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Objective Physics for the JEE Main 2015 Sanjeev Kumar
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A Complete Resource Book in for JEE Main 2019
PHYSICS
Sanjeev Kumar
The aim of this publication is to supply information taken from sources believed to be valid and reliable. This is not an attempt to render any type of professional advice or analysis, nor is it to be treated as such. While much care has been taken to ensure the veracity and currency of the information presented within, neither the publisher nor its authors bear any responsibility for any damage arising from inadvertent omissions, negligence or inaccuracies (typographical or factual) that may have found their way into this book.
This book is sold subject to the condition that it shall not, by way of trade or otherwise, be lent, resold, hired out, or otherwise circulated without the publisher’s prior written consent in any form of binding or cover other than that in which it is published and without a similar condition including this condition being imposed on the subsequent purchaser and without limiting the rights under copyright reserved above, no part of this publication may be reproduced, stored in or introduced into a retrieval system, or transmitted in any form or by any means (electronic, mechanical, photocopying, recording or otherwise), without the prior written permission of both the copyright owner and the publisher of this book.
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ISBN 978-93-530-6216-3 eISBN 9789353063429
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Published by Pearson India Education Services Pvt. Ltd, CIN: U72200TN2005PTC057128.
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Preface
JEE Mains
Preface
About the Series
ACompleteResourceBookforJEEMain series is a must-have resource for students preparing for JEE Main examination. There are three separate books on Physics, Chemistry, and Mathematics; the main objective of this series is to strengthen the fundamental concepts and prepare students for various engineering entrance examinations. It provides class-tested course material and numerical applications that will supplement any ready material available as student resource. To ensure high level of accuracy and practicality, this series has been authored by highly qualified and experienced faculties.
About the Book
It gives me immense pleasure to present this book ACompleteResourceBookinPhysicsForJEEMain2019. This book will help the students in building the analytical and quantitative skills necessary to face the examination with confidence. This title is designed as per the latest JEE Main syllabus, spread across 20 chapters. It has been structured in an user friendly approach such that each chapter begins with topic-wise theory, followed by sufficient solved examples and then practice questions. The brain-map section in every chapter will help the students to revise the important formulae. The chapter end exercises are structured in line with JEE questions; where ample number of questions on single choice correct question (SCQ), multiple-type correct questions (MCQ), assertion and reasoning, column matching, passage based and integer type questions are included for extensive practice. Previous 15 years’ questions of JEE Main and AIEEE are also added in every chapter. HintsandSolutions at the end of every chapter will help the students to evaluate their concepts and numerical applications.
Series Features
• Complete coverage of topics along with ample number of solved examples.
• Includes various types of practice problems with complete solutions.
• Chapter-wise Previous 15 years’ AIEEE/JEE Main questions.
• Fully solved JEE Main 2017 and 2018 questions are included in the book.
• 5 Mock Tests based on JEE Main pattern.
• 5 Free Online Mock Tests as per recent JEE Main pattern.
I dedicate this book to my family for their immense support and love. Special thanks to my parents for their support and encouragement and to my wife Pallawi and my sons Haardik and Saarthak for sustaining me throughout this project. I would like to express my heartfelt gratitude to the Pearson team, without them I would not have been able to bring out this book.
Any suggestions and comments from the readers would be highly appreciated. Please communicate to us if there are any errors, misprints or other such concerns.
Sanjeev Kumar
JEE Mains 2018 PaPEr
1. The density of a material in the shape of a cube is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are respectively 1.5% and 1%, the maximum error in determining the density is (A) 2.5% (B) 3.5% (C) 4.5% (D) 6%
2. All the g raphs below are intended to represent the same motion. One of them does it incorrectly. Pick it up.
4. A par ticle is moving in a circular path of radius a under the action of an attractive potential U
Its total energy is
5. In a collinear collision, a par ticle with an initial speed v0 strikes a stationary particle of the same mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after collision, is
3. Two masses m1 = 5 kg and m2 = 10 kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of m2 to stop the motion is
6. Seven identical circular planar discs, each of mass M and radius R are welded symmetrically as shown. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point P is
7. From a uniform circular disc of radius R and mass 9M, a small disc of radius R 3 is removed as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is
2R R 3
(A) 4MR2 (B) 40 9 2 MR (C) 10MR2 (D) 37 9 2 MR
8. A par ticle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the nth power of R. If the period of rotation of the particle is T, then
(A) T ∝ R3/2 for any n (B) TR n ∝ + /2 1
(C) T ∝ R(n+1)/2 (D) T ∝ Rn/2
12. A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 1012/sec. What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver = 108 and Avogadro’s number = 6.02 × 1023 gm mole–1)
(A) 6.4 N/m (B) 7.1 N/m
(C) 2.2 N/m (D) 5.5 N/m
13. A g ranite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7 × 103 kg/m3 and its Young’s modulus is 9.27 × 1010 Pa. What will be the fundamental frequency of the longitudinal vibrations?
