Vol. XXIII No. 9 September 2015 Corporate Office :
Plot 99, Sector 44 Institutional area, Gurgaon -122 003 (HR). Tel : 0124-4951200 e-mail : [email protected] website : www.mtg.in Regd. Office
406, Taj Apartment, Near Safdarjung Hospital, Ring Road, New Delhi - 110029.
Managing Editor : Mahabir Singh Editor : Anil Ahlawat (BE, MBA)
contents
Physics Musing (Problem Set-26) 8
AIPMT (Re-Exam) 10
Solved Paper 2015
Ace Your Way CBSE XI 25
Series 1
JEE Accelerated Learning Series 31
Brain Map 46
Thought Provoking Problems 58
JEE Workouts 61
Ace Your Way CBSE XII 65
Series 4
Core Concept 73
Physics Musing (Solution Set-25) 78
Exam Prep 80
You Ask We Answer 84
Crossword 85
rial
edit
Send D.D/M.O in favour of MTG Learning Media (P) Ltd.
Payments should be made directly to : MTG Learning Media (P) Ltd, Plot No. 99, Sector 44, Gurgaon - 122003 (Haryana)
We have not appointed any subscription agent.
subscribe online at www.mtg.in
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.
A New star is Born in heaven
T
he name of the new star is Abdul Kalam. Born in the temple town of Rameswaram, he had his higher education in St. Joseph’s College Trichy (Tiruchirapalli). He had taken Physics Hons. as his choice. He, like the other top scientists of India developed the rockets which could carry atomic missiles, fulfilling the dreams of Dr. Bhabha, Vikram Sarabhai and the great scientists like Raja Ramanna and his team.The next step was sending rockets to the moon. Now India is sending simultaneously five satellites for other countries to study various planets simultaneously. Yes, our respected ex-president was more than a great scientist. After his term as president, he had a new dream- igniting the minds of the young to teach them to excel themselves. No film hero had such a following as the star of science.
Our advice to the young is this. Try to study his books when you are grown up. We cannot classify him as an engineer, physicist or astronomer. He was just a great scientist without barriers, inspiring students from the age of five to eighty and more. We pray for the great man who is no more with us physically. May god inspire us to do great things together.
AA lg oh;Za djokogsAA
Anil Ahlawat Editor
individual subscription rates combined subscription rates 1 yr. 2 yrs. 3 yrs. 1 yr. 2 yrs. 3 yrs. Mathematics Today 330 600 775 PCM 900 1500 1900 Chemistry Today 330 600 775 PCB 900 1500 1900 Physics For You 330 600 775 PCMB 1000 1800 2300 Biology Today 330 600 775
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 (a) l l l l l = + 1 2 1 2 (b) l = 2(l1 + l2) (c) l=l1+l2 2 (d) l l l = 1+ 2 4
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
respectively. What force will arise in the plates if the temperature is increased by T°C?(Assume that the plates suffer the same net expansion) (a) 2 2 AY Y T Y Y c s c s s c (a −a ) + (b) AY Y T Y Y c s c s s c (a −a ) + 2 (c) 2AY Y T Y Y c s c s s c (a −a ) + (d) AY Y T Y Y c s c s s c (a −a ) +
3. A disc of radius R has a mass 9m. A hole of radius R
3 is cut from it as shown in figure. The moment of inertia of the remaining part about an axis passing through centre O of the disc and perpendicular to the plane of disc is
(a) 8mR2 (b) 4mR2
(c) 40
9 mR 2 (d) 379 mR2
4. The surface density of a circular disc of radius
a depends on the distance from the center as r(r) = A+ Br. Steel Copper Steel R/3 R O
The moment of inertia about the line perpendicular to the plane of the disc through its centre is
(a) 2 5 6 5 6 π Aa +Ba (b) 2 3 4 3 4 π Aa +Ba (c) 2 4 5 4 5 π Aa +Ba (d) π Aa3 Ba4 3 + 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 that current becomes I
3 is (a) 2
3N (b) N3 (c) N2 (d) N4
nn
Set 26
P
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.
