibrium,
⇒ x = 2a b U(x = ∞) = 0
∴
Physics for you | SEPTEMBER ‘15
= PE + KE = −4 MJ + | PE | = −2 MJ 2
Energy required = 2 MJ 15. (9) : Area of all the six faces, A =(2 × 120 × 120 + 4 × 200 × 120) = 124800 cm2 x = 2 cm, T1 – T2 = 30 – 0 = 30°C, K = 0.0004 cal s–1 cm–1 °C–1 Q = mL =
KA(T1 − T2 )t x
Rate of melting of ice,
m KA(T1 − T2 ) 0.0004 × 124800 × 30 = 9.36 g s–1. = = t xL 2 × 80 ∴ b = 9.36 ≈ 9
Series 4 CHAPTERWISE PRACTICE PAPER : Ray Optics | Wave Optics 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. Which is more readily seen at a distance, a white 2. 3.
4.
5.
blotting paper or a polished mirror? Why? Will a star twinkle if seen from space (say moon or a spacecraft)? When an unpolarised light is incident on a plane glass surface, what should be the angle of incidence so that the reflected and refracted waves are perpendicular to each other? What focal length should the reading spectacles have for a person for whom the least distance of distinct vision is 50 cm? Calculate the distance that a beam of light of wavelength 500 nm can travel without significant broadening , if the diffracting aperture is 3 mm wide. section-b
6. How is a wavefront different from a ray? Draw the
geometrical shape of the wavefronts when (i) light diverges from a point source, and (ii) light emerges out of convex lens when a point source is placed at its focus. 7. How will the angular separation and visibility of fringes in Young’s double slit experiment change
when (i) screen is moved away from the plane of the slits, and (ii) width of the source slit is increased? 8. Show that a light ray will pass through an equilateral glass prism (m = 1.5) if the angle of incidence is greater than 30°. 9. Explain why does a convex lens behave as a converging lens when immersed in water (m = 1.33) and as a diverging lens when immersed in carbon disulphide (m = 1.63). The refractive index of lens material is 1.52. 10. A fish, looking up through the water sees the outside world contained in a circular horizon. If 4 the refractive index of water is and the fish is 3 30 cm below the surface, then calculate the radius of the circle. OR Show that a convex lens produces an N times magnified image when the object distances from f . Here, f is the the lens have magnitudes f ± N magnitude of the focal length of the lens. Hence, find the two values of object distance for which a convex lens of power 2.5 D will produce an image that is 4 times as large as the object? physics for you | september ‘15
65
section-c 11. A slide of 35 mm × 23 mm is placed at a distance
of 21 cm from a convex lens of focal length 20 cm. Where should a screen be placed on the other side of the lens so as to form a sharp image of the slide? What are the dimensions of the image formed on the screen? Also, find the areal magnification. 12. The radius of curvature of the convex surface of a planoconvex lens is 12 cm and its refractive index is 1.5. (a) Find the focal length of planoconvex lens. (b) The plane surface of planoconvex lens is silvered so that it begins to behave as a concave mirror. Find the focal length of the mirror so formed. 13. A parallel beam falls on a solid glass sphere at normal incidence. Prove that the distance of the image from the outer edge in terms of refractive R(2 − m) . index m and radius R of sphere is 2(m − 1)
distance is measured to be 30.0 cm. What is the refractive index of the liquid? 17. Discuss the action of a prism, a convex lens and a concave mirror, when a plane wavefront is incident on each of them. 18. Three polarisers are arranged as shown in figure. Polariser Io
14. An astronomical telescope uses two lenses of powers
10 D and 1 D. (i) State with reason, which lens is preferred as objective and which as eye-piece. (ii) Calculate the magnifying power of the telescope, if the final image is formed at the near point. (iii) How do the light gathering power and resolving power of a telescope change, if the aperture of the objective lens is doubled? 15. A person looking at another person wearing a shirt with a pattern comprising vertical and horizontal lines is able to see the vertical lines more distinctly than the horizontal ones. What is this defect due to? How is such a defect of vision corrected? 16. White light reflected at perpendicular incidence from a soap film has, in the visible spectrum, an interference maximum at 6000 Å and a minimum at 4500 Å with no minimum in between. If m = 4/3 for the film, what is the thickness of the film? OR Figure shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0 cm. The liquid is removed and the experiment is repeated. The new 66
physics for you | september ‘15
19.
20. 21.
22.