(A) 5 kHz (B) 2.5 kHz (C) 10 kHz (D) 7.5 kHz
14. Three concentric metal shells A, B and C of respective radii a, b and c (a < b < c) have surface charge densities +σ, -σ and +σ respectively. The potential of shell B is
, is
9. A solid sphere of radius r made of a soft material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless piston of area a floats on the surface of the liquid, covering entire cross section of cylindrical container. When a mass m is placed on the surface of the piston to compress the liquid, the fractional decrement in the radius of the sphere, dr
(A) Ka mg (B) Ka mg3 (C) mg Ka 3 (D) mg Ka
10. Two moles of an ideal monoatomic gas occupies a volume V at 27°C. The gas expands adiabatically to a volume 2V. Calculate (a) the final temperature of the gas and (b) change in its internal energy.
(A) (a) 189 K (b) 2.7 kJ
(B) (a) 195 K (b) -2.7 kJ
(C) (a) 189 K (b) -2.7 kJ (D) (a) 195 K (b) 2.7 kJ
11. The mass of a hydrogen molecule is 3.32 × 10 -27 kg. If 1023 hydrogen molecules strike, per second, a fixed wall of area 2 cm2 at an angle of 45° to the normal, and rebound elastically with a speed of 103 m/s, then the pressure on the wall is nearly
(A) 2.35 × 103 N/m2 (B) 4.70 × 103 N/m2
(C) 2.35 × 102 N/m2 (D) 4.70 × 102 N/m2
15. A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material of dielectric constant K = 5 3 is inserted between the plates, the magnitude of the induced charge will be (A) 1.2 nC (B) 0.3 nC (C) 2.4 nC (D) 0.9 nC
16. In an a.c. circuit, the instantaneous emf and cur rent are given by e = 100 sin30t it =−
2030 4 sin
In one cycle of a.c., the average power consumed by the circuit and the wattless current are, respectively: (A) 50, 10 (B) 1000 2 10, (C) 50 2 0, (D) 50, 0
17. Two batteries with emf 12 V and 13 V are connected in parallel across a load resistor of 10 Ω. The internal resistances of the two batteries are 1 Ω and 2 Ω respectively. The voltage across the load lies between (A) 11.6 V and 11.7 V (B) 11.5 V and 11.6 V (C) 11.4 V and 11.5 V (D) 11.7 V and 11.8 V
18. An electron, a proton and an alpha par ticle having the same kinetic energy are moving in circular orbits of radii re, rp, rα respectively in a uniform magnetic field B. The relation between re, rp, rα is
(A) re > rp = rα (B) re < rp = rα (C) re < rp < rα (D) re < rα < rp
19. The dipole moment of a circular loop car rying a current I is m and the magnetic field at the centre of the loop is B1. When the dipole moment is doubled by keeping the current constant the magnetic field at the centre of the loop is B2. The ratio B B 1 2 is (A) 2 (B) 3 (C) 2 (D) 1 2
20. For an RLC circuit driven with voltage of amplitude vm and frequency ω0 1 = LC the current exhibits resonance. The quality factor, Q is given by (A) ω0 L R (B) ω0 R L (C) R C () ω0 (D) CR ω0
21. An EM wave from air enters a medium. The electric fields are EExv z c t 101 2
in
and EExkzct 202 2 =−() ˆ cos[ ] in medium, where the wave number k and frequency v refer to their values in air. The medium is non-magnetic. If ∈r1 and ∈r2 refer to relative permittivities of air and medium respectively, which of the following options is correct?
(A) 0° (B) 30° (C) 45° (D) 60°
23. The angular width of the central maximum at a single slit diffraction pattern is 60°. The width of the slit is 1 μm. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young’s fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance? (i.e., distance between the centres of each slit.)
(A) 25 μm (B) 50 μm (C) 75 μm (D) 100 μm
24. An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let λn, λg be the de Broglie wavelength of the electron in nth state and the ground state respectively. Let Λn be the wavelength of the emitted photon in the transition from the nth state to the ground state. For large n, (A, B are constants)