PHYSICS
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) (a) 2hcl (b) hc
3l (c) hc2l (d) hcl
2. The input signal given to a CE amplifier having a voltage gain of 150 is Vi =2cos15t+3p.
The corresponding output signal will be
(a) 2 15 5 6 cos t + p (b) 300 15 4 3 cos t + p (c) 300 15 3 cos t +p (d) 75 15 2 3 cos t + p 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 (a) x=mm2L 1 (b) x m L m m = +2 1 2 (c) x m L m m = +1 1 2 (d) x m m L = 1 2
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 (a) CVd2 (b) C V d 2 2 2 2 (c) C Vd 2 2 2 (d) CVd 2 2
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
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
SOLVED PAPER 2015
AIPMT
AIPMT
Re-Exam held on 25 Julyth
(a) 2 (b) 12 (c) 23 (d) 34
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
(c) 7.76 km s–1 (d) 8.56 km s–1
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=coswt i+sinwt j and
B=coswt i+sinwt j
2 2 are functions of time,
then the value of t at which they are orthogonal to each other is
(a) t =wp (b) t = 0 (c) t =4pw (d) t =2pw
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 (c) 6.66 kg m2 s–2 (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
(a) s ss s1 2 1 2 + (b) s ss s1 2 1+ 2 (c) 2 1 2 1 2 s s s s+ (d) s s s s 1 2 1 2 2 +
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) − + +(2 3i j k) (b) − + +(6 9i j k) (c) − + +(3 5 3i j k ) (d) − + +(6 5 2 i j k)
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) r v r v1× = × (b) 1 2 2 r r v v 1− = −2 1 2 (c) r r r r v v v v 1 2 1 2 2 1 2 1 − − = − − (d) r v r v 1 1⋅ = ⋅ 2 2
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 (c) 8.0 JK–1 mol–1 (d) 7.5 JK–1 mol–1
19. A force F=a 3 6 is acting at a point i+ +j k
r= − −2 6 12 . The value of a for which angular i j k
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
(a) p radian (b) p8 radian
(c) p
4 radian (d)
p 2 radian
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
(c) 14 ms–1 (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 (a) VR n r 2 3 2 (b) V Rnr 2 (c) VR n r 2 2 2 (d) VRnr 2 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
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
(a) L IL I+− (b) LI (c) L
I +1 (d) LI −1
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 initial direction and has a speed v3. The second block’s speed after the collision is
(a) 3
2v (b) 23v
(c) 2 2
3 v (d) 34v
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.
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 (a) E l L0 (b) LE r r r l01 ( + ) (c) LE r lr10 (d) E rr r l L 0 1 ( + ).
35. Two stones of masses m and 2m are whirled in horizontal circles, the heavier one in a radius r
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 minima in the interference pattern, I
Imaxmin is
(a) 121 49 (b) 4
9 (c)
9
4 (d) 12149
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
wavelength of the emitted electron is (a) ≥ 2.8 × 10–9 m (b) ≤ 2.8 × 10–12 m (c) < 2.8 × 10–10 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=4sin(2pt i) +4cos(2pt j)
Where R is in meters, t is in seconds and iand 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.
(d) Magnitude of acceleration vector is v
R 2
, where
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
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) b a 2 (b) 2pb a (c) b a 2 2 (d) ab 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
Kmax1 =hc− 0
l f
and that in the second case is
Kmax2 hc hc 2 2 0 0 = l −f = − l f
But Kmax2 = 3Kmax1 (given) \ 2hc− 0=3hc− 0 l f l f 2 3 3 0 0 hc hc l −f = l − f 3f0 f0 3 2 l l − = hc− hc 2f0= hcl or f0= hc2l 2. (b) : Here,
Input signal, Vi =2cos15t+3p
and voltage gain, Av = 150 As Av VVo
i
=
\ Output signal, Vo = AvVi
Since CE amplifier gives a phase difference of p(=180°) between input and output signals,
\ = + + Vo 150 2cos 15t p p3 =300cos15 4t+ 3p
3. (d) : Current through resistor, i = Current in the circuit
= + = + V R X V R C C 0 2 2 0 2 (1/w )2
Voltage across capacitor, V = iXC
= + × V R C C 0 2 1 2 1 ( /w ) w = + V R C 0 2 2 2w 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
W=1I = m x +m L x− 2 1 2 0 2 1 2 2 2 02 w [ ( ) ]w For W to be minimum, dW dx = 0 i.e. 1 2[2m x1 +2m L x2( − )( )]−1 w20=0 or m1x – m2(L – x) = 0 (... w0 ≠ 0) or (m1 + m2)x = m2L or x=m mm L+2 1 2
5. (d) : Force of attraction between the plates of the parallel plate air capacitor is
F Q
A
= 2
0
2e
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 and C=e0dA or e0A Cd= \ F C V= = Cd CV d 2 2 2 2 2
6. (c) : The P-V diagram of an ideal gas compressed from its initial volume V0 to V20by several processes is shown in the figure.
