Spectrum Physics - March 2016 (gnv64).pdf

FROM THE EDITOR'S DESK

EXAM TACTICS:

Tackling JEE Made Easy The present structure of JEE includes a two tier examination system i.e. JEE Mains and JEE Advanced. Let us explore them one by one

JEE MAINS PAPER In Physics, most of the problems are numerical based along with some graphs and other figure based questions. Although, the pattern or frame of the paper is not fixed, but the analysis of past three examinations suggest following weightage (approximately) of topics in the Mechanics

6-12 questions

Fluid Mechanics

1-2 questions

Heat and Thermodynamics

3-5 questions

Waves and Oscillations

2-4 questions

Electrostat and Electric Current

2-11 questions

Magnetism and Magnetic Effects of Current

2-5 questions

Optics

3-6 questions

EMI and AC

2-4 questions

Modern Physics

2-5 questions

Miscellaneous

2-5 questions

The class coverage wise analysis of 2015 JEE Mains suggests 44% questions from class 11th syllabus and 56% questions from class 12th syllabus. The difficulty level wise analysis of 2015 JEE Mains paper suggests the presence of 26% easy questions, 16% tough questions and 58% medium level questions.

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MANTRAS TO CRACK JEE MAINS The two tier examination pattern was introduced in 2013. The weightage of board percentage in final merit was one of the features of this system. The JEE Mains paper of this examination system is comparable with the IIT JEE screening examination of late nineties and early twenties. For successfully facing this kind of examination try to focus on following points: • One must prepare for final exam (JEE Advanced) only, i.e. do not target Mains examinations first. In my opinion, “Chote exams target karne se kabhi bhi bade results nahin milte”, i.e. targeting smaller exams never gives big results. If you consider JEE Advanced as your target then Mains will be cleared automatically.

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In my opinion, Mains is only a screening to differentiate between serious and non-serious aspirants. Thus, if you are among the serious lot than Mains will be of no problem to you. The analysis of 2013, 2014 and 2015 JEE Mains paper clearly suggests the importance of class 11th syllabus for this paper, i.e. 46% in Physics, 33% in Chemistry and 57% in Mathematics. So, a real grip over class 11th syllabus is required to succeed in JEE Mains paper. Nobody can deny the importance of NCERT Textbooks for this examination. Thus, a good command over NCERT textbooks is required for a great performance in this examination. For capturing NCERT line by line very recently, I saw Master the NCERT. A book, in which objective type questions are framed on each and every line of NCERT. This book can be very helpful to get success in this examination. Choose offline or online paper in accordance of your comfort zone and fine preparation. Both the examinations have the own advantages or disadvantages. The difficulty levels are also more or less same in both the formats of JEE Mains but on analysis it is quite visible that the online paper is framed on a slightly tougher side. Although, the questions paper contains all the three subjects and merit formation includes total marks obtained in the paper but remember subject wise cut off also exists in JEE. So, a balance in preparation is required subject wise to save you from such a subject wise cut off. As all the questions are single answer type MCQs hence, use elimination technique in those questions where you don’t have clue about the correct option, but you have clue about incorrect options. Eliminate 3 incorrect options to get the most probable correct option. As negative marking is there, hence be cautious in answering the questions. Do not take so much of risks.

JEE ADVANCED PAPER The JEE Advanced paper is based upon old IIT JEE paper till 2013. It has 2 papers in total, i.e. first in the morning session (9 am to 12 pm) and second in the afternoon session (2 pm-5pm). Both the papers contain 20 questions per subject, i.e. total 60 questions per paper. The question paper pattern is not at all fixed.

In 2015 Advanced paper following pattern was followed:

PAPER 1 • Section 1 with 8 Integer Type Questions • Section 2 with 10 MCQs having One or More than One Correct Option • Section 3 with 2 Matrix Match Questions. PAPER 2 • Section 1 with 8 Integer Type Questions • Section 2 with 8 MCQs having One or More than One Correct Option • Section 3 with 2 Paragraphs having 2 questions each. In both the papers, section 2 and 3 had a negative marking of 50%. However, in 2014 JEE Advanced questions paper following pattern was followed :

PAPER 1 • 10 Multiple Choice Questions which are one or more than one answer correct. • 10 Integer Type Questions. All the 20 questions with 3 marks each with no negative marking. PAPER 2 • 10 Multiple Choice Questions with only one answer correct. • 3 Paragraphs with 2 question each having only One Answer Correct. • A Matching Type MCQs with one answer correct. • All the questions with 3 marks each and there is the negative marking of 1/3, i.e. +3 for correct answer and –1 for wrong answer. Thus, the uniqueness of JEE Advanced paper lies in 2 facts i.e. Its surprising element Every year JEE Advanced paper incorporates one or more surprise elements e.g. in 2014, negative marking in paper 2 was the surprised element. Similarly, the frame of question paper was changed in 2013 as compared to the one seen in 2012 IIT-JEE paper. Its variety of questions In both 2015 and 2014 JEE Advanced paper following type of questions were asked • Single answer correct (MCQs) • More than one answer correct (MCQs) • Integer type question • Linked comprehension type • Matching type The matrix match type which was the usual component of IIT JEE papers before 2013, become reintroduced in 2015. Likewise comprehension linked questions were 2 per passage in 2015 and 2014, while it was 3 per passage in IIT JEE before 2013.

So, we can say the nature of JEE Advanced paper is very dynamic. No pattern is fixed for the paper. The syllabus wise distribution of marks is more or less same as that seen in JEE Mains paper.

MANTRAS TO CRACK JEE ADVANCED I personally feel that JEE Advanced is one of the toughest exam, among those of same nature we have worldwide. I also believe in the fact that more organised examinations are always easier to crack. So, although it is toughest pattern or frame wise, but if we planned scientifically than we can hit the bull’s eye. For scientific planning following points must be kept in mind : • Dedicate your 2 years study to this particular exam only with keeping in mind that the board weightage is one of the considerations in this examination for final merit hence a dual target planning has to be executed as given in chapter 5. • Phase wise learning and practice is a must to crack these kinds of examinations. After having the analysis of previous year exams some conclusions can be drawn easily like, those about (i) the weightage of class 11th syllabus. (ii) the difficulty level distribution among the questions of previous year papers. (iii) the topic wise, question wise distribution of questions (iv) the topic wise, marks wise distribution Such conclusions will help in strategy development according to your strength and weaknesses. • If you will be approaching phase wise towards this examination, then it is very clear that till Mains you would be having very less opportunities to practice different question types which are the features of JEE Advanced only. Utilise the period between JEE Mains and JEE Advanced for this purpose. A mass of students waste their some of this precious time in the wait of Mains result. Don’t ever commit this kind of mistake. Be optimistic, if you have JEE advanced as the target from day 1 in your mind than you would clear JEE Mains with surety. • The surprise attack is one of the most lethal weapon of this examination, hence, those aspirants who don’t have any kind of premature frame about the pattern of question paper in their mind, has better chances of hitting the target. So, never create the premature frame of question paper in your mind. • The large time span (9am to 5pm) is a very crucial factor in this examination. Such a large time span creates tiredness of all kind (i.e. physical as well as mental) in any person. So during the gap of 2 hrs between Paper 1 and 2 try to relax as much as you can. Remember! The more you relax, better you perform. Last but not the least, a balance between the approaches towards Subjective and Objective formats is required to succeed in IIT as a whole as in present format the final merit also includes the weightage of your Board examinations. Always keep in mind the words of Sir Winston Churchill

“The pessimist sees difficulty in every opportunity. The optimist sees opportunity in every difficulty.”

Formulae at a Glance Gravitation ●

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where, h is height of a satellite. 2π Time period of a satellite, T = R

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Mechanical Properties of Fluids ● ● ● ●

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( R + h )3 g 1/ 3

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⎡T 2 R2 g ⎤ Height of satellite above the earth’s surface, h = ⎢ −R 2 ⎥ ⎣ 4π ⎦ − GMm Total energy of satellite, E = PE + KE = 2( R + h ) where, M is mass of the earth and m is the mass of the satellite. GMm Binding energy of satellite, − E = ( R + h) 2 GM Escape speed, ve = = 2 gR R

Mechanical Properties of Solids ●

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1 1 F × ΔL = × Load × Extension 2 2 Relation between Y , B,G and σ (i) y = 3 B (1 − 2 σ) (ii) y = 2 G (1 + σ) 3B − 2G 9 1 3 (iii) σ = (iv) = + 2G + 6B y B G Thermal stress = y × Thermal strain = y ∝ Δθ where, α is the coefficient of linear expansion of the solid rod. =

G m1 m 2 Newton’s law of gravitation, F = , where G is the universal r2 gravitational constant, m1 and m 2 are point mass bodies. GM Acceleration due to gravity, g = 2 , where M and R are the mass R and radius of the earth. gR 2 ⎡ 2h⎤ and g ′ = g ⎢1 − Effect of altitude, g ′ = , where h is height R ⎥⎦ ( R + h )2 ⎣ of an object. ⎡ d⎤ Effect of depth, g ′ = g ⎢1 − ⎥, where d is depth of the earth. R⎦ ⎣ Effect of rotation of earth, g ′ = g − Rω 2 cos 2 λ GM Intensity of gravitational field, I = 2 , where distance r from the r centre of the body of mass M. Work done (W ) − GM Gravitational potential, V = = Test mass ( M 0 ) r Gravitational potential energy, U = Gravitational potential × Mass of − GM the body = ×m r g Orbital speed of a satellite, vO = R R+h

Normal stress F / A Mg l = = Longitudinal strain Δ l / l πr 2 Δ l Normal stress p − pV Bulk modulus, B = = = Volumetric strain − Δ V / V ΔV 1 − ΔV Compressibility = = Bulk modulus pV Tangential stress F/ A FL F Modulus of rigidity, G = = = = Shearing strain Δ L/L AΔL Aθ If a spring or a wire follows Hooke’s law, then spring constant or F YA force constant is given by K = = , where l is the length of wire Δl l and A is the area of cross-section of the wire. The number of atoms having interatomic distance r0 , in length l of a wire is N = l / r0 Lateral strain − ΔR / R − ΔRl Poisson’s ratio σ = = = Longitudinal strain Δl / l R Δl Elastic potential energy stored per unit volume of a strained body, 1 (Stress )2 Y × (Strain )2 u = × Stress × Strain = = 2 2Y 2 where, Y is the Young’s modulus of elasticity of a solid body. Work done in a stretched wire, 1 W = × Stress × Strain × Volume 2

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Young’s modulus, Y =

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● 1 bar = 10 6 dyne/cm 2 1 torr = 1 mm of mercury 1 pascal = N/m 2 Thrust = Pressure × Area = hρg × A Density of substance Relative density of the substance = Density of water at 4 ° C Apparent weight of the body of density ρ when immersed in a liquid of density σ = Actual weight − Upward thrust 1 Bernoulli’s equation is given by p + ρv 2 + ρgh = constant 2 1 where, p = pressure, ρv 2 = kinetic energy per unit volume and 2 ρgh = potential energy per unit volume. The velocity of efflux is given by v = 2 gh where, h = depth of the hole below the free surface. dv Viscous force is given by F = − η A dy where, η = coefficient of viscosity, A = area of layer of fluid in contact dv and = velocity gradient between the layers. dy πpR 4 Poiseuille’s formula V = 8 ηL where, V = volume of liquid coming out of tube per second L = length of the capillary tube R = radius of the capillary tube p = pressure difference across the tube 2 r 2 (ρ − σ)g Terminal velocity, vT = 9η where, r = radius of the body, ρ = density of material, σ = density of fluid, η = coefficient of viscosity. Surface tension, T = F / l where, F = force acting on one side of imaginary line of length l. 2 S cos θ Ascent formula, h = rρg where, S = surface tension of liquid, θ = angle of contact, r = radius of capillary tube, ρ = density of liquid, g = acceleration due to gravity. 2T Excess pressure in a liquid drop of radius r, p = r where, T = surface tension of the liquid drop. 4T Excess pressure in a soap bubble of radius r, p = r Excess pressure in a soap bubble of radius r, when it is inside a 2T liquid, p = r If two droplets of radii r1 , and r2 in vacuum coalesce under isothermal conditions, then the radius of the big drop is given by r = 3 r13 + r23

The viscous force ( F ) acting on a spherical body of radius r moving with velocity v through a medium of coefficient of viscosity η is given by Stoke’s law as F = 6 π η r v R η Critical velocity of a liquid ( vc ) is given by vc = e ρD where, η = coefficient of viscosity of liquid, ρ = density of liquid and D = diameter of tube and Re = Reynold’s number.