Second polariser
Analyser
(90°-)
(a) What is the intensity of the transmitted light through analyser in terms of the incident light (I0) and the angle (q) between the first and the second polariser ? (b) What happens if the second polariser is rotated ? (c) When is the light not transmitted through the system? (d) For what value of q will the transmitted intensity be maximum? Light of wavelength 589 nm is used to view an object under a microscope. The aperture of the objective has a diameter of 0.900 cm. (a) Find the limiting angle of resolution. (b) Using visible light, what is the maximum limit of resolution for this microscope? (c) Suppose water (m = 1.33) fills the space between the object and objective. What effect would this have on the resolving power? “A diffraction grating can as well be called an interference grating”. Discuss. A lens is held directly above a coin lying on a table and forms an image of it. After the lens has been moved vertically downwards a distance equal to its focal length, it forms another image of the coin equal in size to previous image. If the diameter of the coin is 4.0 cm, what is the diameter of the coin’s image? A thin equiconvex lens of glass of refractive index mg = 3/2 and of focal length 0.3 m in air is sealed into an opening at one end of a tank filled with water, mw = 4/3. On the opposite of the lens a mirror is placed inside the tank, on the tank wall perpendicular to the lens axis as shown in figure. The separation between the lens and the mirror is 0.8 m. A small object is placed outside the tank infront of the lens at a distance 0.9 m from the lens along its axis. Find the position (relative to the lens) of the image of the object formed by the system.
0.9 m =1
0.8 m w = 4/3
O g = 3/2 Lens
Water
section-D 23. Mr. Lohiya, a retired professor of physics was
walking with his grandson. It was last week of December and so it was dark around 5.30 pm. The streetlights were on and the yellow light flooded the area around. The boy asked professor why yellow lights were used when white light were brighter. The professor answered that during foggy days the tiny droplets act as prisms splitting white light into its constituent colours and thus reducing the clarity. (i) What phenomena was the professor referring to? Why does it happen? (ii) Give one application of prism. (iii) What values of the boy reflect from the conversation? section-e
24. Explain the phenomenon of total internal reflection.
State two conditions that must be satisfied for total internal reflection to take place. Derive the relation between the critical angle and the refractive index of the medium. Draw ray diagrams to show how a right angled isosceles prism can be used to (i) deviate ray through 180°, and (ii) to invert it.
OR A spherical surface of radius of curvature R and of refractive index m2 is placed in a medium of refractive index m1 where m1 < m2. The surface produces a real image of an object kept infront of it. Using appropriate assumptions and sign conventions, derive a relationship between the object distance, image distance, R, m1 and m2. Under what conditions this surface diverge a ray incident on it? 25. (a) Using Huygens’ principle, draw a diagram to show propagation of a wavefront originating from a monochromatic point source and explain briefly. (b) Derive Snell’s law of refraction using Huygens’ principle. OR (a) What are coherent sources of light? State two conditions for two light sources to be coherent.
(b) Derive a mathematical expression for the width of interference fringes obtained in Young’s double slit experiment with the help of a suitable diagram. 26. (a) What is plane polarised light? The polaroids are placed at 90° to each other and the transmitted intensity is zero. What happens when one more polaroid is placed between these two, bisecting the angle between them? How will the intensity of transmitted light vary on further rotating the third polaroid? (b) If a light beam shows no intensity variation when transmitted through a polaroid which is rotated, does it mean that the light is unpolarised? Explain briefly. OR What is meant by diffraction? Draw a graph to show the relative intensity distribution for a single slit diffraction pattern. Obtain an expression for the diffraction of the first minimum and first maximum in the diffraction pattern. solutions 1. A white blotting paper is more readily visible from a
distance than a polished mirror because the surface of white blotting paper provides irregular reflection and hence light spreads in all possible directions and one can see the paper from a distance. A polished mirror gives rise to only regular reflection and can be seen only in the direction of reflected rays. Hence, it cannot be seen from a distance easily. 2. No, a star will not appear twinkling if seen from free space because there will be no atmospheric refraction. The star will appear to be shining continuously. 3. The unpolarised light is incident on a plane glass surface at an angle i such that tan i = m, where m is refractive index of glass. In that case, the reflected and refracted waves are perpendicular to each other and the reflected light is completely plane polarised. 4. The reading matter placed at 25 cm from the correct