(A) Λ n n A B ≈+ λ2
(B) Λ nn AB≈+ λ
(C) Λ nn AB 22 ≈+ λ
(D) Λ n 2 ≈ λ
25. If the series limit frequency of the Lyman series is vL, then the series limit frequency of the Pfund series is.
(A) 25vL (B) 16vL (C) vL/16 (D) vL/25
26. It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is Pd; while for its similar collision with carbon nucleus at rest, fractional loss of energy is Pc. The values of Pd and Pc are respectively
27. The reading of the ammeter for a silicon diode in the given circuit is
200 Ω
22. Unpolarized light of intensity I passes through an ideal polarizer A. Another identical polarizer B is placed behind A. The intensity of light beyond B is found to be 1 2 Now another identical polarizer C is placed between A and B. The intensity beyond B is now found to be 1 8 . The angle between polarizer A and C is
3 V
(A) 0 (B) 15 mA (C) 11.5 mA (D) 13.5 mA
28. A telephonic communication ser vice is working at carrier frequency of 10 GHz. Only 10% of it is utilized for transmission. How many telephonic channels can be transmitted simultaneously if each channel requires a bandwidth of 5 kHz?
(A) 2 × 103 (B) 2 × 104 (C) 2 × 105 (D) 2 × 106
29. In a potentiometer experiment, it is found that no current passes through the galvanometer when the terminal of the cell are connected across 52 cm of the potentiometer wire. If the cell is shunted by a resistance of 5 Ω,
a balance is found when the cell is connected across 40 cm of the wire. Find the internal resistance of the cell.
(A) 1 Ω (B) 1.5 Ω (C) 2 Ω (D) 2.5 Ω
30. On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm. The resistance of their series combination is 1 kΩ. How much was the resistance on the left slot before interchanging the resistances?
(A) 990 Ω (B) 505 Ω (C) 550 Ω (D) 910 Ω
Answer keys
1. (C)
(A)
Hints and solutions
= 1.5 + 3 × 1 = 4.5%
Hence, the cor rect option is (C).
2. Incor rect option is (A) because it represents two values of velocity at the same instant, which is impossible. Hence, the cor rect option is (A).
3. For stopping the motion:
Total energy =+ KE PE =−= k r k r 22 0 22
Hence, the cor rect option is (C).
5. Momentum conser vation yields:
⇒= m m m 1 2 µ =− = 5 015 102333.kg
∴ Minimum weight m = 27.3 kg
Hence, the cor rect option is (B).
xii JEE Mains 2018 Paper
Solving the two equations, we get vvv 12 0 2 −=
Hence, the cor rect option is (B).
×+ ()
6. I MR MR MR 0 2 2 2 2 26 2 =+
By parallel axis theorem:
IIMR p =+ 0 2 73()
∴= +
= 181 2 2 MR
Hence, the cor rect option is (D).
7. Mass per unit area = 9 2 M R π Mass of π π π RM R R M 2 2 2 9 9 9 =× =
Moment of iner tia of the remaining disc III MRMR M R =− =−
16. Wattless current = irms sinφ == × io 2 20 2 1 2 sinφ = 10 A
Average power = Virmsrms cosφ =× × 100 2 20 2 1 2 = 1000 2
Hence, the cor rect option is (B).
17. Equivalent emf == + + E ErEr rr 12 21 12
∴= ×+ × + = E 122131 12 37 3 V
Equivalent resistance == + r rr rr 12 12
∴= × + = r 12 12 2 3 Ω
Cur rent == + = i 37 3 2 3 10 37 32 A
ViRR== ×= 37 32 101156.V
Hence, the cor rect option is (B).
18. r mv qB = KE == ⇒= kmvv k m 1 2 2 2
∴= = r m qB K m mK qB 22 r mk eB r mk eB r mk eeB e p p == = 2 2 2 2 ;; α α r mk eB mk eB pp α = × = 24 2 2
∴< = rrr ep α
Hence, the cor rect option is (B).
19. Dipole moment = M = iA
∴= ⇒∝ MirMr π 22 B i r B r o =⇒ ∝ µ 2 1
∴= ⇒= = r r M M M M r r 1 2 1 2 1 2 1 2 2 2 1 2
Also, r r B B 1 2 2 1 1 2 ==
∴= B B 1 2 2
Hence, the cor rect option is (C).
20. Quality factor == Q L R ω0
Hence, the cor rect option is (A).
21. EExv z c t 101 2 =−
cos π or, EEx cz c t 101 2 =− ˆ cos π
EExv z c t 2 =−
=−[] Exkzct 01 ˆ cos( )
We know that, kr = ε
Also, EExkct 202 2 2 =−
coszEExv z c t 101 2 =−
∴= = kkkk 2;12
⇒=
= ε ε r r K K 1 2 2 1 4 2
Hence, the cor rect option is (C).
22. cos2 22 II θ = I I A B 2
cos 2 10θθ =⇒ =
Let the angle between the pass axes of A and C be b I A IC IB C B I 2
I I C = 2 2 cos β
I I B =
2 22coscosββ
∴= ⇒= II 82 1 4 2222 (cos )(cos) ββ
∴= ⇒=coscos 2 1 2 1 2 ββ
∴= ° β 45
Hence, the cor rect option is (C).