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1

General Instructions 1. This test consists of 30 questions. 2. Each question is allotted 4 marks for correct response. 3. 1/4 marks will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated in each question. 4. There is only one correct response for each question. Filling up more than one response in any question will be treated as wrong response and marks for wrong response will be deducted according as per instructions.

1. The speed of ripples (v) on the surface of water depends

on the surface tension (σ), density ( ρ ) and wavelength (λ). The square of speed (v) is proportional to σ (a) ρλ

ρ (b) σλ

λ (c) σρ

(d) ρ λσ

2. Two bodies of masses m1 and m2 are initially at rest at infinite distance apart. They are then allowed to move towards each other under mutual gravitational attraction. Their relative velocity of approaching at a separation distance r is (a) ⎡2G ⎢⎣

1/ 2

( m1 − m2 )⎤ ⎥⎦ r 1/ 2

1/ 2

2G (b) ⎡ ( m1 + m2 )⎤ ⎢⎣ r ⎥⎦ 1/ 2

⎡ ⎤ r (c) ⎢ ⎥ ⎣ 2G ( m1m2 )⎦

2G (d) ⎡ m1 m2 ⎤ ⎣⎢ r ⎦⎥

3. A long horizontal rod has a bead which can slide along its length and initially placed at a distance L from one end A of the rod. The rod is set in angular motion about A with constant angular acceleration α. If the coefficient of friction between the rod and the bead is μ, and gravity is neglected, then the time after which the beads starts slipping is (a)

μ α

(b)

μ α

(c)

1 μα

(d) μα

4. A block of mass m is placed on a smooth wedge of inclination θ. The whole system is accelerated horizontally so that the block does not slip on the

30

wedge. The force exerted by the wedge on the block (g is acceleration due to gravity) will be (a) mg cos θ

(b) mg sin θ

(c) mg

(d)

mg cos θ

5. A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses T (m1 = 0.32 kg) and (m2 = 0.72 kg). Taking T 2 g = 10 m/s , find the work done (in a m1 m2 a joules) by the string on the block of mass 0.36 kg during the first second after the system is released from rest (a) 6 J

(b) 5 J

(c) 8 J

(d) 4 J

6. A body is moved along a straight line by a machine delivering constant power. The distance moved by the body in time t is proportional to (a) t 1 / 2 (c) t

(b) t 3 / 4

3/ 2

(d) t 2

7. The potential energy function for the force between two atoms in a diatomic molecule is approximately a b given by U (x) = 12 − 6 , where a and b are constants x x and x is the distance between the atoms. If the dissociation energy of the molecule is D = [U (x = ∞) − Uat equilibrium ], where D is (a)

b2 6a

(b)

b2 2a

(c)

b2 12 a

(d)

b2 4a

8. Three rods of identical area of cross-section and made

14. Let there be a spherically symmetric charge

from the same metal from the sides of an isosceles triangle ABC, right-angled at B. The points A and B are maintained at temperatures T and 2 T, respectively. In the steady state, the temperature of the point C is TC . T Assuming that only heat conduction takes place, C is T equal to

distribution with charge density varying as 5 r ρ (r)ρ0 ⎛⎜ − ⎞⎟ upto r = R, and ρ (r) = 0 for r > R, where r is ⎝ 4 R⎠ the distance from the origin. The electric field at a distance r (r < R) from the origin is given by

(a)

1

(b)

( 2 + 1)

3 ( 2 + 1)

(c)

1 2 ( 2 − 1)

(d)

1 3 ( 2 + 1)

9. The change in the entropy of a 1 mole of an ideal gas which went through an isothermal process from an initial state (p1 , V1 , T) to the final state (p2 , V2 , T) is equal to (a) zero

(b) R ln T

(c) R ln

V1 V2

(d) R ln

V2 V1

10. Two identical containers A and B with frictionless pistons contain the same ideal gas at the same temperature and the same volume V . The mass of the gas in A is mA and that in B is mB . The gas in each cylinder is now allowed to expand isothermally to the same final volume 2V. The changes in the pressure in A and B are found to be Δp and 1.5 Δp, respectively. Then,

(a) 4 mA = 9 mB (c) 3 mA = 2 mB

(b) 2 mA = 3 mB (d) 9 mA = 3 mB

11. For the circuit shown below E1 = 4.0 V, R1 = 2 Ω, E2 = 6.0 V, R2 = 4 Ω and R3 = 2 Ω. The current I1 is (a) 1.6 A

(b) 1.8 A E1=4 V

(c) 1.25 A

(d) 1.0 A

R1=2 Ω

ρ0r 4ε 0 ρr (c) 0 4ε 0 (a)

between two other fixed similar charges each of magnitude Q placed at a distance 2d apart. The system is collinear as shown in the figure. The particle is now displaced by a small amount x (x < < d) along the line joining the two charges and is left to itself. It will now oscillate about the mean position with a time period (ε0 = permittivity of free space) d Q

E2=6 V

R2=4 Ω

12. A wire of resistance 10 Ω is bent to form a circle. P and Q are P points on the circumference of 3V the circle dividing it into a Q 1Ω quadrant and are connected to a battery of 3 V and internal resistance 1Ω as shown in the figure. The currents in the two parts of the circle are 6 18 A and A 23 23 4 12 (c) A and A 25 25

5 15 A and A 26 26 3 9 (d) A and A 25 25

(b)

13. Two equal negative charge −q are fixed at the fixed

points (0, a) and (0, − a) on the Y-axis. A positive charge q is released from the point (2 a, 0) on the X-axis. The charge q will (a) (b) (c) (d)

Q

q

π Mε 0d π Mε 0d (b) 2 Qq Qq 3

(a) 2

d

2

3

(c) 2

π 3 Mε 0d 3 π 3 Mε 0 (d) 2 Qq Qq d 3

16. The field normal to the plane of a wire of n turns and radius r which carries a current i is measured on the axis of the coil at a small distance h from the centre of the coil. This is smaller than the field at the centre by the fraction 3 h2 2 r2

(b)

2 h2 3 r2

(c)

3 r2 2 h2

(d)

2 r2 3 h2

17. A solenoid has a core of material with relative R3=2 Ω

I2

(a)

(b)

15. A particle of mass M and charge q is at the mid-point

(a) I1

4 π ρ0r ⎛ 5 r⎞ ⎜ − ⎟ 3ε 0 ⎝ 3 R ⎠ 4ρ r 5 r (d) 0 ⎛⎜ − ⎞⎟ 3ε 0 ⎝ 4 R ⎠

⎛5 − r ⎞ ⎟ ⎜ ⎝ 4 R⎠ ⎛5 − r ⎞ ⎟ ⎜ ⎝ 3 R⎠

execute simple harmonic motion about the origin move to the origin and remains at rest move to infinity execute oscillatory but not of simple harmonic motion

permeability 500 and its windings carry a current of1 A. The number of turns of the solenoid is 500 per metre. The magnetisation of the material is nearly (a) 2.5 × 10 3 Am−1 (c) 2.0 × 10 3 Am−1

(b) 2.5 × 10 5 Am−1 (d) 2.0 × 10 5 Am−1

18. A magnet is suspended in the magnetic meridian with an untwisted wire. The upper end of wire is rotated through 180° to deflect the magnet by 30° from magnetic meridian. When this magnet is replaced by another magnet, the upper end of wire is rotated through 270° to deflect the magnet 30° from magnetic meridian. The ratio of magnetic moments of magnets is (a) 1 : 5 (c) 5 : 8

(b) 1 : 8 (d) 8 : 5

19. Each atom of an iron bar ( 5 cm × 1 cm × 1 cm) has a

magnetic moment 18 . × 10 −23 Am −2 . Knowing that the density of iron is 7.78 × 103 kg- m−3 , atomic weight is 56 and Avogadro’s number is 6.02 × 1023 , the magnetic moment of bar in the state of magnetic saturation will be (a) 4.75 Am2 (c) 7.54 Am2

(b) 5.74 Am2 (d) 75.4 Am2

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20. An inductor of inductance L = 400 mH and resistors of

resistances R1 = 2 Ω and R2 = 2 Ω are connected to a battery of emf 12V as shown in the figure. The internal resistance of the battery is negligible. The switch S is closed at t = 0. The potential drop across L as a function of time is

25. Figure shows a circuit in which three identical diodes are used. Each diode has forward resistance of 20 Ω and infinite backward resistance. Resistors R1 = R2 = R3 = 50 Ω. Battery voltage is 6 V. The current through R3 is D1

L

E R1

D2

(a) 50 mA (c) 60 mA

12 −3 t V e t

(b)

−t ⎞ ⎛ (c) 6 ⎜ 1 − e 0. 2 ⎟ V ⎟ ⎜ ⎠ ⎝

(d) 12 e −5 t V

on the capacitor is q. The charge on the capacitor when the energy is stored equally between the electric and magnetic field is q

(c)

2

q

(d)

3

q 3

22. A simple pendulum with bob of mass m and conducting wire of length L swings under gravity through an angle 2θ. The earth’s magnetic field component in the direction perpendicular to swing is B. Maximum potential difference induced across the pendulum is θ

A2 with half-lives of 20 s and 10 s, respectively. Initially, the mixture has 40 g of A1 and 160 g of A2 . The amount of the two in the mixture will become equal after (a) 60 s (c) 20 s

27. Hydrogen atom from excited state comes to the ground

state by emitting a photon of wavelength λ. If R is the Rydberg’s constant, the principal quantum number n of the excited state is (a)

λR λR − 1

(b)

λ λR − 1

(c)

λR 2 λR − 1

(d)

λR λ −1

at the centre is 3 mm. If the speed of light in the material of the lens is 2 × 108 m/s, find the focal length of the lens. (a) 15 cm (c) 30 cm θ (b) BL sin ⎛⎜ ⎞⎟ ( gL ) ⎝2⎠ θ (d) BL sin ⎛⎜ ⎞⎟ ( gL )2 ⎝2⎠

23. An α-particle of 5 MeV energy strikes with a nucleus of uranium at stationary at an scattering angle of 180°. The nearest distance upto which α-particle reaches the nucleus will be of the order of (a) 1 Å

(b) 10 −10 cm

(c) 10 −12 cm

(d) 10 −15 cm

24. When a certain metal surface is illuminated with light

of frequency ν, the stopping potential for photoelectric current is V0 . When the same surface is illuminated by ν V light of frequency , the stopping potential is 0 . The 2 4 threshold frequency for photoelectric emission is ν (a) 6

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ν (b) 3

(b) 80 s (d) 40 s

28. Diameter of a plano-convex lens is 6 cm and thickness

L

h

θ (a) 2 BL sin ⎛⎜ ⎞⎟ ( gL )1/ 2 ⎝2⎠ θ (c) BL sin ⎛⎜ ⎞⎟ ( gL )3/ 2 ⎝2⎠

(b) 100 mA (d) 25 mA

26. A mixture consists of two radioactive materials A1 and

21. For an oscillation of L-C circuit, the maximum charge

(b)

R3

6V

(a) 6 e −5 t V

q 2

R2

D3

+ –

R2

S

(a)

R1

2ν (c) 3

4ν (d) 3

(b) 20 cm (d) 10 cm

29. A car is fitted with a convex side-view mirror of focal length 20 cm. A second car 2.8 m behind the first car is overtaking the first car at a relative speed of 15 m/s. The speed of the image of the second car as seen in the mirror of the first one is 1 m/s 10 (c) 10 m/s (a)

1 m/s 15 (d) 15 m/s (b)

30. In Young’s double slit experiment, the two slits acts as coherent sources of equal amplitude A and wavelength λ. In another experiment with the same set up; the two slits are of equal amplitude A but are incoherent. The ratio of the intensity of light at the mid-point of the screen in the first case to that in the second case is (a) 1 : 2 (c) 4 : 1

(b) 2 : 1 (d) 1 : 1

Answers with Explanation 1. (a) As speed of ripples ( v ) ∝ σ aρb λc Equating dimensions on both sides, we get [M0 L1T −1 ] ∝ [MT −2 ]a [ML−3 ]b [L ]c [M0 L1T −1 ] ∝ [M]a + b [L ]− 3b + c [T]− 2a Equating the powers of M, L and T on both sides, we get a + b = 0,− 3b + c = 1, − 2 a = − 1 Solving above equations, we get a = 1 / 2, b = − 1 / 2, c = − 1 / 2 ∴ v ∝ σ 1/ 2 ρ−1/ 2 λ−1/ 2 σ v2 ∝ ρλ

2. (b) Let velocities of their masses at distance r from each other be v1 and v 2, respectively. By conservation of momentum, m1v1 − m2v 2 = 0 ...(i) ⇒ m1v1 = m2v 2 By conservation of energy, Change in potential energy = Change in kinetic energy G m1m2 1 1 = m1v12 + m2v 22 2 2 r 2 Gm1m2 …(ii) ⇒ m1v12 + m2v 22 = r On solving Eqs. (i) and (ii), we get v1 =

2 Gm22 and v 2 = r ( m1 + m2 )

2 Gm12 r ( m1 + m2 )

∴ Relative velocity of approach 2G vapp = | v1| + | v 2| = ( m1 + m2 ) r

For the condition of equilibrium, g sinθ mg sinθ = ma cos θ ⇒ a = cos θ ∴ Force exerted by the wedge on the block R = mg cos θ + ma sinθ g sinθ ⎞ or R = mg cos θ + m ⎛⎜ ⎟ sinθ ⎝ cos θ ⎠ mg (cos 2 θ + sin2 θ) cos θ mg ⎤ ⎡ R = ⎣⎢ cosθ ⎦⎥ =

5. (c) In the given condition, tension in the string T =

and acceleration of each block ⎛ m − m1 ⎞ ⎛ 072 . − 0.36 ⎞ 10 a=⎜ 2 m/s 2 ⎟g = ⎜ ⎟ × 10 = 3 . + 0.36 ⎠ ⎝ m1 + m2 ⎠ ⎝ 072 Let s is the distance covered by block of mass 0.36 kg in first second. 1 1 10 10 m s = ut + at 2 ⇒ 0 + ⎛⎜ ⎞⎟ × 12 = 2 2⎝ 3⎠ 6 10 = 8J ∴ Work done by the string (W ) = Ts = 4.8 × 6

6. (c) As, we know power

⇒

For critical condition, Frictional force provides the centripetal force, i.e.