lens should produce the virtual image at 50 cm. Therefore, u = – 25 cm, v = –50 cm By thin lens formula, 1 1 1 1 1 = − = − v u −50 −25 f 1 −1 + 2 = = 50 50 or f = + 50 cm physics for you | september ‘15
67
The positive sign shows that the corrective lens must be a convex lens of focal length 50 cm. 5. Here d = 3 mm = 3 × 10–3 m, l = 500 nm = 500 × 10–9 m The distance upto which a beam of light can travel without significant broadenning is called Fresnel distance and its value is given by (3 × 10 −3 )2 d2 = = 18 m. l 500 × 10 −9 6. A wavefront is a surface obtained by joining all points vibrating in the same phase. A ray is line drawn perpendicular to the wavefront in the direction of propagation of light wave. The wavefront of light emerging from a point source are spherical as shown in figure (a). When a point source is placed at the focus of a convex lens, the emerging light has plane wavefronts, as shown in figure (b). DF =
Ray S S
For water (mm = 1.33), (mg – mm) is positive and hence, fm is positive. It means that the lens behaves as a converging lens. However, for carbon disulphide (mm = 1.63), (mg – mm) is negative and hence, fm is negative. It means that the lens behaves as a diverging lens. 4 10. As per question, mw = and h = 30 cm. 3
Dl d
l b = d D (i) When the screen is moved away from the slits, the distance D increases. Fringe width b increases but angular separation q remains unchanged. (ii) The interference pattern becomes less and less sharp. When the source slit becomes so l s wide that the condition < (Here, s is the d S width of the source slit and S its distance from the plane of the two slits) is not satisfied, the interference pattern disappears. But the angular width q remains unchanged. sin i 8. From Snell’s law, m = . sin r If the light is incident at 30°, then physics for you | september ‘15
90° ic
h
Figure (b)
Angular separation, q =
68
glass-air interface. If i < 30°, then i′ > ic and the ray will suffer total internal reflection at the second face and will not emerge through the prism. 9. For a convex lens (mg = 1.52), focal length f has positive sign. When it is immersed in a medium of refractive index mm, its focal length changes to fm, where (m g − 1) ⋅ m m ⋅f fm = (m g − m m )
r
Figure (a)
7. Fringe width, b =
0. 5 sini sin 30° = = = 0.33 1 .5 m 1. 5 –1 or r = sin (0.33) ≈ 19° Angle of incidence at second face, i′ = A – r = 60° – 19° = 41° which is just less than the 1 1 critical angle 42° = sin −1 = sin −1 for m 1. 5
sinr =
ic
As shown in figure the fish can see the outside world contained in a circular cone of semi-vertical angle 1 1 ic = sin–1 or sinic = mw mw r But sinic = 2 r + h2 1 r \ = 2 2 m w r +h ⇒
r=
\
r=
h m 2w − 1 30 cm 2
4 3 −1
=
30 cm
OR
7 9
= 34.0 cm
Linear magnification produced by a lens is given by
v f = u u+ f Here, m = ±N (the positive sign is for virtual image and negative sign for real image). f \ ±N = u+ f m=
f N f f ⇒ u = –f ± or |u| = f ± N N Thus, the desired result is proved. As power, P = 2.5 D, 1 1 = m = 0.4 m = 40 cm and N = 4 \ P 2. 5 \ Two values of the object distance, f 40 u=f± = 40 ± = 50 cm and 30 cm N 4 u+f=±
11. Here, f = + 20 cm, u = – 21 cm and size of the slide,
h1 × h2 = 35 mm × 23 mm 1 1 1 From the lens formula, − = , we have v u f
1 1 20 − 21 1 1 1 1 = + = = = + u f 420 (−21) (+20) (−21 × 20) v ⇒ v = 420 cm or 4.2 m Let the dimension of the image formed on the screen be h1′ × h2′. h′ v Then, from the relation, m = = , we have h u 420 cm v h′1 = × h1 = × 35 mm (−21 cm) u = –700 mm = – 70.0 cm 420 cm v and h′2 = × h2 = × 23 mm (−21 cm) u = –460 mm = – 46.0 cm Thus, the dimensions of image formed on the screen is 70 cm × 46 cm. The negative sign signifies that the image is inverted. h′h′ Areal magnification, m′ = 1 2 h1h2 (−70 cm) × (−46 cm) = = 400 (35 mm) × (23 mm) 12. (a) For the plano-convex lens, we have
R1 = +12 cm, R2 = ∞ and m = 1.5 Hence, its focal length f is given by 1 = (m – 1) 1 − 1 f R1 R2
1 1 0. 5 1 = (1.5 – 1) = − = 24 ( + 12 ) ∞ 12 ⇒ f = +24 cm (b) If plane surface of the lens is silvered, then a light ray after passing through the lens will be incident on plane polished 12 cm surface (behaving as a plane mirror), which will reflect it back through the lens again. Hence, power P of the entire arrangement is given by P = PL + PM + PL where PL is the power of the lens and PM is the power of the plane mirror. 1 1 1 1 Now PL = = = = D 24 cm fL 0.24 m 0.24 1 1 and PM = – =– =0 ∞ fM 1 1 2 1 +0+ \ P= = = D 0.24 0.24 0.24 0.12 As the arrangement behaves as a concave mirror of focal length F, 1 Hence, P = – F 1 1 ⇒ = – or F = – 0.12 m = –12 cm F 0.12 13. Figure shows the refraction of a parallel beam of
light through a glass sphere of refractive index m at normal incidence. For refraction at first surface from air to glass, m1 = 1, m2 = m, u = ∞, R = +R Air, = 1 S Glass A
C
Air, = 1 B