23. 26030 θθ =° ⇒= °
Condition for diffraction minima: d sinsin . θλ λ =⇒ =° =× 1030 05 10 66
For double slit: β λ = D d
⇒= ×× 10 05 10 05 2 6 .. d
⇒= d 25 µm
Hence, the cor rect option is (A).
24. KE == = k n P me 136 2 2 2
⇒= = P k n h λ ; where k is a constant
∴= = λn nh k nc; where c is a constant
A B n + λ2
Hence, the cor rect option is (A). 25. 11 1 1 2 2 2 λ =−
R nn v C LR=−
⇒=vRLC (1) For Pfund
vCPR=−
∴= = v CRv P L 2525
Hence, the cor rect option is (D).
26. Velocity of neutron after the collision is: V mm mm u 1 12 12 = + () Fractional loss in KE
For collision with deuterium, mmmm 12 2 == ;
∴= ×× + == P mm mm d 42 2 8 9 089 2 ()
For collision with carbon, mmmm 12 12 == ;
P mm mm C = ×× + == 42 12 48 169 028 2 ()
Hence, the cor rect option is (A).
27. For a silicon diode the barrier potential is 0.7 volts
∴=
×= i 30 7 200 10115 3 .mA.
Hence, the cor rect option is (C).
28. 10% of 10 GHz =× ×= 1010 10 100 10 99 Hz
∴ Number of channels = × =× 10 510 210 9 3 5
Hence, the cor rect option is (C).
29. Let the voltage gradient be λ
∴ Terminal voltage = 40λ
401 40521 λ λλ E r Rr r Rr ⇒ + = R Rr 10 13 ⇒ + =⇒ = 5 5 10 13 15 r r Ω
Hence, the cor rect option is (B).
30. R x R x 12 100 = ()
R x R x 21 10110 = ()
Also, RR12 1000 += (3)
Solving these equations, we get RR 21450550 ==ΩΩ ;
Hence, the cor rect option is (C).
JEE Mains 2017 PaPEr
1. An obser ver is moving with half the speed of light towards a stationary microwave source, emitting waves at frequency 10 GHz. What is the frequency of the microwave measured by the observer?
2. The following observations were taken for determining surface tension T of water by capillary method: Diameter of capillary, D = 1.25 × 10 -2 m; rise of water, h = 1.45 × 10 -2 m. Using g = 9.80 m/s2 and the simplified relation, T rhg =× 2 103 N/m, the possible error in surface tension is closest to (A) 1.5% (B) 2.4% (C) 10% (D) 0.15%
3. Some energy levels of a molecule are shown in the figure. The ratio of the wavelengths, r = λλ12/, is given by
Cp – Cv = a, for hydrogen gas; Cp – Cv = b, for nitrogen gas. The correct relation between a and b is
(A) a = b (B) a = 14b
(C) a = 28b (D) ab = 1 14
6. The moment of inertia of a uniform cylinder of length l and radius R about its perpendicular bisector is I. What will be the ratio of l to R, such that the moment of inertia is minimum? (A)
4. A body of mass, m =10 -2 kg is moving in a medium and experiences a frictional force, F = -kv2. Its initial speed is, v0 = 10 m/s. If, after 10 seconds, its energy is 1 8 0 2 mv , then the value of k will be
5. Cp and Cv are specific heats at constant pressure and constant volume, respectively. It is observed that
7. A radioactive nucleus A with a half life T, decays into a nucleus B. At t = 0, there is no nucleus B. At some value of t, the ratio of the number of B to that of A is 0.3. Then, t is given by
(A) tT = log. log 13 2
(B) t = T log(1.3)
(C) t T = log(13.)
(D) t T = 2 2 13 log log.
8. Which of the following statements is false?
(A) In a balanced wheatstone bridge if the cell and the galvanometer are exchanged, the null point is disturbed.
(B) A rheostat can be used as a potential divider.
(C) Kirchhoff ’s second law represents energy conservation.
(D) Wheatstone bridge is the most sensitive when all the four resistances are of the same order of magnitude.
9. A capacitance of 2 μF is required in an electrical circuit across a potential difference of 1.0 kV. A large number of 1 μF capacitors are available which can withstand a potential difference of not more than 300 V. The minimum number of capacitors required to achieve this, is
(A) 16 (B) 24
(C) 32 (D) 2
10. In the given circuit diagram, when the current reaches steady state in the circuit, the charge on the capacitor of capacitance C will be
r1 r2 r C E (A) CE r rr 1 2 () + (B) CE r rr 2 2 () + (C) CE r rr 1 1 () + (D) CE 11. 2 V 2 V 2 V 2 V2 V 2 V 1 Ω 1 Ω 1 Ω
In the above circuit the current in each resistance is (A) 0.25 A (B) 0.5 A (C) 0 A (D) 1 A
12. In amplitude modulation, sinusoidal car rier frequency used is denoted by ωc and the signal frequency is denoted by ωm. The bandwidth (∆ωm) of the signal is such that, ∆ωm << ωc Which of the following frequencies is not contained in the modulated wave?