⇒

P P v2 dt = vdv ⇒ ×t = 2 m m 2P⎞ v = ⎛⎜ ⎟ ⎝ m⎠

⇒ Now, s = L

mω 2L = μR = μm × at = μLmα ⇒ m (αt )2 L = μmLα Time taken by bead to start slipping, μ t = ⇒ α

mg sin θ θ

mg

mg cos θ + ma sin θ

∫ v dt = ∫

∴

2P⎞ s = ⎛⎜ ⎟ ⎝ m⎠

⇒

s ∝ t 3/ 2

1/ 2

⎛ 2P⎞ ⎟ ⎜ ⎝ m⎠

1/ 2

t 1/ 2 dt

⎡ 2 t 3/ 2 ⎤ ⎢ 3 ⎥ ⎣ ⎦

a b − 6 x12 x dU 12 a 6b F =− = 13 − 7 dx x x

[As, ω = αt]

ma cos θ

ma

(t )1/ 2

U ( x) =

pseudo force (ma) works on a block towards right.

θ

1/ 2

7. (d) As, potential energy function

4. (d) When the whole system is accelerated towards left, then

θ

2 m1m2 . 2 × 0.36 × 072 ⋅g = × 10 = 4.8 N . 108 ( m1 + m2 )

dv P = Fv = mav = m ⎛⎜ ⎞⎟ ⎝ dt ⎠

3. (a) Let the bead starts slipping after time t .

R

[Q sin2 θ + cos 2 θ = 1]

a

⇒

2a x = ⎛⎜ ⎞⎟ ⎝ b⎠

1/ 6

U( x = ∞ ) = 0 −b2 b a = − Uequilibrium = 2 4a ⎛ 2a⎞ ⎛ 2a⎞ ⎜ ⎟ ⎜ ⎟ ⎝ b⎠ ⎝ b⎠ ⎛ − b2 ⎞ b2 . ∴ U ( x = ∞ ) − Uequilibrium = 0 − ⎜ ⎟ = ⎝ 4a ⎠ 4a

33

8. (b) Q TB > TA ⇒ Heat will flow B to A via two paths (i) B to A (ii) and along BCA as shown Rate of flow of heat in path BCA will be same, i.e. Q ⎛Q⎞ = ⎛⎜ ⎞⎟ ⎜ ⎟ ⎝ t ⎠ BC ⎝ t ⎠ CA ⇒

TC 3 = T ( 2 + 1)

⇒

10 = 2.5 Ω and 4 3 Resistance of part PMQ, R 2 = × 10 = 7.5 Ω 4 Resistance of part PNQ, R1 =

(T) A

K ( 2T − TC ) A K (TC − T ) A = a 2a

12. (a) In the following figure,

i2

a√2

a

i

B √2 TB

a

9. (d) The change in entropy of an

ΔQ …(i) T In isothermal process, temperature does not change, i.e. internal energy which is a function of temperature will remains same, i.e. ΔU = 0. First law of thermodynamics gives ΔU = ΔQ − W or 0 = ΔQ − W or ΔQ = W i.e., ΔQ = Work done by gas in isothermal process which went through from ( p1, V1,T ) to ( p2, V2, T ) ⎛V ⎞ …(ii) ΔQ = μRT loge ⎜ 2 ⎟ ⇒ ⎝ V1 ⎠

For 1 mole of an ideal gas, μ = 1, so from Eq. (i) and Eq. (ii), we get ⎛V ⎞ ⎛V ⎞ ΔS = R loge ⎜ 2 ⎟ = R ln ⎜ 2 ⎟ ⎝ V1 ⎠ ⎝ V1 ⎠

10. (c) As, process is isothermal. Therefore, T

is constant, ⎛ p ∝ 1 ⎞ volume is increasing, therefore, pressure will ⎟ ⎜ ⎝ V⎠

decreases. In chamber A

Req =

R1 × R 2 2.5 × 7.5 15 = = Ω R1 + R 2 (2.5 + 7.5) 8

Main current, i =

μ A RT μ A RT μ A RT − = V 2V 2V

μ B RT μ B RT μ B RT − = V 2V 2V μA 1 2 From Eqs. (i) and (ii), we get = = . 3 μ B 15 mA / M 2 = ⇒ ⇒ 3 mA = 2 mB mB / M 3

13. (d) By symmetry of problem, the components of force on Q due to charges at A and B along Y-axis will cancel each other along X-axis will add up and will be along CO. Under the action of this force change q will move towards O. If at any time, charge q is at a distance x from O. Net force on charge q. − qq 1 x Fnet = 2 F cos θ = 2 . × 4 πε0 ( a 2 + x 2 ) ( a 2 + x 2 )1 / 2 2qqx −1 ⋅ 4 πεo ( a 2 + x 2 )3 / 2

...(i)

A –q a

...(ii)

E1=4 V I1

1

R1=2 Ω

F θ

O

B

i1

–q

Therefore, the restoring force Fnet is not linear, motion will be oscillatory (with amplitude 2 a) but not Simple Harmonic Motion. r r 5 r 14. (c) Total charge, q = ∫ ρ dv = ∫ ρ0 ⎛⎜ − ⎞⎟ 4 π r 2dr 0

I2

2 E2=6 V

r

i2

= 4 π ρ0 E= …(i) ...(ii)

=

⎛ 5r 2

∫ ⎜⎝ 0

R2=4 Ω

For Loop (1), −2 i 1 − 2 ( i 1 − i 2 ) + 4 = 0 ⇒ 2 i1 − i 2 = 2 For Loop (2), −2 ( i 1 − i 2 ) + 4i 2 − 6 = 0 ⇒ − i 1 + 3i 2 = 3 On solving Eq (i) and Eq (ii), we get i 1 = 18 . A

34

C

2a

(i1–i2) R3=2 Ω

q

F

x

a

11. (b) Applying Kirchhoff’s law for the loops (1) and (2) as shown in the figure.

24 18 6 A − = 23 23 23

and i 2 = ( i − i 1 ) =

In chamber B

1.5 Δp = pi − pf =

24 3 A = ⎛ 15 + 1⎞ 23 ⎟ ⎜ ⎠ ⎝ 8

⎛ R 2 ⎞ 24 ⎛ 7.5 ⎞ 18 So, i 1 = i × ⎜ A ×⎜ ⎟ = ⎟ = ⎝ R1 + R 2 ⎠ 23 ⎝ 2.5 + 7.5 ⎠ 23

i.e. Fnet =

Δp = pi − pf =

Q

3 V, 1 Ω

C (TC)

ideal gas ΔS =

M

P i1 N

4

0

−

⎝4

R⎠

⎡ 5r 3 r3 ⎞ r4 ⎤ − ⎟ dr = 4 π ρ0 ⎢ 4R ⎥⎦ R⎠ ⎣ 12

⎡ 5r 3 Kq 1 r4 ⎤ = − . 4 π ρ0 ⎢ 2 2 4R ⎥⎦ r 4 πε0 r ⎣ 12 ρ0r ⎡ 5 r − ⎤ 4 ε0 ⎢⎣ 3 R ⎥⎦

15. (c) Restoring force on displacement of x

19. (c) The number of atoms per unit volume in a specimen is

⎡ Qq ⎡ ⎤ 1 1 Qq ⎤ F =K ⎢ − = KQq ⎢ − 2 2⎥ 2 2⎥ (d + x ) ⎦ (d + x ) ⎦ ⎣ (d − x ) ⎣ (d − x ) ⎡ 4 dx ⎤ 4 dx = K Qq ⎢ 2 = K Qq ⎡ 4 ⎤ if (d > > x ) 2 2⎥ ⎢ ⎣ d ⎥⎦ ⎣ (d − x ) ⎦ 4x = K Qq ⎡ 3 ⎤ ⎢⎣ d ⎥⎦ Acceleration, a =

4KQq F 4KQq = x ⇒ ω2 = M Md 3 Md 3

Time Period, T =

2π = 2π ω

Md 3 π 3Md 3ε0 =2 Qq 4KQq

μ 0 2 πin μ 0 n1 × = ⋅ 4π 2 r r Field at a distance h from the centre 2 π nir 2 μ B2 = 0 . 2 4 π ( r + h2 )3/ 2

16. (a) Field at the centre B1 =

μ = 0⋅ 2

nir 2 ⎛ h2 ⎞ r ⎜1 + 2 ⎟ r ⎠ ⎝ 3

⎛ 3 h2 ⎞ = B1 ⎜ 1 − . 2 ⎟ 2 r ⎠ ⎝

3/ 2

⎛ h2 ⎞ = B1 ⎜ 1 + 2 ⎟ r ⎠ ⎝

ρ NA A For iron ρ = 7.8 × 10 3 kgm−3 NA = 6.02 × 10 26 /kg mol A = 56 7.8 × 10 3 × 6.02 × 10 26 n= = 8.38 × 10 28 m−3 ⇒ 56 Total number of atoms in the bar is N0 = nV = 8.38 × 10 28 × ( 5 × 10 −2 × 1 × 10 −2 × 1 × 10 −2 ) N0 = 419 . × 10 23 The saturated magnetic moment of bar = 419 . × 10 23 × 18 . × 10 −23 2 = 7.54 Am n=

20. (d) Electric field across BC, E LdI1 + R 2 I2 dt 12 V I2 = I0 (1 − e −t /t 0 ) E 12 I0 = = = 6A R2 2 =

−3 / 2

400 × 10 −3 L = R 2 = 02 . s ∴ I2 = 6 (1 − e −t / 0. 2 ) Potential drop across L = E − R 2 I2 = 12 − 2 × 6 (1 − e −t / 0. 2 ) = 12 e −t / 0. 2 = 12 e −5 t V τ = t0 =

[By Binomial theorem]

Hence, B2 is less than B1 by a fraction =

3h 2 . 2r2

17. (b) Here, n = 500 turns/m I = 1 A , μ r = 500 As μ r = 1 + χ, where χ is the magnetic susceptibility of the material or χ = (μ r − 1) Magnetisation (M) = χ H = (μ r − 1) ⋅ H = ( 500 − 1) × 500 Am −1 = 499 × 500 Am −1 = 2.495 × 105 Am −1 ~ − 2.5 × 105 Am −1

18. (c) Let M1 and M 2 be the magnetic moments of magnets and H is the horizontal component of earth’s field. We have τ = MHsinθ. If φ is the twist of wire, then τ = c φ, c being restoring couple per unit twist of wire. ⇒ cφ = MH sinθ π rad Here, φ1 = (180° − 30° ) = 150° = 150 × 180 π rad φ 2 = (270° − 30° ) = 240° = 240 × 180 So, c φ1 = M1H sinθ [For deflection θ = 30° of I magnet] c φ 2 = M 2H sinθ [For deflection θ = 30° of II magnet] φ M Dividing 1 = 1 φ2 M 2 π 150 × M1 φ 180 = 15 = 5 = 1 = M 2 φ 2 240 × π 24 8 180 ⇒ M1 : M 2 = 5 : 8