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.
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.
But as here i= °45 and sinic=1 m \ sin45° >1 1 > 2 1 m or 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
M RT P = r as P RT M = r
where P, T and r are the pressure, temperature and
density of the gas respectively and R is the universal gas constant.
\ The molecular weight of A is
M RT P A A A A =r and that of B is M RT P B B B B =r
Hence, their corresponding ratio is
M M T T P P A B A B A B B A = r r Here, rrA B A B A B T T P P =1 5 3= = = 2 1 2 . , and \ M = = MAB 32 1 12 3 4 ( )
9. (c) : The orbital speed of the satellite is
v R g
R h
o= ( + )
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 10 9 8 6 38 10 0 25 10 6 2 6 6 m m s m m = 7.76 × 103 m s–1 = 7.76 km s–1 (... 1 km = 103 m) 10. (c) : As l =hcE
where the symbols have their usual meanings. Here, E = 15 keV = 15 × 103 V and hc = 1240 eV nm \ = × = l 1240 15 103 0 083 eVnm eV . nm
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
K q B R m = 2 2 2 2 as R mv=qB = qBmK 2
where q and m are the charge and mass of the particle and R is the radius of circular orbit.
K q B R m p p p p = 2 2 2 2
and that by the alpha particle is
K q B R m a a a a = 2 2 2 2 Thus, K K q q m m R R p p p p a a a a = 2 2 or K K qq m m R R p p p p a a a a = 2 2 Here, Kp qq mm p p =1 =2 =1 4 MeV, a a , and R Rap =1 \ Ka =(1 )( )2 1( ) 4 1 2 2 MeV = 1 MeV
12. (a) : Two vectors A and are orthogonal to each B
other, if their scalar product is zero i.e.A B ⋅ = 0. Here, A=coswt i+sinwt j
and B=coswt i+sinwt j
2 2
\ A B ⋅ =(coswti+sinwt j) cos⋅ wt i+sinwt j
2 2
=cos cosw +sin sin
w w w
t t t t
2 2
( i i j j⋅ = ⋅ =1and i j j i⋅ = ⋅ =0) =cos w w t− 2t
(... cos(A – B) = cosAcosB + sinAsinB) But A B ⋅ = 0 (as Aand are orthogonal to each other) B
\ cos w w t− t = 2 0 coswt−wt =cosp 2 2 or w w p t− t= 2 2 wt p 2 =2 or t = p w
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,
v =54 − =54 5× − = −
18 15
1 1 1
km h m s m s
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
wi v R = =15 − = − = − 0 45 1500 45 100 3 1 1 1 m s m rad s rad s .
and its final angular speed is
wf = 0 (as the vehicle comes to rest) \ The angular retardation of the wheel is a w w = f − i = − = − − t 0 100 3 15 100 45 2 s rad s
The magnitude of required torque is t=I | | (a = 3 )100 −
45
2 2
kg m rad s
=203 kg m s2 2− =6 66. kg m s2 2−
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
R l A 1 1 = s ... (i)
and that of the second wire is
R l A 2 2 = s ... (ii)
As they are connected in series, so their effective resistance is Rs = R1 + R2 = + l A l A
s1 s2 (using (i) and (ii))
= + l A 1 1 1 2 s s ... (iii)
If seff is the effective conductivity of the combination, then R l A s = 2s eff ... (iv)
Equating eqns. (iii) and (iv), we get
2 1 1 1 2 l A l A seff = s +s 2 2 1 1 2 s s s s s eff = + s s s s s eff = 2 +1 2 1 2
16. (d) : The electric field E and potential V in a region
are related as E V x i V y j V z k = − ∂∂ + ∂ ∂ + ∂∂ Here, V(x, y, z) = 6xy – y + 2yz
\ = − ∂∂ − + + ∂ ∂ − + E x(6xy y 2yz i) y(6xy y 2yz j) + ∂ ∂z(6xy y− +2yz k) = −[( ) (6y i+ 6x− +1 2z k) ( ) ]+ 2y k At point (1, 1, 0), E= −[( ( )) ( ( )6 1 i+ 6 1 1 2 0− + ( )) ( ( )) ]j+ 2 1 k = − + +(6 5 2i j k)
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.
r v t r v t1+ 1 = +2 2 r r v t v t 1− =2 2 −1 =(v2−v t1) ... (i) Also, | r r1− 2| |= v2−v t1| or t r r v v = − − | | | | 1 2 2 1
Substituting this value of t in eqn. (i), we get
r r v v r r v v 1 2 2 1 1 2 2 1 − = − − − ( )| | | | or r r r r v v v v 1 2 1 2 2 1 2 1 − − = − − | | ( ) | |
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
v RT
M
= g
where g is the ratio of two specific heats, R is the universal gas constant and T is the temperature of the gas.