(A) ωc (B) ωm + ωc (C) ωc - ωm (D) ωm
13. In a common emitter amplifier circuit using an n-p-n transistor, the phase difference between the input and the output voltages will be:
(A) 90° (B) 135° (C) 180° (D) 45°
14. A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75°C. T is given by
(Given: Room temperature = 30°C, specific heat of copper = 0.1 cal/gm°C)
(A) 885°C (B) 1250°C (C) 825°C (D) 800°C
15. In a Young’s double slit experiment, slits are separated by 0.5 mm, and the screen is placed 150 cm away. A beam of light consisting of two wavelengths 650 nm
and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide, is
(A) 7.8 mm (B) 9.75 mm (C) 15.6 mm (D) 1.56 mm
16. An electric dipole has a fixed dipole moment p, which makes an angle θ with respect to X-axis. When subjected to an electric field, EEi 1 = ˆ , it experiences a torque, Tk 1 = τ ˆ . When subjected to another electric field, EEj 21 3 = ˆ , it experiences a torque, TT 21 =− Then, the angle θ is (A) 45° (B) 60° (C) 90° (D) 30°
17. A slender unifor m rod of mass M and length l is pivoted at one end, so that it can rotate in a vertical plane (see figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle θ with the vertical, is
18. An external pressure P is applied on a cube at 0°C, so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and α is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by (A) P K α (B) 3α PK (C) 3PKα (D) P K 3α
19. A diverging lens with magnitude of focal length 25 cm is placed at a distance of 15 cm from a converging lens of magnitude of focal length 20 cm. A beam of parallel light falls on the diverging lens. The final image formed is
(A) Virtual and at a distance of 40 cm from convergent lens.
(B) Real and at a distance of 40 cm from the divergent lens.
(C) Real and at a distance of 6 cm from the convergent lens.
(D) Real and at a distance of 40 cm from convergent lens.
20. An electron beam is accelerated by a potential difference V to hit a metallic target to produce X-rays. It produces continuous as well as characteristic X-rays. If λmin is the smallest possible wavelength of X-rays in the spectrum, the variation of log λmin with log V is correctly represented in
(A) log minλ
V
log minλ
V (C) log minλ
(D) log minλ
V
21. The temperature of an open room of volume 30 m3 increases from 17°C to 27°C due to the sunshine. The atmospheric pressure in the room remains 1 × 105 Pa. If ni and nf are the number of molecules in the room before and after heating, then nf – ni will be
22. In a coil of resistance 100 Ω, a current is induced by changing the magnetic flux through it, as shown in the figure. The magnitude of change in flux through the coil is
(amp)
(A) 225 Wb (B) 250 Wb (C) 275 Wb (D) 200 Wb
23. When a cur rent of 5 mA is passed through a galvanometer having a coil of resistance 15 Ω, it shows full scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into a voltmeter of range 0–10 V is (A) 2.045 × 103 Ω (B) 2.535 × 103Ω (C) 4.005 × 103 Ω (D) 1.985 × 103Ω
24. A time dependent force, F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, then the work done by the force during the first 1 second will be (A) 22 J (B) 9 J (C) 18 J (D) 4.5 J
25. A magnetic needle of magnetic moment 6.7 × 10–2 Am2 and moment of inertia 7.5 × 10–6 kgm2 is performing simple harmonic oscillations in a magnetic field of 0.01T. Time taken for 10 complete oscillations is (A) 8.89 s (B) 6.98 s (C) 8.76 s (D) 6.65 s
26. The variation of acceleration due to gravity g with distance d from centre of the earth, is best represented by (R = Earth’s radius) (A) g
27. A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time?
28. A par ticle A of mass m and initial velocity of v collides with a particle B of mass m 2 which is at rest. The collision is head on, and elastic. The ratio of the de Broglie wavelengths λA to λB after the collision is (A) λ λ A B = 2 (B) λ λ A B = 2 3 (C) λ λ A B = 1 2 (D) λ λ A B = 1 3
29. A par ticle is executing simple harmonic motion with a time period T. At time t = 0, it is at its position of equilibrium. The kinetic energy-time graph of the particle will look like
30. A man g rows into a giant such that his linear dimensions increase by a factor of 9. Assuming that his density remains same, the stress in the leg will change by a factor of
(A) 1 9 (B) 81
(C) 1 81 (D) 9
1. (B)
1. ′ = + ffcu
2.