I
v
I u
v
m − m1 m m2 – 1 = 2 R u v mR 1 m m −1 \ – = or v = m −1 ∞ v R This image formed at I acts as virtual object for second surface. For refraction at second surface from glass to air, u′ = BI = AI – AB = v – 2R mR mR − 2R(m − 1) 2 − m = – 2R = = R m − 1 m −1 m −1 As
physics for you | september ‘15
69
Now m1 = m, m2 = 1, R = – R m 1− m 1 – = \ −R v ′ u′ m −1 m (m − 1) 1 or – = R R (2 − m) v′ 1 m −1 m (m − 1) or = + v′ R R (2 − m) 2(m − 1) (m − 1)(2 − m + m) = = R(2 − m) R(2 − m) R(2 − m) \ v′ = 2(m − 1) 14. (i)The lens of power 1 D should be used as objective
because of its larger focal length and the lens of 10 D should be used as eyepiece because of its smaller focal length. 1 (ii) Here, f0 = = 1 m = 100 cm, 1D 1 fe = = 0.1 m = 10 cm 10 D
f0 f 1+ e = 100 1 + 10 = 14. fe D 10 25 (iii) Light gathered ∝ Area of the objective of a telescope πD 2 i.e., Q∝ or Q ∝ D2 4 When aperture (D) is doubled, light gathering capacity increases 4 times. Resolving power of a telescope ∝ D When aperture (D) is doubled, resolving power also gets doubled. 15. The given defect is called astigmatism. It arises because the curvature of the cornea plus eye lens refracting system is not the same in different planes. The eye lens is usually spherical, i.e., has the same curvature in different planes but the cornea is not spherical in case of an astigmatic eye. In the present case, the curvature in the vertical plane is enough, so sharp images of vertical lines can be formed on the retina. But the curvature is insufficient in the horizontal plane, so horizontal lines appear blurred. This defect can be corrected by using a cylindrical lens with its axis along the vertical. Clearly, parallel rays in the vertical plane will suffer no extra refraction, but those in the horizontal plane can get the required extra convergence due to refraction by the curved surface of the cylindrical lens if the curvature of the cylindrical surface is chosen appropriately. \
70
|m| =
physics for you | september ‘15
16. Here l1 = 6000 Å = 6 × 10–7 m
4 3 For normal incidence, the condition for (n + 1)th maximum is 2mt 1 1 2mt = n + l1 or n + = …(i) l1 2 2 The condition for (n + 1)th minimum is 2mt 2mt = (n + 1)l2 or n + 1 = …(ii) l2 Subtracting (i) from (ii), we get l − l2 1 = 2mt 1 − 1 = 2mt 1 l 2 l 1l 2 2 l1
l2 = 4500 Å = 4.5 × 10–7 m, m =
\
t=
6 × 10 −7 × 4.5 × 10 −7 l 1l 2 = 4 4m(l1 − l 2 ) 4 × × (6 − 4.5) × 10 −7 3
27 × 10–7 m = 3.375 × 10–7 m. 8 OR Distance of the needle from the lens in the first case = focal length F of the combination of the convex lens and plano-concave lens formed by the liquid i.e., F = 45 cm and in the second case = focal length of the convex lens i.e., f1 = + 30 cm The focal length f2 of the plano-concave lens is given by 1 1 1 = + F f1 f 2 =
or
1 1 2−3 1 1 1 1 = − = = =– − F f1 90 90 f2 45 30
\ f2 = – 90 cm Now for the equiconvex lens, we have R1 = R, R2 = – R, f1 = 30 cm, m = 1.5 Using Lens maker’s formula 1 1 1 = (m – 1) − f R1 R2 1 1 1 2 = (1.5 – 1) + = 0.5 × R R 30 R or R = 0.5 × 2 × 30 cm = 30 cm For plano-convex lens, f2 = – 90 cm For concave surface, R1 = – R = – 30 cm, For plane surface, R2 = ∞ 1 As = (m′ – 1) 1 − 1 R R f2 1 2 \
1 1 1 = (m′ – 1) − −90 −30 ∞ 1 −30 or m′ – 1 = =+ 3 −90 1 or m′ = 1 + = 1.33. 3 17. Behaviour of prism : Since the speed of light in glass is smaller than that in air, therefore, the lower part C of the plane wavefront which travels through the greatest thickness of the glass prism is slowed down the most and the upper part A, which travels through the minimum thickness of the glass prism, is slowed down the least. This explains the tilting of a plane wavefront after refraction through a glass prism. \
Incident wavefront A B
Refracted wavefront A
C
B
Behaviour of a convex lens : The central part B of the plane wavefront travels through the greatest thickness of the lens and is, therefore, slowed down the most. The marginal parts A and C of the wavefront travel through a minimum thickness of the lens and are, therefore, slowed down the least. So the emerging wavefront is spherical and converges to a focus F. Refracted wavefront A B C
B C
F
Behaviour of concave mirror : The central part B of the
incident wavefront has to travel the greatest distance before getting reflected compared to the marginal parts A and C. Therefore, the central portion B of the reflected wavefront is closer to the mirror than the marginal portions A′ and C′. Hence the reflected wavefront is spherical and converges to a focus. Incident wavefront A B C
Refracted wavefront F
A B C
polariser, I1 = I0/2. Intensity of transmitted light through second polariser I2 = I1cos2q = (I0/2)cos2q Intensity of transmitted light through analyser, i.e., I3 = I2cos2(90° – q) = [(I0/2)cos2q]cos2(90° – q) = (I0/2)cos2qsin2q = (I0/8)sin22q (b) Since I3 depends on q, rotating the second polariser will change the light intensity. (c) If q = 0° and 90°, then I3 = 0 (as sin 0° = cos90° = 0) Thus, for q equal to 0° and 90°, no light will be transmitted through the system. (d) I3 is maximum when q = 45°.