\ =g Mv
RT 2
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=( . × − − −− )( − ) = ( . )( ) . 4 0 10 952 8 3 273 1 6 3 1 1 2 1 1 kg mol ms JK mol K By definition, g =C C p v or Cp = gCv
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= − −2 6 12 and i j k F= + +a 3 6 i j k \ = − − t a i j k 2 6 12 3 6 = − +i( 36 36) (−j12 12+ a) (+k 6 6+ a) = −j(12 12+ a) (+k 6 6+ a) 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.
In figure A and B represent the edges of the slit AB of width a and C represents the midpoint of the slit.
asinq = l ... (i) where l is the wavelength of light.
The path difference between the wavelets from A to
C is Dx a= = a 2 1 2 sinq ( sin )q = l 2 (using (i)) The corresponding phase difference Df is
Df pD l p l l p =2 =2 × = 2 x 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) = 20.4 × 103 N/m2
\ Power of the heart =PVt
=( .20 4 10× )(5 10× − )= .
60 1 70
3N/m2 3 3m
s 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
i e. . 1mv mv mgh 2 1 2 2 02 − =
(where m is the mass of the ball) or v v2− 02=2gh ... (i) Now, when the ball collides with the ground, 50% of its energy is lost and it rebounds to the same height h. \ 50 = 100 1 2mv2 mgh 1 4v2=gh or v2 = 4gh
Substituting this value of v2 in eqn. (i), we get 4gh v− 02=2gh
or v02=4gh−2gh=2gh
or v0= 2gh
Here, g = 10 ms–2 and h = 20 m \ v0= 2 10( ms−2)(20m)=20ms−1
23. (d) : Let the speed of the ejection of the liquid through the holes be v. Then according to the equation of continuity, pR2V = npr2v or v R V n r VR nr =p = p 2 2 2 2
24. (b) : For a string fixed at both ends, the resonant frequencies are
un nv
L
=
2 where n = 1, 2, 3, ...
The difference between two consecutive resonant frequencies is Dun= un+1 – un= +(n )v− L nv L 1 2 2 = v2L
which is also the lowest resonant frequency (n = 1).
Thus the lowest resonant frequency for the given string
= 420 Hz –315 Hz = 105 Hz
25. (c) : [vc] = [hxryrz] (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
Kinetic energy of thorium nucleus is KTh pmTh
Th
= 2
2 and that of helium nucleus is
K p m He He He = 2 2
\ = K K p p m m Th He Th He He Th 2 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 As 1 1 1 v u f− = \ − − + = 1 1 1 v (fo fe) fe 1 1 1 v f f f f f f f f f f f f f e o e o e e e o e o e o e = − + = + − + = + ( ) ( ) or v f f f f e o e o = ( + ) Thus, LI vu f f f f f f f f e o e o o e e o = = + + = ( ) ( ) or f f L I o e = ... (i) \ The magnification of the telescope in normal adjustment is m f f L I o e = = (using (i))
30. (c) : The circuit is shown in the figure.
Resistance of the ammeter is
RA= + ( )( ) ( ) 480 20 480 20 W W W W = 19.2 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
I V R = =30 = = 60 1 2 0 5 V A A W .