A

B I2

I1

L R1 R2

S D

C

21. (b) Maximum energy stored in the capacitor q2 2C The energy is stored equally in electric and magnetic fields. 1 ⎛ q2 ⎞ So, energy in electric field, E = ⎜ ⎟ 2 ⎝ 2C ⎠ Umax =

q′ 1 q2 = 2C 2 2C Net charge inside a capacitor, q ⇒ q′ = 2

Now,

22. (a) From the given figure, ⇒ h = L (1 − cos θ) Maximum velocity at equilibrium is given by v 2 = 2 gh = 2 gL (1 − cos θ) θ = 2 gL ⎛⎜ 2 sin2 ⎞⎟ ⎝ 2⎠ θ 2 Thus, maximum potential difference

⇒

…(i) θ

L

h

v = 2 gL sin

θ Vmax = B × L = B × 2 gL sin L 2 θ = 2 BL sin⎛⎜ ⎞⎟ ( gL )1/ 2 ⎝2⎠

35

23. (c) At closest distance of approach

28. (c) According to lens formula,

i.e. Kinetic energy = Potential energy 1 ( Ze )( Ze ) 5 × 1016 × 16 . × 10 −19 = × 4 πε0 r

⎡ 1 1 1⎤ = (μ − 1) ⎢ − ⎥ f ⎣ R1 R 2 ⎦ The lens is plano-convex, i.e. R1 = R and R 2 = ∞ 1 (μ − 1) R Hence, = ⇒f = ( μ − 1) f R

For uranium, Z = 92, so r = 5.3 × 10 −12 cm

24. (b) According to Einstein’s photoelectric emission Kmax = hν − φ 0 where, ν is the incident frequency and φ 0 is the work done of the metal. As, Kmax = eV0, where V0 is the stopping potential. …(i) Therefore, eV0 = hν − φ 0 ν V0 …(ii) and = h − φ0 e 4 2 From Eqs. (i) and (i), we get hν φ 0 hν − φ 0 hν hν or φ 0 = − = = − 4 4 2 4 2 4 hν 3 hν or φ 0 = φ0 = ⇒ 3 4 4 Therefore, threshold frequency (ν 0) hν 1 ν φ × = . ν0 = 0 = h 3 h 3

25. (a) As diode is conducting in forward bias condition and now conducting in reverse bias condition. Diode D1 is in forward bias, and diode D2 is in forward bias but diode D3 is reverse bias. So, the figure can be drawn as Here, 20 Ω , R1 Ω and R 3 Ω are in series. Equivalent resistance = 50 + 50 + 20 = 120 Ω R1 20 Ω V 6 I= = ∴ R 120 1 A = + – 20 R3 6V ⇒ I = 50 mA

Speed of light in medium of lens, v = 2 × 10 8 m/s c 3 × 10 8 3 . μ = = = = 15 v 2 × 108 2 A R r O

If r is the radius and Y is the thickness of lens (at the centre), the radius of curvature R of its curved surface in accordance with the figure is given by R 2 = r 2 + ( R − y)2 ⇒ r 2 + y 2 − 2 Ry = 0 ( 6 / 2 )2 r2 Neglecting y 2, we get R = = = 15 cm 2 y 2 × 0.3 Hence, focal length of the lens, 15 f = = 30 cm 15 . −1

29. (b) According to mirror formula, 1 1 1 + = v −280 20 1 1 1 1 14 + 1 = + ⇒ = v 20 280 v 280 280 ⇒ v = 15 2.8

we have

⇒

27.

⎞

− 2⎟ ⎜ 40 160 = ⇒ 2 t / 20 = 2 ⎝ 10 ⎠ (2 ) t / 20 (2 )t /10 t t t t t = −2⇒ − = − 2 or = 2 ⇒ t = 40 s 20 10 20 10 20

⎡1 1 1⎤ (a) According to Rydberg’s formula, = R ⎢ 2 − 2 ⎥ λ ni ⎦ ⎣ nf Here, nf = 1, ni = n 1− 1 1 = R ⎡ 2 2⎤ ⇒ ∴ ⎢⎣ 1 n ⎥⎦ λ

1 1 = R ⎡1 − 2 ⎤ ⎢⎣ λ n ⎥⎦

Multiplying Eq. (i) by λ on both sides, we get 1 1 1 1 1 1 = λR ⎡1 − 2 ⎤ ⇒ = 1− 2 ⇒ 2 = 1− ⎢⎣ λR λR n ⎥⎦ n n ⇒

36

15

N01 N02 , i.e. N1 = N2 , N2 = (2 ) t / 20 (2 ) t /10 ⎛t

⇒

λR − 1 1 = ⇒n = λR n2

λR λR − 1

(R–y) y

C

26. (d) According to radioactive decay of half-life of an element, N1 =

B

m

f

=

20

cm

2

v v I = − ⎛⎜ ⎞⎟ v 0 ⎝u⎠ ∴ Speed of the image 2

⎛ 280 ⎞ 1 −15 m/s m/s = − vI = − ⎜ ⎟ 15 = 15 15 × 15 ⎝ 15 × 280 ⎠

30. (b) Resultant intensity, I = I1 + I2 + 2 I1 I2 cos ϕ …(i)

At central position with coherent source (and I1 = I2 = I0 ) …(i) Icoh = 4I0 In case of incoherent at a given point, φ varies randomly with time so (cos φ)av = 0 ...(ii) ∴ Iincoh = I1 + I2 = 2 I0 Icoh 2 Hence, = Iincoh 1

More than 10,000 Practice Questions as a whole

24

arihantbooks.com

2 Questions to Measure Your Problem Solving Skills General Instructions 1. This test consists of 30 questions. 2. Each question is allotted 4 marks for correct response. 3. 1/4 marks will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated in each question. 4. There is only one correct response for each question. Filling up more than one response in any question will be treated as wrong response and marks for wrong response will be deducted according as per instructions.

1. The SI unit of energy is J = kgm2 s - 2 , that of speed v is

ms - 1 and of acceleration a is ms - 2 . Which one of the formula for kinetic energy given below is correct on the basis of dimensional arguments? [Given, m stands for mass of the body] (b) K = ma 1 (d) K = mv 2 2

(a) K = m v 1 mv 2 + ma 2 2 2

(c) K =

2. A body of mass 5 ´ 10 -3 kg is launched upon a rough inclined plane making an angle of 30° with the horizontal. Find the coefficient of friction between the body and the plane if the time of ascent is half of the time of descent. (a) 0.346

(b) 0.436

(c) 0.463

(d) 0.364

3. A uniform chain of length L and mass M over hangs a horizontal table with its two-third part on the table. The friction coefficient between the table and the chain is m. The work done by the friction during the period, the chain slips off the table is (a) -

1 2 4 6 m MgL (b) - m MgL (c) - m MgL (d) - m MgL 4 9 9 7

4. A neutron travelling with velocity u and kinetic energy K collides head on elastically with the nucleus of an atom of mass number A at rest. The fraction of its kinetic energy retained by the neutron even after the collision is æ1 - A ö (a) ç ÷ è A + 1ø

38

2

æ A + 1ö (b) ç ÷ è A - 1ø

2

A - 1ö (c) æç ÷ è A ø

2

A + 1ö (d) æç ÷ è A ø

2

5. One end of a uniform rod of mass m 1 , uniform area of cross-section A is suspended from the roof and the mass m2 is suspended from the other end. What is the stress at the mid-point of the rod?

( m1 + m2 ) . g A ( m1 / 2 ) + m2 )ù é (c) .g úû êë A

(a)

( m1 - m2 ) . g A m1 + ( m2 / 2 )ù é (d) .g úû êë A (b)

6. Two moles of an ideal monoatomic gas at 27°C occupies a volume of V. If the gas is expanded adiabatically to the volume 2V, then the work done by 5 the gas will be æç Y = , R = 8.31 J / mol-K ö÷ è ø 3 (a) + 2767.23 J

(b) 2627.23 J

(c) 2500 J

(d) - 2500 J

7. Use Moseley’s law with b = 1 to find the frequency of

the Ka X-rays of La (Z = 57) if the frequency of . ´ 1018 Hz Ka X-rays of Cu(Z = 29) is known to be 188

(a) 7.52 ´ 1018 Hz (c) 8.51 ´ 1019 Hz

(b) 3.25 ´ 1016 Hz (d) 9.1 ´ 1015 Hz

8. A small particle of mass m is

projected at an angle q with the X-axis with an initial velocity v0 in the X-Y plane as shown in v sin q , the figure. At a time t < 0 g the angular momentum of the particle is

Y v0

θ

X

1 mg v 0 t 2 cos q $i 2 (c) mg v t cos q k$ (a)

0

(b) - mg v 0 t 2 cos q $j (d) -

æ 3ö 1 -1 æ 2 ö -1 æ 1 ö (a) sin- 1 æç ö÷ (b) sin- 1 ç ÷ (c) sin ç ÷ (d) sin ç ÷ è2 ø è 3ø è 3ø è 2 ø

1 mgv 0 t 2 cos q k$ 2

16. When two tunning forks (fork 1 and fork 2) are

9. A ball falls on an inclined plane of inclination q from a height h above the point of impact and makes a perfectly elastic collision where will it hit the inclined plane again (a)

8h sin q

(b) 4h sin q

(c) 8h sin q

(d)

4h sin q

(a) 200 Hz (c) 196 Hz

10. A thin uniform annular disc of mass M has outer radius 4R and inner radius 2R. The work required to take a unit mass from point P on its axis to infinity is

the origin under the action of an electric field, E = E0 $i and B = B0 $i with a velocity, v = v0 $j. The speed of the particle will becomes 5 / 2 v0 after a time

4R

- 2GM ( 4 2 - 5) 7R 2GM (d) ( 2 - 1) 5R

2GM ( 4 2 - 5) 7R GM (c) 4R

(a)

(a)

(b)

from the same metal form the sides of square. The temperature of two diagonally opposite points are T and 2 T respectively, in the steady state. Assuming that only heat conduction takes place, what will be the temperature difference between other two points? ( 2 + 1) T 2

(b)

2 . T (c) 0 ( 2 + 1)

(d) None of these

12. An object 2.4 m in front of a lens forms a sharp image on a film 12 cm behind the lens. A glass plate 1 cm thick of refractive index 1.50 is interposed between lens and film with its plane faces parallel to film. At what distance (from lens) should object shifted to be in sharp focus on film? (a) 7.2

(b) 2.4

(c) 3.2

(d) 5.6

13. A sphere and a cube of same material and same volume. One heated upto the same temperature and allowed to cool in the same surroundings. The ratio of the amounts of radiation emitted will be (a) 1 : 1

(b)

4p :1 3

1/ 3

p (c) æç ö÷ è 6ø

:1

(d)

1 æ 4p ö ç ÷ 2è 3 ø

2/ 3

:1

14. In Young’s double slit experiment, when violet light of a wavelength 4358 Å is used, then 84 fringes are seen in the field of view, but when sodium light of certain wavelength is used, then 62 fringes are seen in the field of view, the wavelength of a sodium light is (a) 6893 Å

15. A

(b) 5904 Å

(c) 5523 Å

mv 0 qE

(b)

mv 0 2qE

(c)

3mv 0 2qE

(d)

5mv 0 2qE

18. The current in the primary circuit of a potentiometer

11. Four rods of identical cross-sectional area and made

(a)

(b) 202 Hz (d) 204 Hz

17. A particle of charge q and mass m starts moving from

4R 2R

sounded simultaneously, 4 beat s - 1 are heard. Now, some tape is attached on the prong of the fork 2. When the tunning forks are sounded again, 6 beats s -1 are heard. If the frequency of fork 1 is 200 Hz, then what was the original frequency of fork 2?