19. (a) Here, l = 589 nm = 589 × 10–9 m
C
Incident wavefront A
18. (a) Intensity of transmitted light through the first
a = 0.900 cm = 0.900 × 10–2 m 589 × 10 −9 m l Dq = 1.22 = 1.22 rad 0.900 × 10 −2 m a = 7.98 × 10–5 rad (b) To obtain the smallest angle corresponding to the maximum limit of resolution, we have to use the shortest wavelength in the visible spectrum. Violet light (l = 400 nm) gives us a limiting angle of resolution of 400 × 10 −9 m Dq = 1.22 rad 0.900 × 10 −2 m = 5.42 × 10–5 rad (c) Wavelength of light in water, 589 nm l lw = a = = 443 nm m 1.33 443 × 10 −9 m Dq = 1.22 rad 0.900 × 10 −2 m
= 6.00 × 10–5 rad Since Dq in this case is less than that in case (a), resolving power increases. 20. A diffraction grating is an arrangement consisting of a large number of parallel slits of the same width separated by equal opaque spaces. It consists of a well-polished metal or glass surface upon which equidistant, fine and parallel lines are ruled by a diamond point. The number of lines may vary from 15,000 to 40,000 per inch. The problem of studying the diffraction of a grating is essentially the problem of finding the resultant due to interference of a large number of secondary distrubances issuing from physics for you | september ‘15
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the various transparent slits separated by opaque equidistant fine diamond rulings. Thus, we can call a diffraction grating as an interference grating. 21. Case I : Object distance = –u, focal length = f and image distance = v 1 1 1 = − By lens formula, f v u 1 1 u− f 1 1 = + or = v f uf v u So, m1 =
\
v=
uf u− f
f v = u− f u
Case II : If lens has been moved by f vertically down, then object distance, u′ = –(u – f), f 1 1 1 = + \ v ′ (u − f ) f or
1 u− f − f = v′ f (u − f )
⇒
f (u − f ) v′ = (u − 2 f )
So,
m2 =
As \
v′ f = (u − f ) (u − 2 f )
m1 = –m2 f f =– (u − 2 f ) (u − f ) u – 22ff = –u + f
3 2u = 33ff or u = f 2 f \ m1 = =2 3 f−f 2 Hence, diameter of coin’s image, D′ = 2 × 4 = 8 cm 22. Let us trace the path of light from the object. First it is refracted at surface S1 (air to glass), next refraction takes place at surface S2 (glass to water), then reflection at the mirror. It is reflected back and refracted through surfaces S2 and S1 once again in the order of their appearance. The given parameters are u = – 0.9 m, m = 1, mg = 3/2, mw = 4/3 Using lens maker’s formula, mg − m 1 1 1 = − m R R f 1 2 1 1. 5 − 1 2 = or R = 0.3 m 1 R 0. 3 72
physics for you | september ‘15
As \ or
m − m mw − m g mw m − = g + v u R R (3 / 2) − 1 (4 / 3) − (3 / 2) 4/3 1 = + − 0. 3 (− 0.3) v −0.9 v = 1.2 m
The image formed is real, to the right of the lens and behind the mirror. It is a virtual object at a distance 0.4 m for the mirror. Now the mirror forms an image in front of it. This image is an object for the lens system. m − mw m − m g m mw = g − + v ′ u′ R R (3 / 2) − (4 / 3) 1 − (3 / 2) 1 4/3 = + − 0. 3 (− 0.3) v ′ (−0.4) \
v′ = – 0.9 m = – 90 cm
23. (i) Dispersion, which happens as speed of each
colour is different when they enter glass. (ii) Studying and analysing the spectrum of distant light sources. (iii) Curiosity, research mindedness, awareness. 24. Refer point 6.4 page no. 371 (MTG Excel in Physics). OR Refer point 6.5(5((i), (ii))) page no. 372 (MTG Excel in Physics). 25. Refer point 6.11(1, 5) page no. 444 (MTG Excel in Physics). OR Refer point 6.13(7, 9) page no. 448 (MTG Excel in Physics). 26. Refer point 6.15 page no. 453 (MTG Excel in Physics). OR Refer point 6.14 page no. 449 (MTG Excel in Physics). nn Learning gives creativity, creativity leads to thinking, thinking provides knowledge, knowledge makes you great -A P J Abdul Kalam
To be able to apply this Work Energy Theorem (WET) we need to understand first of all, what actually work stands for! Let us begin by taking two small examples of motion – 1. An object moves under the action of a force acting along the direction of motion. The path obviously is a straight line.