Hence, the cor rect option is (B).
Answer keys
Hence, the cor rect option is (A).
Hints and solutions
Dividing Eq. (1) by Eq. (2), we get
1 2 1 3 =
Hence, the cor rect option is (C).
4. F
⇒ k = 10 -4 kg/m.
Hence, the cor rect option is (B).
5. For H2,
N2,
From Eqs. (1) and (2), we get a b = 14
a = 14b
Hence, the cor rect option is (B).
6. I MRMl =+ 22 412
πρRlM 2 = l M R = πρ 2 I MRMM R =+ + 22 24 2 412 πρ
For I to be minimum, dI
dR = 0 2 4 3 2 22 5 R M R = πρ R l R = 2 3 2
⇒= l R 3 2 R
Hence, the cor rect option is (D).
7. AB at at → = =− tN ttNNN 00 0 0 , , NN N 0 03 =
N0 = 1.3N
N0 = 1.3 N0e -λt
⇒= λt ln(.13)
t = 1 13 λ ln(. )
tT = log( .) log() 13 2
Hence, the cor rect option is (A).
8. Conceptual.
Hence, the cor rect option is (A).
9. To get capacitance of 2 μF, such that potential difference across any one should not be more than 300 V, can be arranged as follows:
∴ The minimum number of capacitors required = 4 × 8 = 32. Hence, the cor rect option is (C).
10. In steady state no cur rent will pass through capacitor.
Cur rent through cell, i E rr = + () 2
Potential difference across capacitor = Potential difference across r2
Vir = 2 = + Er rr 2 2 ()
∴ Charge on capacitor = CV = + CEr rr 2 2 ()
Hence, the cor rect option is (B).
Taking voltage of point A as = 0
Then voltage at other points can be written as shown in figure
Hence voltage across all resistance is zero.
Hence cur rent = 0
12. Refer NCERT Page No. 526 Three frequencies are contained ωω ωω ω mccmc +− , and
13. Conceptual
Hence, the cor rect option is (C).
14. From principle of calorimetry, Heat lost by copper sphere = Heat gained by calorimeter + Heat
by water
⇒ T – 75 = 810
⇒ T = 885°C.
Hence, the cor rect option is (A).
15. Let n1th order maxima of wavelength 520 nm coincides with n2th order maxima of wavelength 650 nm.
That is, n1 × 520 = n2 × 650
⇒= n n 1 2 5 4
So, 5th order maxima of wavelength 520 nm coincides with 4th order maxima of wavelength 650 nm. Then, the least distance from central maxima is,
Y D d = 5λ
=× ×× × × 5 5201015010 510 92 4 ()
= 7.8 mm.
Hence, the cor rect option is (A).
16. ττ12 =
⇒= ×−PEPEsinsin() θθ 390
⇒= tan θ 3
⇒ θ = 60° Y X O P θ
Hence, the cor rect option is (B).
17. Mg M l
sin θ l 2 Mg lMl 23 2 sin θα =
αθ = 3 2lg sin.
Hence, the cor rect option is (D). 18. dV = –V(3
From Eqs. (1) and (2), we get dT P K = 3 α
Hence, the cor rect option is (D).
19. I2 I1 L1 L2 f = 25 cm f = 20 cm 15 cm
For ‘L1,’ image will be formed at a distance of 25 cm from ‘L1’ (left)
For ‘L2,’ u = –40 cm
f = +20 cm
v = +40 cm (using lens formula)
⇒ Real image will be formed at a distance of 40 cm from L2, i.e., converging lens (right of L2).
Hence, the cor rect option is (D).
20. hc e λmin = V 1 λmin = e hc V nVn e hc 1 λmin
=+ n
−= + nnVn e hc () minλ nnVn e hc () minλ =−
It is a straight line with -ve slope.
21. Using, PV = NRT (N = Number of moles), we get 105 × 30 = NiR × 290 and 105 × 30 = NfR × 300
⇒− = ×
NNfiR 10301 300 1 290 5 = ×− ×
103010 300290 5 R = × 10 290 5 R
⇒ For molecules, nnNNN fifi −= 0 ()
= −× × × 60231010 8314290 23 5
= –249.8 × 1023
−× 25 10 25
Hence, the cor rect option is (C).
22. ε = iR
d dt iR φ = dRidt φ = ∫∫
100 1 2 10 1 2
= 250 Wb.
Hence, the cor rect option is (B).
23. 5 mA 15 Ω R G
() R +× = 15 5 1000 10
R + 15 = 2000
R = 1985 Ω. Hence, the cor rect option is (D).