(d) 6429 Å

transparent solid cylinderical rod has a θ refractive index of 2 / 3 . It is surrounded by air. A light ray is incident at the mid-point of one end of the rod as shown in the figure. The incident angle q for which the light ray grazes along the wall of the rod is

is 0.2 A. The specific resistance and cross-section of the potentiometer wire are 4 ´ 10 - 7 W m and 8 ´ 10 - 7 m2 , respectively. The potential gradient will be equal to (a) 0.2 V/m (c) 0.3 V/m

(b) 1 V/m (d) 0.1 V/m

19. Two straight long conductors AOB and COD are perpendicular to each other and carry currents i1 and i2 . The magnitude of the magnetic induction at a point P at a distance a from the point O in a direction perpendicular to the plane ABCD, is

m0 ( i1 + i 2 ) 2 pa m (c) 0 ( i 12 + i 22 )1/ 2 2 pa

(a)

m0 ( i1 - i 2 ) 2 pa i1 . i 2 m (d) 0 2 pa ( i 1 + i 2 )

(b)

20. Two conductors have the same resistance of 0°C, but their temperature coefficients of resistance are a1 and a2 . The respective temperature, coefficients of their series and parallel combinations are nearly

( a1 + a2 ) , ( a1 + a2 ) 2 a . a2 (c) ( a1 + a2 ), 1 ( a1 + a2 ) (a)

( a1 + a2 ) 2 ( a1 + a2 ) ( a1 + a2 ) , (d) 2 2 (b) ( a1 + a2 ),

21. A

parallel plate capacitor C with plates C of unit area and the d R d/3 separation d is filled with a liquid of dielectric constant K = 2, the level of liquid is d/3, initially. Suppose, the liquid level decreases at a constant speed v, the time constant as a function of time t is

6 e 0R 5 d + 3 vt 6 e 0R (c) 5 d - 3vt

(a)

(b) (d)

(15 d + 9vt ) e0R 2d 2 - 3dvt - 9v 2t 2 (15 d - 9vt ) e0R 2d 2 + 3 dvt - 9v 2t 2

39

22. A 10 mF capacitor and a 20 mF capacitor are connected in series across 200 V supply line. The charged capacitors are then disconnected from the line and reconnected with their positive plates together and negative plates together and no external voltage is applied. What is the potential difference across each capacitor? (a)

800 V 9

(b) 400 V

(c)

800 V 3

27. Two inductors L1 and L2 are connected in parallel and a time varying current flows as shown in the figure. i The ratio of current, 1 at any time is i2

i

V

(a)

Blv Blv ,l= R R Blv (d) l1 = l2 = l = R

(c) 6

1 2qq (a) 4pe0 a 1 2qq (c) 4pe0 a

1 ö æ ÷ ç1 è 5ø 1 ö æ ÷ ç1 + è 5ø

V

R

t I

(c)

(d) t

t

29. The graph given below represents the I-V characteristics of a Zener diode. Which part of the characteristic curve is most relevant for its operation as a voltage regulator? I (mA)

a Forward bias

Reverse bias VZ

d e Current

(d) 8

1 2qq 4pe0 a

L22 ( L1 + L2 )2

D

t

c

b

V

I (μA)

(a) ab (c) cd

(b) bc (d) de

30. A fully charged capacitor C with initial charge q0 is connected to a coil of self-inductance L at t = 0. The time at which the equation is stored equally between the electric and magnetic field is

(b) zero (d)

(d)

(b)

26. Two positive charges of magnitude q are placed at the end of a side 1 of a square of side 2a. Two negative charges of the same magnitude are kept at the other corners. Starting from rest, if a charge q moves from the middle of side 1 to the centre of square, its kinetic energy at the centre of square is

L12 ( L1 + L2 )2

I

I

in its volume with a charge density r = kr a , where k and a are constants and r is the distance from its R 1 centre. If the electric field at r = is times that are 2 8 r = R, then the value of a is (b) 4

(c)

(a)

25. A solid sphere of radius R has a charge q distributed

(a) 2

L1 L2

I

I has a sliding connector PQ of v RΩ RΩ RΩ length l and a resistance R W and it I I2 is moving with the Q I1 speed v as shown. The set up is placed in the uniform magnetic field going into the plane of the paper. The three currents I1 , I2 and I are

(b) l1 = - l2 =

(b)

in the figure can act as a rectifier. An alternating current source (V ) is connected in the circuit. The current (I) in the resistor (R) can be shown by

P

24. A rectangular loop

L2 L1

28. A p-n junction (D) shown

(a) equal to 5 W (b) greater than 5 W (c) less than 5 W (d) greater or less than 5 W depending upon its material

Blv 3R 2 Blv 3R

L2

i2

moving coil voltmeter V and a moving coil R ammeter A and resistor A C D R as shown in the figure. If the voltmeter reads 10 V and the ammeter reads 2 A, then R is

Blv ,l= 6R Blv ,l= (c) l1 = l2 = 3R

i

(d) 200 V

23. A candidate connects a

(a) l1 = l2 =

L1

i1

2 ö æ ÷ ç1 è 5 ø

(a)

p 4

LC

(b) p

LC

(c) 2 p LC

(d) LC

Answers 1. (d) 11. (c) 21. (a)

40

2. (a) 12. (d) 22. (a)

3. (b) 13. (c) 23. (c)

4. (a) 14. (b) 24. (c)

5. (c) 15. (d) 25. (a)

6. (a) 16. (c) 26. (a)

7. (a) 17. (b) 27. (a)

8. (a) 18. (d) 28. (b)

9. (c) 19. (c) 29. (d)

10. (a) 20. (d) 30. (a)

Paper 1 One or More Than One Option Correct Type 1. A solid sphere of radius R and density r is attached to one end of a massless spring of force constant k. The other end of the spring is connected to another solid sphere of radius R and density 3r. The complete arrangement is placed in liquid of density 2r and is allowed to reach equilibrium. Choose the correct statement (s). 4pR 3rg 3k 8pR 3rg (b) The net elongation of spring is 3k (c) The light sphere is partially submerged

(a) The net elongation of spring is

4 pR 3rg (d) The elongation of spring due to small sphere is 3 k

2. In a modified Young’s double slit experiment, a monochromatic uniform and parallel beam of light of wavelength 6000 Å and intensity (10/p) W/m2 is incident normally on two circular apertures A and B of radii 0.001 m and 0.002 m, respectively. A perfectly transparent film of thickness 2000 Å and refractive index 1.5 for wavelength 6000 Å is placed in front of aperture A . Calculate the power (in watts) received at the focal spot F of the lens. The lens is symmetrically placed with respect to the L A aperture. Assume that 10% of F the power received by each aperture goes in the original B direction and is brought to the focal point. Also, find the phase difference.

(a) The power received at the focal spot F of the lens is 7 ´ 10 -6 W (b) The value of phase difference is p/3 (c) The power received at the focal spot F of the lens is 9.6 ´ 10 -7 W (d) None of the above

3. Calculate the angular frequency of the system shown in the figure. Friction is absent everywhere and the threads, spring and pulleys are massless. Given that mA = mB = m

A k B

(a) w = 3k / 4m (c) w = 2 k / 3m

(b) w = (d) w =

4k / 5m 6k / 5m

4. A wooden rod weighing 25 N is mounted on a hinge below the free surface of water as shown. The rod is 3 m long and uniform in cross-section and the support is 1.6 m below the free surface. The cross-section of rod is 9.5 ´ 10 -4 m2 in area. The density of water is 1000 kg/m 3 . Assume buoyancy to act at centre of immersion g = 9.8 m/s2 . Also, find the reaction on the hinge in this position. 3m 1.6 m

(a) 1.5 N

(b) 1.1 N

α

(c) 2.6 N

(d) 1.6 N

41

5. Determine

the maximum M horizontal force F that may be R applied to the plank of mass m for m F which the solid sphere does not slip as it begins to roll on the plank. The sphere has a mass M and radius R. The coefficient of static and kinetic friction between the sphere and the plank are m s and m k , respectively. (a) The linear acceleration for sphere will be m s g / M 7 (b) The maximum horizontal force is F = m s g æç M + mö÷ è 2 ø 3 ö æ (c) The maximum horizontal force, F = m s g ç M + m÷ è 2 ø (d) None of the above

6. A plano-convex lens of focal length 20 cm has its place side silvered. (a) The radius of curvature of its curved surface of plano-convex lens is equal to half of radius of curvature of a surface of equi-convex lens of focal length 20 cm (b) An object placed at 15 cm on the axis on the convex side of silvered plano-convex lens gives rise to an image at a distance of 30 cm from it (c) An object placed at a distance of 20 cm on the axis on the convex side of silvered plano-convex lens gives rise to an image at 40 cm from it (d) Silvered plano-convex lens acts as a concave mirror of focal length of 10 cm

7. A particle of mass m and

y

charge -q has been projected from the ground as shown in the figure. Which of the following statement(s) is/are correct?

E (Uniform electric field).

θ

(a) current through r is zero 2 Bw a2 (b) current through r is 5r (c) direction of current in external resistance r is from centre to circumference (d) direction of current in external resistance r is from circumference to centre

10. Two circular plates A and B of a parallel plate air

capacitor have a diameter of 0.1 m and are 2 ´ 10 -3 m apart. The plates C and D of similar capacitor have a diameter of 0.12 m and are 3 ´ 10 -3 m apart. Plate A is earthed. Plates B and D are connected together. Plate C is connected to the positive pole of a 120 V battery whose negative pole is earthed. Calculate the potential energy stored in it. (a) 0241 . mJ

(b) 01224 . mJ

(c) 2.43 J

(d) 4.34 mJ

Integer Answer Type 11. What is the velocity (in cm/s) of image in situation as

shown below. (O = object, f = focal length). Object moves with velocity 10 cm/s and mirror moves with velocity 2 cm/s as shown. 20 cm/s

8. Two small balls A and B of mass

O

10 cm/s

H B(3M)

Principal axis

10 cm

F =10 cm

12. A galvanometer has internal resistance of 50 W and current required for full scale deflection is 1mA. The resultant series resistance required is n ´ 104 W as a voltmeter with different ranges. The value of n is G

(a) If collision is perfectly elastic, ball B will rise to a height H/4 (b) If the collision is perfectly elastic, ball A will rise upto height H/4 (c) If the collision is perfectly inelastic, the combined mass will rise to a height H/16 (d) If the collision is perfectly inelastic, the combined mass will rise to a height H/4

42

conducting ring of negligible B resistance and radius a. PQ is a uniform rod of resistance r. It is P Q hinged at the centre of the ring r and rotated about this point in R clockwise direction with a uniform angular velocity w . There is a uniform magnetic field of strength B pointing inwards and r is a stationary resistance. Then,

x

(a) The path of the motion of particle is parabolic (b) The path of the motion of particle is a straight line 2u sin q (c) Time of flight of particle is g (d) Range of motion of the particle can be less than, greater than u 2 sin 2 q or equal to g

M and 3M hang from the ceiling by string of equal length. The M ball A is drawn a side so that it is A raised to a height H. It is then released and collides with ball B. Select the correct option(s).

9. For the given figure, R is a fixed

R1

R2

1V

R3

10 V

100 V

13. Two separate air bubbles (radii 0.004 m and 0.002 m) formed of the same liquid (surface tension 0.07 N/m) come together to form a double r2 bubble. The radius of air bubble is -3 m and the sense of n ´ 10 P2 P curvature of the internal film r1 P1 surface common to both the bubbles. Find the value of n.

14. In a transition from state n to a state of excitation energy 10.19 eV, a hydrogen atom emits a 4890 Å photon. After calculating the binding energy of the initial state. Determine the value of n.

15. A sphere of radius 0.1 m and mass 8 p kg is attached to the lower end of a steel wire of length 5 m and diameter 10 -3 m. The wire is suspended from 5.22 m high ceiling of a room. When the sphere is made to swing as a simple pendulum, it just grazes the floor at its lowest point. Calculate the velocity of the sphere at the lowest position (in m/s). Young’s modulus of steel is 1.994 ´ 1011 N/m2 .

16. A small quantity of solution containing

24

Na radio nuclide (half-life 15 h) of activity/micro curie is injected into the blood of a person. A sample of the blood of volume/cm3 taken after 5 h shows an activity of 296 disintegration min-1 . Determine the total volume of blood (in L) in the body of the person. Assume that the radioactivity solution mixes uniformly in the blood of the person. [1 curie = 3.7 ´ 1010 disintegration s -1 ]

17. A series L-C-R circuit containing a resistance of 120 W has angular resonance frequency 4 ´ 105 rads -1 . At resonance, the voltage across the resistance and inductance are 60 V and 40 V, respectively. So, the frequency is n × 105 rad/s at which the current in the circuit lags the voltage by 45°. The value of n is

18. A silver ball of radius 4.8 cm is suspended by a thread in a vacuum chamber. A ray of ultraviolet light of wavelength 200 nm is incident on the ball for sometime during which a total light energy of 1 ´ 10 -7 J falls on the surface. Assuming that on the average, one photon out of 10000 photons is able to eject a photoelectron, find the value of n, as electric potential is n ´ 10 -1 V at the surface of the ball, assuming zero potential at infinity.

19. Water level is maintained in a cylindrical vessel placed on horizontal floor upto a fixed height H. A tiny hole of area A is punched in the side wall at a height from the bottom of the vessel equal to y. The emerging stream strikes the ground at a horizontal distance (y ¹ 0) from cylinder. To maintain the level in the vessel at H, the 2 gH rate (volume/sec) of addition of water is A . Find n the value of n.