For object A the speed (hence KE) increases whereas for object B the speed decreases. 2. An object moves under the action of a force of constant magnitude but always perpendicular to path, as in uniform circular motion (UCM).
⇒ Fdr cosq = m vdv ⇒ F ⋅ dr = m vdv Integrating both sides we get, r2 v ∫ F ⋅ dr = m ∫ vdv r1
u
1 1 ...(i) = mv 2 − mu2 = change in KE = DKE 2 2 Hence the quantity on left, i.e., the line integral of the dot product of force with instantaneous displacement is said to be the work done by force. \ Work done by any force (F ) in moving a mass from r1 to r2 is, r2
WF = ∫ F ⋅ dr r1
Such a force which is incapable of changing the KE of the object is always perpendicular to the direction of motion. Now, whenever we talk of work being done by a force, we check whether the force is capable of changing the kinetic energy of the mass (or system of masses) which can happen only if the applied force has a component which is parallel to the direction of motion. Now with this understanding, let us see how do we define work. Consider an object moving in a curved path and at an instant, the net force acting on it is F, when the velocity is v.
The tangential component of F, i.e., Ft = Fcosq is responsible for changing the KE. \ Ft = mat (where at = tangential acceleration) dv dv ⇒ F cos q = m = mv dt dr
Coming back to the equation derived, F was the vector sum of all forces acting on the object. Hence equation (i) becomes work done by all forces = change in kinetic energy ⇒ Wall forces = DKE This equation is work energy theorem. Before we proceed further with this theorem, let us see further examples of work done by forces in various situations. Work done by a constant force A force is constant only if its magnitude as well as direction remains constant which is possible only if all the three components of force are each individually constants. ^ \ If F = Fx i^ + Fy ^j + Fz k and, ^ dr = dx i^ + dy ^j + dzk , then, dWF = F ⋅ dr = Fx dx + Fy dy + Fz dz \ WF = ∫ dWF = WF =
r2 = x2 ^ i + y2 j^ + z2^ k
F ⋅ dr
∫
r1 = x1^ i + y1 j^ + z1k^
x2
y2
z2
x1
y1
z1
∫ Fxdx + ∫ Fydy + ∫ Fzdz
Contributed By: Bishwajit Barnwal, Aakash Institute, Kolkata
physics for you | SEPTEMBER ‘15
73
But since, Fx, Fy and Fz are constants, they can be taken out of integration. \ WF = Fx
x2
y2
x1
y1
∫ dx + Fy
∫
z2
dy + Fz ∫ dz z1
WF = FxDx + FyDy + FzDz Apart from memorizing the result, let us try to understand the above expression. If we treat the components of force Fx, Fy and Fz as three independent forces, then work done by Fx is Fx Dx which is force multiplied with displacement in the direction of the force or displacement multiplied with force in the direction of displacement. Let us try to apply this for a physical situation where an object is moved from A to B in a curved path and one constant force F acts on the object at all instants at an angle q with the line AB
Before we start solving questions, we need to take a look at the classification of forces once.
Conservative ForCes • These are those forces for which the work done is independent of path and depends only on initial and final positions.
Hence for all the 3 paths, if its a conservative force, W1 = W2 = W3 • The work done by these forces in a closed path is always necessarily zero. We always associate a scalar quantity known as potential energy (PE) with these forces and the work done by conservative forces is the negative of the change in PE. WC = – DPE 74
as shown.
physics for you | SEPTEMBER ‘15
\ WF = (Fcosq) l or F(lcosq) Now, to be able to apply WET we follow these basic steps.