24. a = 6t dV dt t = 6 dvtdt vt = ∫∫ 6 0 0 v t t == 6 2 3 2 2 v = 3 m/s Wmv =∆ == KE J 1 2 45 0 2
Hence, the cor rect option is (D).
25. T I MB ==20 665 π s
For 10 oscillations, time = 10T = 6.65 s.
Hence, the cor rect option is (D).
26. For d ≤ R, g GM R d = 3
For d ≥ R, g GM d = 2
Hence, the cor rect option is (C).
27. v = u – gt t v
Hence, the cor rect option is (B).
28. Rest v
Adding Eqs. (1) and (2), we get 2 2 3
4 3
From Eq. (1) , v v v =− 4 3 1 v v v v 1 4 33 =− = Pmv mv A == 1 3
λ λ B A = 1 2
∴= λ λ A B 2.
Hence, the cor rect option is (A).
29. t = 0 mean
v = vmax cos ωt KE = 1 2 22 mvt max cos ω
T T/4 T/2
Hence, the cor rect option is (C).
30. Stress, σ ρ ρ == == F A mg A LgA ALg
Now when length increases,
′ = ′ = ′ ′ = σ ρ ρ mg A ALg ALg 9 9.
Hence, the cor rect option is (D).
CHAPTER
Unit and Dimension 1
Chapter Highlights
Physics, technology and society, SI units, Fundamental and derived units. Least count, Accuracy and precision of measuring instruments, Errors in measurement, Significant figures. Dimensions of Physical quantities, Dimensional analysis and its applications. Basic Mathematics.
PhYSiCAl QUAntitieS
The quantities which can be measured by an instrument and by means of which we can describe the laws of physics are called physical quantities. Till class X we have studied many physical quantities.
For example, length, velocity, acceleration, force, time, pressure, mass, density etc.
Physical quantities are of three types
Fundamental or Basic Quantities
Derived Quantities
Fundamental (Basic) Quantities
Supplementary Quantities
These are the elementary quantities which covers the entire span of physics.
Any other quantities can be derived from these. All the basic quantities are chosen such that they should be different, that means independent of each other. (i.e., distance, time and velocity cannot be chosen as basic quantities as v = d t ). An International Organization named CGPM: General Conference on weight and measures chose seven physical quantities as basic or fundamental.
These are the elementary quantities (in our planet) that’s why chosen as basic quantities.
In fact any set of independent quantities can be chosen as basic quantities by which all other physical quantities can be derived.
i.e.,
Can be chosen as basic quantities (on some other planet, these might also be used as basic quantities) But (L) Length (A) Area (v) Velocity cannot be used as basic quantities as Area = (Length)2 so they are not independent.
Derived Quantities
Physical quantities which can be expressed in terms of basic quantities (M, L, T....) are called derived quantities. i.e., Momentum
P = mv
= (m) displacemen t time = ML T M1 L1 T – 1
Here [M1 L1 T –1] is called dimensional formula of momentum, and we can say that momentum has 1 Dimension in M (mass) 1 Dimension in L (meter) and –1 Dimension in T (time)
The representation of any quantity in terms of basic quantities (M, L, T....) is called dimensional formula and in the representation, the powers of the basic quantities are called dimensions.
Supplementary Quantities
Besides seven fundamental quantities two supplementary quantities are also defined. They are
• Plane angle (The angle between two lines)
• Solid angle θ
DimenSionS
• Height, width, radius, displacement etc. are a kind of length. So we can say that their dimension is [L] [Height]
[L] [Width] [Radius] [Displacement]
here [Height] can be read as “Dimension of Height”
• Area = Length × Width
[Area] = [Length] × [Width]
= [L] × [L]
= [L2]
For circle Area = pr 2
[Area] = [p] [r2]
= [1] [L2]
= [L2]
Here p is not a kind of length or mass or time so p shouldn’t affect the dimension of Area.
Hence its dimension should be 1 (M0L0T 0) and we can say that it is dimensionless. From similar logic we can say that all the numbers are dimensionless.
[200]
[M 0 L0 T 0 ] = 1 Dimensionless [–1] [3] 1 2
• [Volume] = [Length] × [Height]
= L × L × L
= [L3] For sphere Volume = 4 3 pr 3
[Volume] = 4 3 p ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ [r3]
= (1) [L3] = [L3]
So dimension of volume will be always [L3] whether it is volume of a cuboid or volume of sphere.
Dimension of a physical quantity will be same, it doesn’t depend on which formula we are using for that quantity.