20. A long container has air enclosed inside at room temperature and atmospheric pressure (105 Pa). It has a volume of 20000 cc. The area of cross-section is and force constant of spring is 100 cm2 k spring = 1000 N/ m. We push the right piston isothermally and slowly till it reaches the original position of the left piston which is movable. Final length of air column is found to be 25h cm. Assume that spring is initially relaxed. Find the value of h.

Paper 2 Only One Option Correct Type 1. A person is pulling a mass m from the ground on a rough hemispherical surface upto the top of the hemisphere with the help of a light R inextensible string as μ m shown in the figure given alongside. The radius of the hemisphere is R. Find out the work done by the tension in the string. 1ö æ (a) mgR ç1 + ÷ è mø

(b) mgR (1 - m )

(c) mgR (1 + m )

ö æ1 (d) mgR ç - 1÷ èm ø

2. AB is a cylinder of length 1m fitted with a thin flexible diaphragm C at the middle and the two others thin flexible diaphragms A and B at the ends. The portions AC and BC contain hydrogen and oxygen gases, respectively. The diaphragms A and B are set into vibrations of same frequency. What is the minimum C B frequency of these vibrations A for which the diaphragm C is a H2 O2 node? Under the condition of the experiment, the velocity of sound in hydrogen is 1100 m/s and in oxygen is 300 m/s. (a) 1750 Hz

(b) 1600 Hz

(c) 1800 Hz

(d) 1650 Hz

3. An earth satellite is revolving in a circular orbit of radius a with velocity v0 . A gun is in the satellite and is aimed directly towards the earth. A bullet is fired from v the gun with muzzle velocity 0 . Neglecting resistance 2 offered by cosmic dust and recoiling of gun. Calculate the maximum and minimum distance of bullet from the centre of the earth during its subsequent motion. (a) 2 a ,

2a 3

(b) 2 a, a

(c) a,

a 3

(d) a,

a 2

4. A uniform disc of radius r0 lies on a smooth horizontal plane. A similar disc spinning with the angular velocity w0 is carefully lowered on to the first disc. How soon do both discs spin with same angular velocity, if the friction coefficient between them is equal to m? 8 mg 3r0 w 0 3r w (c) 0 0 8 mg (a)

4r0 w 0 3 mg 2 mg (d) 5r0 w 0 (b)

5. A thin biconvex lens of refractive index 3/2 is placed on a horizontal plane mirror as shown in the given figure. The space between the lens and the mirror is then filled with water of refractive index 4/3. It is found that when a point object is placed 15 cm above the lens on its principal axis, the object coincides with its own image.

43

On representing with another liquid, the object and the image again coincide at a distance 25 cm from the lens. Calculate the refractive index of the liquid.

(a) 1.5

(b) 1.24

(c) 1.6

(d) 1.33

6. A neutron with an energy of 4.6 MeV collides

slowly by 0.1 m. Calculate the final temperature of the gas and the heat supplied by the heater. The force constant of the spring is 8000 N/m, atmospheric pressure 1 ´ 105 N / m2 . The cylinder and piston are thermally insulated. The piston is massless and there is no friction between the piston and the cylinder. Neglect heat loss through the lead wires of the heater. The heat capacity of the heater coil is negligible. Assume the spring to be massless.

elastically with protons and is retarded. Assuming that upon each collision the neutron is deflected by 45°, find the number of collisions which will reduce its energy to 0.23 eV. (b) 25

(c) 23

Heater

(a) 26

Open atmosphere

(d) 24

Rigid support Piston

7. A mass M is uniformly distributed over the rod of length L as shown in the figure. The linear mass density of rod is l, then R

M P

(a) 800 K, 718.6J (c) 400 K, 618.6 J

(b) 553 K, 666.6 J (d) 478 K, 753.8 J

10. Consider a body at rest in the L-frame which explodes α β

in two fragments of masses m1 and m2 . Calculate energies of the fragments of the body.

Q

y

(a) if a = b, then Iy = 0 (Iy is gravitational field in y-direction) 2Gl if length of rod is infinite (b) Ix = R (c) gravitational potential is not defined at P if length of rod is infinite (d) if a = b, then Ix = 0 and Iy = 0

8. Two blocks A and B each of mass m, are connected by a massless spring of natural length L and spring constant k. The blocks are initially resting on the smooth horizontal floor with the spring at its natural length as shown in figure. A third identical block C, also of mass m moves on the floor with a speed v along the line joining A and B and collides elastically with A. Then, v

C

L A

B

(a) the kinetic energy of AB system at maximum compression of the spring is zero (b) the kinetic energy of AB system at maximum compression mv 2 of the spring is 4 m (c) the maximum compression of the spring is v æç ö÷ èkø (d) the maximum compression of the spring is v

m 2k

9. An ideal monoatomic gas is confined in a cylinder by a

spring loaded piston of cross-section 8 ´ 10 -3 m2 . Initially, the gas is at 300 K and occupies a volume of 2.4 ´ 10 -3 m3 and the spring is in its relaxed (unstretched, uncompressed) state. The gas is heated by a small electric heater until the piston moves out

44

m1 p1

m2 O p2=0

y

x

O

p1 m1′

p2=0

x

m′2 p2 (Before collision)

(After collision)

( m1 - m2 )Q ( m2 - m1 )Q m1 Q m2Q , (b) , ( m1 + m2 ) ( m1 + m2 ) m1 + m2 m1 + m2 m2 Q m1 Q m1Q m2Q (c) , (d) , ( m1 - m2 ) ( m1 + m2 ) ( m1 - m2 ) ( m1 - m2 ) (a)

Comprehension Type This section consists of 3 paragraphs, each describing theory, experiments, data etc. Six questions related to the three paragraphs with two questions on each paragraph. Each question has only one correct answer among the four given options (a), (b), (c) and (d). Paragraph I A thin, 50 cm long metal bar with mass 750 g rests on, but is not attached to two metallic supports in a uniform 0.450 T magnetic field as shown in the given figure. A battery and a 25 W resistor in series are connected to the supports.

V

R

B

11. What is the largest voltage, the battery can have without breaking the circuit at the supports? (a) 817 V (c) 325 V

(b) 412 V (d) 160 V

12. The battery voltage has the maximum value calculated in above question. If the resistor suddenly gets partially short-circuited, then its resistance decreasing to 2.0 W. Find the initial acceleration of the bar.

(a) 113 ms -2 (b) 55 ms -2 (c) 180 ms -2 (d) 12.4 ms -2

Paragraph III Two closed identical conducting containers are found in the laboratory of an old scientist. For the verification of the gas, some experiments are performed on the two boxes and the results are noted.

Gas A

Paragraph II An unmanned satellite A and a spacecraft B are orbiting around the earth in the same circular orbit as shown in figure.

A B

The spacecraft is ahead of the satellite by sometime. Let us consider that some technical problem has arisen in the satellite and the astronaut from B has made it correct. For this to be done docking of two (A and B) is required (in Layman terms connecting A and B). To achieve this, the rockets of A have been fired in forward direction and docking takes place as shown in the figure below. Rocket fired A

A B

A A

B

B Rocket retrofired

Mono (molar mass)

He 4g

Ne 20 g

Ar 40 g

Kr 84 g

Xe 131 g

Di (molar mass)

H2 2g

F2 19 g

N2 28 g

O2 32 g

Cl 2 71 g

(b) Dia, Dia (d) Dia, Mono

16. Identify the name of gas filled in the containers A and Docking over

Take mass of the earth = 5.98 ´ 1024 kg Radius of the earth = 6400 km Orbital radius = 9600 km Mass of satellite A = 320 kg Mass of spacecraft = 3200 kg Assume that initially, spacecraft B leads satellite A by 100 s, i.e. A arrives at any particular position after 100 s of B’s arrival.

B, respectively (a) N 2, Ne

(b) He, H 2

(c) O 2, Ar

(d) Ar, O 2

Matching Type 17. Three travelling sinusoidal waves are on identical strings having same tension. The mathematical form of waves are y1 = Asin(3 x - 6t ), y2 = Asin(4 x - 8t ) and y3 = Asin(6 x - 12t ). Column I A. Speed of each wave is

13. The initial total energy and time period of satellite are

Column II p. y x

respectively, (a) -6.65 ´ 1010 J, 9358 s (b) -6.65 ´ 10 9 J, 9358 s (c) -6.65 ´ 1010 J, 9140 s (d) -6.65 ´ 10 9 J, 9140 s

14. From a base station, rocket system of A has been operated, so that its rocket has been fired in forward direction, thus reducing the speed of A by 0.5%, then which of the following will happen? (a) (b) (c) (d)

Rd 222 g

15. Identify the type of gas filled in containers A and B, (a) Mono, Mono (c) Mono, Dia

A

New orbit of A

Experiment I When the two containers are weighed wA = 225 g, wB = 160 g and mass of evacuated container wC = 100 g. Experiment II When the two containers are given same amount of heat, same temperature rise is recorded. The pressure changes found to be DpA = 2.5 atm, DpB = 1.5 atm. Required data for unknown gas

respectively.

B

A

Gas B

Its orbit becomes elliptical with semi-major axis 9505.3 km Its total energy becomes - 6714 ´ 10 9 J . Its new time period becomes 9219.67 s All of the above

B. y1 is best represented by

q. y x

C. y2 is best represented by r.

y x

D. y3 is best represented by s. 2 m/s Codes A B C D (a) (s) (p) (q) (r) (c) (q) (r, s) (p) (r, p)

A B C (b) (p,q) (r) (s) (d) (p, q, r)(r, s) (p)

D (p) (q)

45

18. In a photoelectric experimental arrangement, light of

Column I

frequency f is incident on the metal target whose work hf as shown. In Column I, KE of function is f = 3 photoelectron is mentioned at various locations/ instants and in Column II, the corresponding values.

T

R and L

p.

I

B.

2 R and L

q.

II

C.

R and 2 L

r.

III

D.

2 R and 2 L

s.

IV

Codes A B C D (a) (p, q) (q,s) (p, q) (s) (b) (r) (p) (s) (q) (c) (p, q, r) (r) (p) (q, r, s) (d) (r, s) (p) (q) (s)

Collector Vacuum P

Column II

A.

C

Target

Column I A. Maximum KE of photoelectrons just after emission from the target B. KE of photoelectrons just after emission from target C. KE of photoelectrons when they are half-way between the target and collector D. KE of photoelectrons as they reach the collector

20. In Column I, some statements are given while in

Column II p. Zero

Column II some neutral objects are mentioned. Now, a point charge q is placed in front of the objects in Column II and statements in Column I are related to the effects on objects in Column II.

q. hf 3 r. hf 2 s. 2 hf 3

Column I A.

Codes A B C D (a) (p, q) (r, s) (q, r, s) (s) (b) (s) (p, q, r, s) (q, r, s) (p, r, s) (c) (p, q, r) (r, s) (p) (q, r) (d) (r, s) (p, q) (p, s) (r, s)

19. The switch S in the circuit is connected with point a for a very long time, then it is shifted to position b. The resulting current through the inductor is shown by curves in the graph for four sets of values for the resistance R and inductance L (given in Column I). Which set correspond with which curve?

Column II

Nature of distribution of p. charge is same on

B. Charge distribution on object is non-uniform

q.

C. The object remains neutral

r.

D. The interior of the object remains charge free

s.

+ E –

Solidconducting cubeofsidea

Ahollow/solid conductingsphere

t.

i a

Hollow conducting cube of side a

Infinitethickplane conductingsheet Hollowconducting cylinder

S

b

R

IV L

III

II I

t

Codes A B C D (a) (p) (q) (s, r, t) (p, q, r, s, t) (b) (q, r) (r, s, t) (p, q) (p, q, r, s) (c) (p, q) (p, q, r, s, t) (p, q, r, s, t) (p, q, r, s, t) (d) (q) (r) (p, q) (r, s, t)

Answers Paper 1 1. (a,d) 11. (0)

2. (a,b) 12. (9)

3. (b)

4. (d)

13. (4)

14. (4)

15. (9)

5. (a,b)

16. (6)

6. (a,c,d)

7. (b,c) 17. (8)

8. (a,b,c) 18. (3)

9. (b,d) 19. (5)

10. (b) 20. (4)

Paper 2 1. (c)

2. (d)

3. (a)

4. (c)

5. (c)

6. (d)

11. (a)

12. (a)

13. (b)

14. (d)

15. (c)

16. (d)

46

7. (a,b,c) 17. (a)

8. (d)

9. (a)

10. (b)

18. (b)

19. (b)

20. (c)

Arjun Sharma

•

A photon can take 40000 years to travel from the core of the Sun to its surface, but only 8 min to travel the rest of the way to Earth. Do you know? A photon travels on an average a particular distance (d) before being briefly absorbed and released by an atom, which scatters it in a new random direction. From the core to the Sun's surface (696000 kms) where it can escape into space, a photon needs to make a huge number of drunken jumps. The calculation is a little tricky, but the conclusion is that a photon takes between many thousands and many millions of years to drunkenly wander to the surface of the Sun. In a way, the light that reaches us today is energy produced may be millions of years ago.