PE can be seen as a stored form of energy which depends on the relative configuration of the system. Example: Gravitational force (mg), spring force, electrostatic force etc. Gravitational Potential energy
If an object of mass m is at a height h (> 1 L2 1 2 5. (c) : I AB = MR 2 ICD = ? Applying the theorem of parallel axes, I CD = IAB + MR 2 1 3 = MR2 + MR2 = MR2 2 2 \
A
C
R B
rA
D
4 3 4 πRA = rB πR3B 3 3 1/3
RA r B = RB rA 82
Physics for you | SEPTEMBER ‘15
2
(... MA = MB)
2 /3
(Using(i))
7. (b) : According to the theorem of parallel axes, moment of inertia of disc about an axis passing through K and perpendicular to plane of disc, is 1 3 = MR2 + MR2 = MR2 2 2 R O
R
K
Total moment of inertia of the system 3 = MR2 + m(2R)2 + m( 2R)2 + m( 2R)2 2 R2 = (3M + 16m) 2 8. (b) : w = a – bt At time t = 0, w = w0 = a dw a= =−b dt Using w2 = w20 + 2aq
O
6. (a) : Moment of inertia of a solid sphere of mass M and radius R about its diameter is 2 I = MR2 5 2 2 2 IA 5 M ARA M ARA \ = = I B 2 M R2 M B RB2 5 B B Mass(M ) Density, r = Volume(V ) 4 3 M = rV = r πR 3 As per question, MA = MB \
IA RA = IB RB
... (i)
0 = w20 + 2aq w 2 a2 q=− 0 = 2a 2b 9. (d) : Acceleration of a body rolling down a fixed inclined plane of inclination q is or
a=
g sin q
…(i) k2 1+ 2 R As cylinder P has most of its mass concentrated near its surface, while cylinder Q has most of its mass concentrated near the axis, therefore, radius of gyration, kP > kQ From (i), aP < aQ Therefore, at bottom of incline, vP < vQ or (wPR) < (wQR) or wP < wQ or wQ > wP i.e., cylinder Q reaches the ground with larger angular speed.
10. (c) : Rod POQ of length l = 1 m = 100 cm is bent at its midpoint O so that ∠POQ = 90° (see figure). The mass of part PO of length l/2 can be taken to be concentrated at its midpoint A whose coordinates are (0, l/4) and of part OQ of length l/2 at its midpoint B whose coordinates are (l/4, 0). The centre of mass of these two equal masses is at midpoint C between A and B. The coordinates of C are (l/8, l/8). 2
2
l l OC = (OE)2 + (CE)2 = + 8 8 100 cm l = = = 17.7 cm 32 32 11. (b) : Angular momentum, L = constant, when τ=r ×F =0 i.e. (2i − 6j − 12k ) × (ai + 3j + 6k ) = 0 \
or 12. (d) :
i j k 2 −6 −12 = 0, which gives a = –1 a 3 6 AB =2 BC
\
AB = DC =
l 3
l 6 m Similarly, mAB = mDC = 3 m and mBC = mAD = 6 \ Moment of inertia of the wire frame about the given axis is I = IAB + IAD + IDC + IBC and
BC = AD =
2
2
2
1 m l m l 1 m l = + + +0 6 3 3 3 3 3 3 3 7 ml 2 ml 2 ml 2 + + = ml 2 81 54 81 162 13. (c) : Let M be the mass of circular plate with centre O. M Mass per unit area = π(28)2 \ Mass of circular portion removed, with centre O1 9 M × π (21)2 = M M1 = 2 16 π(28) =
9 7 M= M 16 16 Let O2 be its centre of mass, where OO2 = x. Now, M1 × OO1 = M2 × OO2 9 7 M ×7 = M ×x 16 16 Mass left, M2 = M – M1 = M −
\
x = 9 cm
14. (b) : For a particle moving in a circle with constant angular speed, velocity vector is always tangent to the circle and the acceleration vector always points towards the centre of the circle or is always along radius of the circle. Since, tangential vector is perpendicular to radial vector therefore velocity vector will be perpendicular to the acceleration vector. But in no case acceleration vector is tangent to the circle. 15. (a) : w = w0 + at or w = 0 + at 15 w rad s −2 = t 0.270 15 \ a = ra = 0.81 × = 45 m s −2 0.270
or a =
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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.
Q1. When water flows past two adjacent boats facing upstream, why do the boats tend to pull together? – Aditya Kashyap (Haryana)
Ans. When the water is forced into the confining space between the boats, it increases speed of the water waves. The only way it can get the energy required for the increased speed is to take it from the internal energy associated with pressure. As a result, the water pressure between the boats decreases. With normal pressure on the exterior sides of the boats and this decreased pressure on the interior sides, the boats are drawn together. Q2. If you inspect a light bulb that has been operating for a long time, a dark region appears on the inner surface of the bulb. What is the origin of this dark region? – Kirti Gupta (U.P.)