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 2 0 12 3 1 2 2 2 2 Mv + = M v + Mv′ v v v 2 2 2 9 = + ′ or v′ =2 v2−v2 = v2−v2 = v2 9 9 9 8 9 v′ = 8v = v= v 9 8 3 2 2 3 2
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
I V R = = 3 100 V W = 3 × 10–2 A = 30 × 10–3 A = 30 mA 34. (d) :
The current through the potentiometer wire is
I E
r r
= +01
( )
and the potential difference across the wire is
V Ir E r
r r
= = +01
( )
The potential gradient along the potentiometer wire is k V L E r r r L = = +01 ( )
As the unknown e.m.f. E is balanced against length
l of the potentiometer wire,
\ E kl E r r r l L = = +01 ( )
35. (c) : Let v be tangential speed of heavier stone. Then, Centripetal force experienced by lighter stone is
( )F m nv( )
r c lighter=
2
and that of heavier stone is ( ) ( / ) F mv r c heavier=2 2 2
But (Fc)lighter = (Fc)heavier (given)
\ m nv = r mv r ( ) ( / ) 2 2 2 2 n mv r mv r 2 2 4 2 = n2 = 4 or n = 2
36. (c) : As, intensity I ∝ width of slit W Also, intensity I ∝ square of amplitude A
\ I = = I W W A A 1 2 1 2 12 22 But W W12 = 125 (given) \ A = = = A A A 12 22 1 2 1 25 1 25 1 5 or \ = + − = + − I I A A A A A A A A max min ( ) ( ) 1 2 2 1 2 2 1 2 2 1 2 2 1 1 = + − = − = = 1 1 5 1 5 1 6 5 4 5 36 16 9 4 2 2 2 2 37. (d) :
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
DLs W LY As s = as Y W A= L L / / D and that in the brass wire is
DLb W LY Ab b = But DLs = DLb (given) \ W L= = Y A W L Y A W W Y Y s s b b s b s b or As Y Ybs = 2 (given) \ W = Wbs 2 1
38. (c) : The coefficient of performance of a refrigerator is a = − T T T1 2 2
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 = ? \ = − 5 253 253 1 K K T 5T1 – 5(253 K) = 253 K 5T1 = 253 K + 5(253 K) = 6(253 K) T1 6 5 253 = ( K = 303.6 K = 303.6 – 273 ) = 30.6°C ≈ 31°C
39. (a) : According to Einstein’s photoelectric equation, the maximum kinetic energy of the emitted electron is
Kmax =hc−
l f0
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 \ Kmax=1240 − . 500 2 28 eV nm nm eV = 2.48 eV – 2.28 eV = 0.2 eV
The de Broglie wavelength of the emitted electron is lmin max = h mK 2
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 \ = × × × × − − − lmin . ( )( . . ) 6 6 10 2 9 10 0 2 1 6 10 34 31 19 J s kg J = × = × − − 6 6 2 4 10 9 2 8 10 9 . . m . m 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
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
u u= − ° 0 v v v 60 scos = − − − − ( ( . ) 100 330 330 19 4 1 2 1 1 1 Hz) ms ms ms = − − − − ( . 100 330 330 9 7 1 1 1 Hz) ms ms ms = − − ( . 100 330 320 3 1 1 Hz) ms ms = 103 Hz
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 rT0 = − D1 g T or g r r DT= −1 T 0 Thus, r r r g 0 0 − T = DT Here, g = 5 × 10–4 K–1 and DT = 40°C = 40 K \ The fractional change in the density of glycerin = − = = × − − = r r r g 0 0 4 1 5 10 40 0 020 T DT ( K )( K) .
42. (a) : Here, R=4sin(2pt i)+4cos(2pt j) The velocity of the particle is
v dR dt d dt t i t j = = [ sin(4 2p ) +4cos(2p ) ] =8pcos(2pt i)−8psin(2pt j) Its magnitude is | |v = ( cos(8p 2pt)) (2+ −8psin(2pt))2 = 64p2cos (2 2pt)+64p2sin (2 2pt) = 64p2[cos (2 2pt) sin (+ 2 2pt)]
= 64p (as sin2 2q + 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,
\ ms =tanq=tan30° = 1 = .
3 0 6
If a is the acceleration produced in the block, then
ma = mgsinq – fk
(where fk is force of kinetic friction) = mgsinq – mkN (as fk = 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 s= 1at 2 2 (as u = 0) or a s t = 22
But s = 4.0 m and t = 4.0 s (given)
\ a=2 4 0 = − 4 0 1 2 2 2 ( . ) ( . ) m s ms
Substituting this value of a in eqn. (i), we get 1 2 10 12 3 2 2 2 ms− = ms− − ( ) mk 1 10= −1 3mk or 3 1 1 10 9 10 0 9 mk = − = = . mk= = 0 9 3 0 5 . .
44. (b) : The wavelength of a spectral line in the Lyman series is 1 1 1 1 2 3 4 2 2 lL =R −n ,n= , , , ...
and that in the Balmer series is
1 1
2
1 3 4 5
2 2
lB =R −n ,n= , , , ...
For the longest wavelength in the Lyman series,
n = 2 \ = − = − = − = 1 1 1 1 2 1 1 1 4 4 1 4 3 4 2 2 lL R R R R or lL R = 4 3
For the longest wavelength in the Balmer series,
n = 3 \ = − = − 1 1 2 1 3 1 4 1 9 2 2 lB R R =R9 4− = R 36 5 36 or lB R = 36 5 Thus , l lLB R R R R = = × = 4 3 36 5 4 3 5 36 5 27
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 b w w w = 2A= A
\ Time period of vibration is
T =2p= 2 =2 w p a b pb a ( / ) nn
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
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-height-time graph.