•

Do you know the fact that an individual blood cell takes about 60 s to make a complete circuit of the body? Explain.

when its container is spun. No longer a mere liquid, the helium has become a super fluid-a liquid that flows without friction atoms in the liquid will collide with one another and slow down. But if we did that with helium at low temperature and came back a million years later, it would still be moving. •

The human genome, the genetic code in each human cell, contains 23 DNA molecules each containing from 500 thousands to 2.5 millions nucleotide pairs. DNA molecules of this size are 1.7 cm to 8.5 cm long when uncoiled or about 5 cm on average. There are about 37 trillion cells in the human body and if you'd uncoil all of the DNA encased in each cell and put them end to end, then these would sum to a total length of 2×1014 m or enough for 17 Pluto round trips (1.2×1013 m/Pluto round trip). •

We have about 5 L of blood in your body and the average heart pumps about 70 mL of blood out with each beat. Also, a healthy heart beats around 70 times a minute. So, if we multiply the amount of blood that the heart can pump by the number of beats in a minute, we actually get about 4.9 L of blood, which is almost our whole body’s worth of blood. In just a minute, the hearts pumps the entire blood volume around our body.

Do you know the fact that faster you move, heavier you get? Explain. If we run really fast, we gain weight. Not permanently or it would make a mockery of diet and exercise plans, but momentarily and only a tiny amount. Light speed is the speed limit of the Universe. So if something is travelling close to the speed of light, and we give it a push, it can't go very much faster. But we have given it extra energy, and that energy has to go somewhere. Where it goes is mass. According to relativity, mass and energy are equivalent. So, the more energy you put in, greater the mass becomes. This is negligible at human speeds-Usain Bolt is not noticeably heavier when running than when still - but once we reach an appreciable fraction of the speed of light, our mass start to increase rapidly.

•

When helium is cooled to almost absolute zero (−460°F or −273°C), the lowest temperature possible, it becomes a liquid with surprising properties: It flows against gravity and will start running up , why? When helium is just a few degrees below its boiling point of – 452 °F (− 269°C), it will suddenly be able to do things that other fluids can't like dribble through molecule-thin cracks, climb up and over the sides of a dish, and remain motionless

Do you know, on an average a human body carries ten times more bacterial cells than human cells? For one thing, bacteria produces chemicals that help us harness energy and nutrients from our food. Germ-free rodents have to consume nearly a third more calories than normal rodents to maintain their body weight and when the same animals were later given a dose of bacteria, their body fat levels spiked, even if they didn't eat any more than they had before. The gut bacteria is also very important to maintaining immunity.

• •

Do you know, there is enough DNA (Deoxyribonucleic Acid) in an average person's body to stretch from the Sun to Pluto and back 17 times?

Do you know, anti-matter is a composition of antiparticles that have same mass but with opposite charge and spin? All matter in the Universe is built up from a relatively small number of elementary particles, which include quarks (the constituents of protons and neutrons) and electrons (which together with protons and neutrons, make up atoms). Associated with each elementary particle is an anti-particle which also occurs in nature. The anti-particle has the same mass as the particle, but with opposite charge. A particle and an anti-particle can combine (or annihilate) to produce a photon or a particle of light. Conversely, a particle-anti-particle pair can be produced from a photon. So, there is a type of symmetry between particles and anti-particles in these processes. More specifically, the sub-atomic particles of anti-matter have properties opposite those of normal matter. The electrical charge of those particles is reversed.

•

Do you know, the fact behind the cause of electromagnetic pollution? Explain. Electromagnetic pollution is a buzzword describing the excessive exposure to electromagnetic radiation or electromagnetic fields (EMF) emitted by electronic devices like cell phones, cordless phones, Wi-Fi routers or Bluetooth-enabled equipment. The energy emitted by electronics is non-ionizing radiation, meaning it does not have the capability to break chemical bonds. In other words, it doesn't damage your DNA, which is a potential cause of cancer. While electromagnetic radiation cannot cause immediate damage, it does interact with our bodies, which can potentially lead to indirect damage, especially following long term exposure.

MARCH 2016

47

NUCLEUS

•

A tiny, positively charged object which lies at the centre of atom, its size is much smaller than the size of atom but constitutes most of its total mass. • Nucleons Protons and neutrons which are present in the nuclei of the atoms are collectively known as nucleons. • Atomic Number Total number of protons present inside a nucleus is called atomic number. It is denoted by Z. • Mass Number Total number of protons and neutrons in a nucleus is called mass number. It is denoted by A. • Number of protons = Number of electrons = Z • Number of nucleons = A, Number of neutrons, N = A − Z.

NUCLEAR MASS Total mass of proton and neutron in a nucleus. • Nuclide A specific nucleus of an atom characterised by its atomic number Z and mass number A, i.e. AZ X, where, X = symbol of the element, Z = atomic number, A = mass number. • Isotopes Nuclei having same number of protons but different number of neutrons, e.g. isotopes of carbon 126 C, 137 C, 146 C. • Isotones Nuclei having with the same neutrons number but different atomic numbers, e.g. 146 C, 15 16 17 7 N, 8 O, 9 F. • Isobars Nuclei having same mass number but different atomic numbers, e.g. 166 C, 716 N, 816 O.

SIZE OF NUCLEUS •

•

Nuclear density Mass per unit volume of a nucleus is called nuclear density. Mass of nucleus i.e. Nuclear density (ρ ) = Volume of nucleus

ATOMIC MASS UNIT(AMU) It is defined as 1/12 of the mass of 6 C 12 atoms. Its value is given by 1 amu = 1.66 × 10 −27 kg = 931.5 MeV • Electron volt The energy required by an electron when it is accelerated through a potential difference of 1 volt. −19 • It is denoted by eV. (1eV = 1. 602 × 10 J).

NUCLEAR FORCES The strong attractive forces in a nucleus which holds the protons and neutrons together are called nuclear forces. (i) Nuclear forces are short range forces and only exist within the range of 10 −15 m. (ii) Nuclear forces are strongest forces in nature. (iii) Nuclear forces are always attractive and stabilise the nucleus. (iv) Nuclear forces are charge independent. (v) Nuclear forces are non-central forces. (vi) Nuclear forces are exchange forces. • Yokawa proposed that the nuclear force between the two nucleons is a result of the exchange of particles mesons which are of 3 types. Positive π meson ( π + ), negative π meson ( π − ) and neutral π meson ( π 0 ).

•

NUCLEAR STABILITY Out of known 1500 nuclides only 260 are stable and rest of them are unstable which used to decay to form other nuclides by emitting α, β and γ-rays which is called radioactivity. Stability of the nucleus is determined by few factors. (i) Neutron-proton ratio ( N / Z ) • For lighter nuclei to be stable, N / Z =1. • For heavy nuclei to be stable, N / Z >1.

Nuclear radius The radius of nucleus is given by R ∝ A 1 / 3 R = R0 A 1 / 3 , where A is mass number of nucleus, R0 = 1. 2 × 10 −15 m. Nuclear volume The volume of nucleus is given by 4 4 V = πR 3 = πR03 A ⇒ V ∝ A 3 3 Difference between α, β and γ -particles 1.

Feature

α-particle

β-particle

γ-rays

Identity

Helium nucleus or doubly ionised helium atom ( 2He 4 )

Fast moving electron

Photons (EM waves)

(−β 0 or β − )

2.

Charge

+ 2e

−e

Zero

3.

Mass

4 mp (mp = mass of proton = 187 . × 10−27)

me

Massless

4.

Speed

= 107 m/s

1% to 99% of speed of light

Speed of light

5.

Range of kinetic energy

4 MeV to 9 MeV

All possible values between a minimum certain value to 1.2 MeV

Between a minimum value to 2.23 MeV

6.

Penetration power ( γ,β, α )

1 (Stopped by a paper)

100 (100 times of α)

10000 (100 times of β upto 30 cm of iron (or Pb) sheet)

7.

Ionisation power ( α >β >γ )

10000

100

1

8.

Effect of electric or magnetic field

Deflected

Deflected

Not deflected

9.

Energy spectrum

Line and discrete

Continuous

Line and discrete

10. Mutual interaction with matter

Produces heat

Produces heat

Produces photoelectric effect, compton effect, pair production.

11. Effect on photographic plate and ZnS phosphor.

Strong

Less

Least

(ii) Even or Odd number of Z or N . • It is found that even-even nucleus (even Z and even N) is more stable. • odd-odd nucleus (odd Z and odd N) is less stable. • Even-odd nucleus (even Z and odd N) or odd-even nucleus (odd Z and even N) is found to be least stable. (iii) Binding energy per nucleon The stability of a nucleus is also determined by the value of its binding energy per nucleon. Greater the binding energy per nucleon, then greater will be stability.

MASS DEFECT The difference in the mass of the nucleus and sum of the masses of nucleons is known as mass defect, i.e. Δm = [Zm p + ( A − Z )m n − m], where Z = atomic number, A = atomic mass, m p = mass of proton, m n = mass of neutron, m = mass of the nucleus.

•

2 BE [Zm p + ( A − Z ) m n − m] c = A A Few nuclei with mass number, A < 20 have large binding energy per nucleon than their neighbour. e.g. 2 He4 , 4 Be8 , 6 C12 and 10 Ne20 .

•

BE per nucleon is maximum of 26 Fe56 , i.e. 8. 8 MeV.

i.e.

B=

•

•

PACKING FRACTION Mass defect per nucleon is called packing fraction. Δm M − A i.e. f = = Α A where, M = mass of nucleus and A = mass number.

•

Note Smaller the value of packing fraction, larger is the stability of nucleus.

RADIOACTIVITY The spontaneous transformation of an element into another with emission of radiation such as α, β and γ-rays is called radioactivity. • Radioactivity of a sample cannot be controlled by any physical (pressure, temperature, electric or magnetic field.) • All the elements with atomic numbers Z > 82 are naturally radioactive. • The conversion of lighter elements into radioactive elements by the bombardment of fast moving particles is called artificial or induced radioactivity. • When a nucleus decays, all the conservation laws are observed like. • Conservation of mass energy → Conservation of linear momentum. • Conservation of angular momentum→ Conservation of charge.

•

Nuclear fusion When two or more nuclei combine to form a single nucleus which leads to the release of tremendous amount of energy due to mass defect is known as nuclear fusion 2 2 4 1 H + 1 H → 2 He + 24 MeV Chain reaction Three neutrons are produced in nuclear fission, under favourable condition these neutrons can be utilised for further fission of other nuclei later producing large number of neutrons until the whole of the uranium is consumed. Multiplication Factor The chain reaction will be steady accelerating or retarding and depends on the factor called critical factor (k). Rate of production of neutrons k= Rate of loss of neutrons If k =1, the chain reaction will be steady. If k >1, the chain reaction will be accelerating, i.e. grows. If k 0, then reaction is exothermic or exoergic and this reaction is energetically possible even if the particles were at rest. • If Q < 0, then reaction is endothermic or endoergic and this reaction cannot take place until we provide a minimum amount of energy to nucleus X.