Ans. The dark region is tungsten that vaporized from the filament of the lightbulb and collected on the inner surface of the glass. Many lightbulbs contain a gas that allows convection to occur within the bulb. The gas near the filament is at a very high temperature, causing it to expand and float upward, due to Archimedes principle. As it floats upward, it carries the vaporized tungsten with it, so that the tungsten collects on the surface at the top of the lightbulb. Q3. Suppose a point charge + Q is in empty space. We surround the charge with a spherical conducting shell so that the charge is at the centre of the shell. What effect does this have on the field lines from the charge? – Shan Doshi (WB) Ans. When the spherical shell is placed around the charge, the charges in the shell adjust so as to satisfy the rules for a conductor in equilibrium and Gauss’s law. A net charge of – Q moves to the interior surface of the conductor, so that the electric field within the conductor is zero (a spherical gaussian surface totally within the shell encloses no net charge). A net charge of + Q resides on the outer surface, so that a Gaussian surface outside the sphere encloses a net charge of + Q, just as if the shell were not there. Thus, the only change in the field lines from the initial situation is the absence of field lines within the conducting shell. Q4. What makes evaporation different from boiling, in terms of the latent heat of vaporisation? How is the boiling point of water dependent upon atmospheric pressure? – Shreya Singh (Delhi) 84
physics for you | SEPTEMBER ‘15
Ans. It might be useful to take both the questions together. The heat required to convert liquid water into a gas is referred to as latent heat. It represents the energy required to pull molecules of water away from each other, overcoming their mutual attraction. At low temperatures, only a few molecules manage to escape, those whose thermal velocities are at the upper end of the energy distribution of molecules in the liquid. This is what causes evaporation. As the temperature is raised, more and more molecules acquire energies sufficient to escape from the liquid, and therefore the rate of evaporation increases. When a temperature that is high enough for most of the molecules to escape the liquid is reached, the liquid begins to boil. The temperature, which is a measure of kinetic energy of the molecules, cannot increase any further, because any increase in the energy of the molecules leads to their escape from the system. At this point, heat just serves to overcome the mutual attraction of the molecules. The molecules attempting to escape the force of others in the liquid have to face the pressure of the atmosphere above, including that exerted by their companions that escaped before them. This pressure keeps them close to the liquid surface and many of them return. If, on the other hand, the pressure outside is low, the molecules escape to greater distances. Lowering the pressure enables even those molecules to leave that have not yet attained velocities that would be considered high enough at normal pressure. Thus, at low pressure, the liquid begins to boil at lower temperatures, because molecules that are not very energetic can also overcome the barrier. This is the reason that the boiling point of water decreases at low pressure and, conversely, is raised when the pressure is increased.
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across Cut Here An electrical conductor held at a constant voltage, used to carry high currents between different circuits in a system. (6) 7 4. Tiny solid or liquid particles suspended among the molecules of atmospheric 14 gases. (8) 8. The branch of mechanics concerned with forces that change or produce the motions of bodies. (8) 14. Interstellar clouds. (7) 15. A very strongly magnetized neutron 23 star that emits a uniform series of equally speed electromagnetic pulses. (6) 26 21. A heat engine that runs backward. (12) 23. A device for continuous removal of heat. (9) 25. Disk shaped region of small icy bodies which ranges from about 30 to 100 AU from Sun. (6, 4) 26. A collision in which kinetic energy is not conserved. (9) 27. A measure of internal friction within a fluid. (9) 28. A unit of luminance equal to 1 candela per cm2. (5) 29. Unit of radiation dose. (3) 30. The collective name given to long-lived baryons other than the proton and neutron. (7) 3.
Down
1. 2. 5. 6. 7. 9.
Unit of heat equivalent to a million billion. (4) Smooth dark areas of moon. (5) A copper-tin alloy used for metal, mirrors and reflection diffraction gratings. (8) A succession of waves of limited duration. (4, 5) The peak of a wave disturbance. (5) A mineral widely used for electrical insulation. (4)
1
2 5
4
3 6 8
9
10
11
12
13 15
16
17 18
19
20
21 22
24
25
27 28 29
30
10. A quantum number used in the theory of quarks 11. 12. 13. 16. 17. 18. 19. 20. 22. 24.
and hadrons. (5) A hypothetical particle of time approximately equal to 10–24 s. (7) Acronym for balanced unbalanced. (5) The size of a vector quantity. (9) An inverted U tube with one limb longer than the other. (6) The slow permanent deformation of a crystal or other specimen under sustained stresses. (5) The maximum value of stress that must be applied to a material in order that it shall flow. (5, 5) A term used to describe the way materials responds to forces. (10) An access point in an electronic circuit or network. (4) The prefix for 10–18. (4) A type of aerial used in radio astronomy. (4)
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Physics for you | september ‘15