• Density = mass volume
[Density] = [mass] [volume] = M L3 = [M1L– 3]
• Velocity (v) = displacemen t time
[v] = [] displacemen t [time] = L T = [M0L1T –1]
• Acceleration (a) = dv dt
[a] = dv dt → → kind of velocity kind of time = LT T LT = 1 2
• Momentum (P) = mV
[P] = [M] [V]
= [M] [LT –1]
= [M1L1T –1]
• Force (F) = ma
[F ] = [m] [a]
= [M] [LT –2]
= [M1L1T –2]
• Work or Energy = force × displacement [Work] = [force] [displacement]
= [M1L1T –2] [L]
= [M1L2T –2]
• Power = work time [Power] = [work] [time] = MLT T 12 2 = [M1L2T – 3]
If light of frequency u is falling, energy of a photon is given by E = hu
Here h = Planck’s constant
[E] = [h] [u]
u = frequency = 1 Time Period
⇒ [u] = 1 [TimePeriod] = 1 T ⎡ ⎣ ⎢ ⎤ ⎦ ⎥
so M1L2T – 2 = [h] [T –1]
[h] = M1 L2 T –1
Some Special Features of Dimensions
Suppose in any formula, (L + a) term is coming (where L is length). As length can be added only with a length, so a should also be a kind of length.
So [a] = [L]
Similarly, consider a term (F – b) where F is force. A force can be added or subtracted with a force only and give rise to a third force. So b should be a kind of force and its result (F – b) should also be a kind of force.
A third force and its dimension will also be M1L1T 2 β β should be a kind of force ⇒ [ ] = M1L1T 2
Rule No. 1: One quantity can be added / subtracted with a similar quantity only and give rise to the similar quantity.
SolveD exAmPleS
1. a t 2 = Fv + b x 2
Find dimension formula for [a] and [b] (here t = time, F = force, v = velocity, x = distance)
Solution:
Since [Fv] = M1L2T –3 , so b x 2 ⎡
⎥ should also be M1L2T –3
b x 2 = M1 L2T –3
[b] = M1L4T –3
Rule No. 2: Consider a term sin (q)
Here q is dimensionless and sinq Perpendicular Hypoteneous
is also dimensionless.
Whatever comes in sin (......) is dimensionless and entire [sin (.......)] is also dimensionless.
sin(- - -)
Dimensionless
Similarly:
Dimensionless
cos(- - -)
Dimensionless Dimensionless
- -)
Dimensionless Dimensionless
⎣ ⎢
⎥ b 2 will also have dimension M1L2T –3
and Fv x +
a t 2 = M1L2T –3 [a] = M1L2T –1
2. For n moles of gas, Van der Waal’s equation is
– b) = nRT
Find the dimensions of a and b, where P is gas pressure, v = volume of gas T = temperature of gas
Solution:
⎟ 2 (V – b) = nRT
should be a should be a kind of pressure kind of volume
So [] [] a V 2 = M1L– 1T –2
So [b] = L3
[] [] a L3 2 = M –1 L–1 T –2 ⇒
[a
Dimensionless
Dimensionless
(- - -)
Dimensionless Dimensionless
3. a = F V 2 sin (bt)
e(- - -)
(here v = velocity, F = force, t = time)
Find the dimension of a and b
Solution: sin ( t) β
Dimensionless
Dimensionless
[ ] [t] = 1
] = [ T 1]
4. a = Fv 2 2b loge 2 2 pb v ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ where F = force, v = velocity
Find the dimensions of a and b
Solution:
Uses of Dimensions
(1) To Check the Correctness of the Formula
If the dimensions of the L.H.S and R.H.S are same, then we can say that this equation is at least dimensionally correct. So this equation may be correct.
But if dimensions of L.H.S and R.H.S is not same then the equation is not even dimensionally correct. So it cannot be correct.
ei A formula is given centrifugal force
Fe = mv r 2 (where m = mass, v = velocity, r = radius)
we have to check whether it is correct or not.
Dimension of L.H.S is
Dimension of R.H.S is
So this equation is at least dimensional correct.
⇒ We can say that this equation may be correct.
Pr = 3 2 22
tx p (where F = force, v = velocity, t = time, x = distance)
Solution:
Dimension of L.H.S = [Pr] = M1L–1T –2
Dimension
5. Check whether this equation may be correct or not
Dimension of L.H.S and R.H.S are not same. So the relation cannot be correct.
Sometimes a question is asked which is beyond our syllabus, then certainly it must be the question of dimensional analyses.
6. A Boomerang has mass m surface Area A, radius of curvature of lower surface = r and it is moving with velocity v in air of density r. The resistive force on it should be r
Solution: Only C is dimensionally correct.
(2) We can Derive a New Formula Roughly
If a quantity depends on many parameters, we can estimate, to what extent, the quantity depends on the given parameters!
SolveD
exAmPleS
7. Derive the formula of time period of simple pendulum if