8. At what angle the two forces A + B and A – B act so that their resultant is 3A2+B2 ?
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.
GENERAL INSTRUCTIONS (i) All questions are compulsory.
(ii) Q. no. 1 to 5 are very short answer questions and carry 1 mark each. (iii) Q. no. 6 to 10 are short answer questions and carry 2 marks each. (iv) Q. no. 11 to 17 are also short answer questions and carry 3 marks each. (v) Q. no. 18 is a value based question and carries 4 marks.
(vi) Q. no. 19 and 20 are long answer questions and carry 5 marks each. (vii) Use log tables if necessary, use of calculators is not allowed.
Series 1
Chapterwise Unit test : Units and Measurement | Kinematics
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 (u2 v2)
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.
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.
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 dimensional analysis, that T= Rk rg3, where k is 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?
Speed Time 1 2 3 A B C D 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.
Ratio of two velocities = °
° = = tan30 / 60 1 3 3 1 3 tan
6. As L.H.S. represents distance, every term on R.H.S. must represent distance.
\
[ ]
bt =[ ]
x[ ]
b =[ ]
[ ]
x =[ ]
[ ]
= −t
or L [LT ]
T 1
Therefore, b represents velocity and hence its unit is m s–1.
7. 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.
8. 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
or cosq = 1 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.
Velocity of object, v dx= dt = 3kt2 and acceleration of object, a dv= dt =3k×2t=6kt
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 Percentage error =±0 03× = ± 1 28 100 2 34 . . . %
Hence, difference in lengths = (1.28 ± 0.03) cm = 1.28 cm ± 2.34 % 11. Given, 50 VSD = 49 MSD
⇒ 1 VSD = 4950 MSD 1 MSD = 0.5 mm In vernier callipers,
Minimum inaccuracy in the measurement of distance by vernier callipers = vernier constant = 1 MSD – 1 VSD =1 −49 50 1 50 MSD MSD = MSD = 1 × 50 0 5. mm = 0.01 mm
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,
h v t= y + 1gt 2 2 ⇒ 9 0= × + ×1 × 2 10 2 t t ⇒ 5 =9 ⇒ = 9 = 5 1 34 2 t t . s
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.
13. Initial speed of the ball, u = 10 m s–1 = 36 km h–1 The ball is thrown at angle of 60° with the horizontal.
Horizontal speed of the ball, ux = ucos60°
=36 1× = −
2 18 km h 1
Vertical speed of the ball, uy = usin60°
=36× 3= −
2 18 3 km h 1
Speed of the car, v = 18 km h–1
u 60° ux= 5 uy m s–1 uy uy
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 = = × × = − − − 126 126 1000 60 60 35 1 1 1 kmh ms ms ; v=0,s=200m,a=?andt=? We know, v2 = u2 + 2as \ 0 = (35)2 + 2 × a × 200 or a = −
( )
×35 = − = − − 2 200 49 16 3 06 2 2 . ms As, v = u + at \ = +− 0 35 49 16 t or t =35 16× = = . s 49 80 7 11 43Negative sign shows that acceleration is negative, which is called retardation i.e. car is uniformly retarded at a = 3.06 m s–2.
15. If a projectile is projected with velocity u, making an angle q with the horizontal direction, then Horizontal range, R u= g2 sin q2
and maximum height, H u
g
= 2 2
2 sin q
Case (i) : If q = a, let R = R1 and H = H1, then
R u g 1 2 2 = sin a …(i) and H u sin g 1 2 2 2 = a …(ii)
Case (ii) : If q = (90° – a), let R = R2 and H = H2, then R u g 2 2 2 90 = sin ( ° −a) =u ° − = g u g 2 2 180 2 2
sin( a) sin a …(iii)
H u g u g 2 2 2 2 2 2 90 2
= sin ( ° −a)= cos a …(iv)
From (i) and (iii), R1 = R2 From (ii) and (iv);
H H u g u g 1 2 2 2 2 2 2 2 + = (sin a+cos )a =
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 the object in going from p to A=tp
2 and the time taken by the object in going from q to A=tq
2 . Taking vertical upward motion of object from
p to A, we have u = u′, v = 0, a = −g t, =tp 2 As, v = u + at \ 0= ′ + − 2 u ( )g tp or u′ = gtp 2 …(i)