•

LAWS OF RADIOACTIVE DISINTEGRATION

BINDING ENERGY Energy required to break up a nucleus into its constituent proton and neutron is called binding energy. (i) Binding energy of a nucleus AZ X is given by ΔE b = Δmc 2 = [Zm p + ( A − Z )m n − m]c 2 . Nuclear binding energies are striking very high. The quantity in bracket in above equation is termed as mass defect. • If we supply energy less than E b to a nucleus then nucleus stays together when the supplied energy is more than E b , the extra energy appears as kinetic energy of neutrons and protons as they fly apart. (ii) Binding energy per nucleon The average energy required to extract one nucleon from nucleus is known as binding energy per nucleon. • •

At any instant the rate of decay of radioactive atom is directly proportional to the number of atoms present at − dN dN that instant ∝N ⇒ = − λN , i.e. N = N o e − λt dt dt where, N = Number of atoms remains undecayed after time t, N 0 = Number of atoms initially present and λ = Decay constant. • Decay constant It is defined as the ratio of the amount of the substance disintegrated in unit time to the amount of the substance present. N N 1 i.e. N = N 0 e − λt ⎜⎛ t = ⎟⎞ ⇒ N = 0 = 0 = 0. 368 N 0 ⎝ λ⎠ e 2.718 The reciprocal of time after which the number of atoms of a radioactive substance decreases to 0.368 of their number initially present. • Nuclear fission The process of splitting of a heavy nucleus into two daughter nuclei of comparable masses is called nuclear fission. 238 + 0 n 1 → 92 U 236 → 56 Ba 141 + 36 Kr 92 + 0n 1 + Q 92 U unstable

Fraction of atoms Remaining fraction of active decayed (N0 − N ) N0 atoms (N / N0 ) probability of probability of survival decay

•

•

•

10

= 99.9%

n

⎧ ⎛ 1⎞ n⎫ ⎨1− ⎜⎝ ⎟⎠ ⎬ 2 ⎭ ⎩

Activity The number of atoms decaying per second is known as activity of a material. A = − dN / dt ⇒ A = A0 e − λt where, A0 = activity at t = 0 , A = activity after time t. units It has three units, becquerel ( B q ), curie (Ci) and rutherford ( Rd ). 1 becquerel = 1 disintegration/s 1 rutherford =10 6 disintegration/s 1 curie = 3.7 × 10 10 disintegration/s. half-life The time interval during which half of the radioactive substance will disintegrate is known as half-life. log 2 0.693 i.e. T1 / 2 = e = λ λ Probability that a particular nucleus decay in n 1 half-lives is ⎛⎜ 1 − n ⎞⎟ . ⎝ 2 ⎠ Average or mean life time It may be defined as the ratio of the combined age of all the atoms of the total number of atoms present in the given sample. 1 T i.e. τ = = 1 / 2 = 1.44 T1 / 2 λ 0. 693

@CLASS XII SYLLABUS

J

Final Touch Electromagnetic Waves

CHARACTERISTICS OF ELECTROMAGNETIC WAVES A changing electric field produces a changing magnetic field and vice-versa y which gives rise to transverse wave known x as electromagnetic z y E waves. The time varying electric and magnetic fields are z B mutually perpendicular c to each other and also x perpendicular to the direction of Electromagnetic waves propagation of wave.

G

Characteristics of Electromagnetic Waves

G

G

G

G

G

The electromagnetic waves are transverse in nature whose speed is same as that speed of light. The two fields E and B have the same frequency of oscillation and they are in same phase with each other. The electric vector E is responsible for the optical effects of an electromagnetic wave and is called the light vector. E and B are such that E × B [is always in the direction of propagation of wave] z

G

G

G

Maxwell’s Equations Maxwell gave the basic laws of electricity and magnetism in the form of four fundamental equations, which are known as Maxwell’s equations. It is based on experimental observations followed by all electromagnetic phenomena, may be stated in the integral form as given below:

Gauss’s Law in Electrostatics G

B

y G

E

Wave propagation

50

x

The electromagnetic wave propagating in the positive x-direction may be represented by E = Ey = E0 sin (kx − ωt ), B = Bz = B0 sin (kx − ωt ) where, E(Ey ), B(Bz) are the instantaneous values of the fields. E0 , B0 are amplitude of the fields and k is angular wave number. Electromagnetic waves obey the superposition principle because the differential equations involving E and B are linear equations, e.g. we can add two waves with the same frequency simply by adding the magnitudes of the two electric field algebraically. Electromagnetic waves show the properties of reflection, refraction, interference, diffraction and polarisation. The velocity of electromagnetic waves in dielectric is less than c ( = 3 × 108 m/s). The EM waves carry energy which is divided equally between electric field and magnetic field vectors.

This law gives total electric flux in terms of charged enclosed by the closed surface. This law states that electric lines of force start from positive charge and end at negative charge, i.e. the electric lines of force do not form a continuous closed path. q i.e. ∫s E ⋅ dS = ε0

Gauss’s Law in Magnetism G

This law shows that number of magnetic lines of force entering a closed surface is equal to the number of magnetic lines of force leaving that closed surface.

G

It tells that the magnetic lines of force form a continuous closed path.

G

It also predicts that the isolated magnetic monopoles does not exist.

∫S B ⋅ dS = 0

i.e.

Faraday’s Laws of Electromagnetic Induction G

G

This law gives a relation between electric field and a changing magnetic flux. It states that changing magnetic field is the source of electric field. − dφB ∫ E ⋅ dl = dt

Displacement Current It is that current which comes into existence (in addition of conduction current) whenever the electric field and hence the electric flux changes with time. It is equal to ε0 times the rate of change of electric flux through a given surface. ID = ε0

Modified Ampere’s Circuital Law It states that the line integral of the magnetic field B over a closed path is equal to μ 0 times the sum of conduction current I C and the displacement current I D threading between the capacitor plates during charging process. Hence, modified law is as below: dφE ⎞ ⎛ ∫ B ⋅ dl = μ0 (IC + ID) or ∫ B ⋅ dl = μ0 ⎜⎝ IC + ε0 dt ⎠⎟

Importance of Maxwell’s Equations Maxwell predicted the following conclusions : (i) An accelerated charge is the source of electromagnetic waves. (ii) The electromagnetic waves can propagate through the space with the speed of light (3 × 10 8 m/s) . (iii) The electromagnetic wave is of transverse nature. (iv) Light itself is an electromagnetic wave as it is transverse in nature and it moves in space with a speed (3 × 10 8 m/s) .

Ampere’s Circuital Law According to this law, the line integral of magnetic field along any closed path or circuit is μ 0 times the total current flowing through a surface enclosed in that loop.

∫ B ⋅ dl = μ 0 I

i.e.

Inconsistency of Ampere’s Law Maxwell explained that Ampere’s law is valid only for steady current or when the electric field does not change with time. To see this inconsistency, consider a parallel plate capacitor being charged by a battery. During the charging time varying current flows through connecting wires. +q

I

I1

–q

I2

dφE dt

Hertz’s Experiment Hertz was the first scientist to experimentally demonstrate production of electromagnetic waves employing a crude form of an oscillatory L - C circuit. EM waves are produced by accelerated charge (or oscillating charge). The frequency of the EM waves is same as the frequency of oscillation of the charge in electric field E or oscillation of 1 charge in magnetic field B, i.e. ν = , where L is 2 π LC G

G

inductance and C is capacitance of a circuit. K

G

Applying Ampere’s law for loop I1 and loop I2 , we get

∫I B ⋅ dl = μ0 I

G

1

But, ∫ B ⋅ dl = 0 (since no current flows through the region I2

between the plates). But practically it is observed that there is a magnetic field between the plates. Hence, Ampere’s law fails. i.e.

∫I

1

B ⋅ dl ≠ μ 0 I

G

The EM waves are transverse in nature. They do not require any material medium for their propagation. He discovered that if one of the spark gap terminal is connected to an antenna and the other terminal is earthed, the electromagnetic waves radiated could go upto several kilometers. Later on Jagdish Chandra Bose produced EM waves of wavelength ranging from 5 mm to 25 mm. Marconi succeeded in transmitting EM waves over distances upto few kilometres.

51

1. Consider the two following statements regarding a linearly polarised plane electromagnetic waves. I. Electric field and the magnetic field have equal average values. II. Electric energy and the magnetic energy have equal average values. (a) I is true (c) Both are true

(b) II is true (d) Both are false

2. An EM wave radiates outwards from a dipole antenna, with E0 as the amplitude of its electric field vector. The electric field E0 which transports significant energy from the source falls off as 1 r3

(b)

1 r2

(c)

1 r

(d) remains constant

3. Instantaneous displacement current of 10 . A in the space between the parallel plate of 1 μF capacitor can be established by changing the potential difference of (a) 10

−6

Vs

Hint As,

−1

6

(b) 10 Vs

q CV or i D = C = t t

−1

(c) 1 Vs

−1

(d) 01 . Vs

⎛ V⎞ ⎜ ⎟ where, i D = displacement current ⎝t⎠

monoxide molecule into carbon and oxygen atoms. The minimum frequency of the appropriate electromagnetic radiation to achieve the dissociation lies in (b) infrared region (d) microwave region

5. An electric field E and magnetic field B exist in a region. If these fields are not perpendicular to each other, then the electromagnetic waves (a) will not pass through the region (b) will pass through the region (c) may pass through the region (d) nothing is definite Hint The electromagnetic wave being packets of energy moving with speed of light may pass through the region.

6. A cube of edge a has its edges parallel to X , Y and Z-axis of rectangular coordinate system. A uniform electric field E is parallel to Y-axis and a uniform magnetic field is parallel to X-axis. The rate at which flows through each face of the cube is (a)

a 2 EB parallel to XY-plane and zero in other 2 μ0

(b)

a 2 EB parallel to XY-plane and zero in other μ0 2

52

(c)

a EB from all faces 2 μ0

(d)

a 2 EB parallel to YZ-plane and zero in other 2 μ0

It will be in negative Z-direction. It shows that the energy will be flowing in faces parallel to XY-plane and is zero in all other faces. Total energy flowing per second from a face in XY-plane 1 EBa 2 . = ( EB sin 90 ° ) a 2 = μ0 μ0

point between the capacitor plates expressed in terms of the rate of change of the electric field strength, i.e. dE between the plates is dt (a)

μ 0i 2 πr

(b)

ε0μ 0r dE . 2 dt

(c) zero

(d)

μ 0i dE 2 π dt

8. The magnetic field between the plates of a capacitor

where r >R is given by (where, r is the distance from the axis of plates and R is the radius of each plate of capacitor) (a)

−1

4. One requires 11eV of energy to dissociate a carbon

(a) visible (c) ultra violet region

1 [ E × B ]. μ0

7. An expression for the magnetic field strength B at a

Hint Electric and magnetic fields and energies have equal average values.

(a)

Hint Energy flowing per second unit area from a face is

μ 0i D 2 πR 2

(b)

μ 0. i D 2π R

(c)

μ 0. i D 2 πr

(d) zero

9. An electromagnetic wave travels along Z-axis. Which of the following pairs of space and time varying fields would generate such a wave? (a) E x , By

(b) E y , Bz

(c) E z , Bx

(d) E y , Bx

Hint E x and By would generate a plane electromagnetic wave travelling in Z-direction. E, B and k from a right handed system k$ is along Z-axis. (As, $i × $j = k$ ) ⇒ E $i + B $j = c k$ . x

y

i.e. E is along X-axis and B is along Y-axis.

10. The charge of a parallel plate capacitor is varying as

q = q0 sin 2 π ft . The plates are very large and close together (area = A, separation = d). Neglecting edge effects, the displacement current through the capacitor is (a)

d A ε0

d sin 2 πft ε0 2 π fq 0 (d) cos 2 πft ε0 (b)

(c) 2 πfq 0 cos 2 πft

11. The magnetic field between the plates of radius 12 cm separated by distance of 4 mm of a parallel plate capacitor of capacitance 100 pF along the axis of plates having conduction current of 015 . A is (a) zero

(b) 15 . T

(c) 15 T

(d) 015 . T

Hint As, B ∝ r, since the point is on the axis, where r = 0, so B = 0.

12. A circular ring of radius r is placed in a homogeneous magnetic field perpendicular to the plane of the ring. The field B changes with time according to the equation, B = kt , where k is a constant and t is the time. The electric field in the ring is (a)

kr 4

(b)

kr 3

(c)

kr 2

(d)

k 2r

PROPERTIES OF ELECTROMAGNETIC WAVES Speed G

G

In

a free space, its speed is E 1 c= = 0 = 3 × 108 m/s, where μ 0 B0 μ 0 ε0

G

given

by

is absolute

permeability and ε0 is absolute permittivity in a free space or vacuum. 1 , where μ and ε are absolute In medium, v = με permeability and absolute permittivity of that medium.

Energy G

G

G

G

G

The energy in electromagnetic waves is divided equally between the electric and magnetic fields. 1 Energy density of electric field, UE = ε 0 E 2 2 The electric vector is responsible for the optical effects of an electromagnetic wave and is called as Light vector. 1 B2 Energy density of magnetic field, Um = 2 μ0 1 B2 Total energy per unit volume, U = UE + Um = 2 μ0

Also, average density (electric or magnetic field) B2 1 Uav = ε 0 E 20 = 0 2 2μ 0

Intensity The energy crossing per unit area per unit time perpendicular to the direction of propagation of electromagnetic waves. Total EM energy Total energy density × Volume i.e. I = = Surface area × Time Surface area × Time

G

Radiation Pressure It is the momentum imparted per second per unit area on which the light falls. For a perfectly reflecting surface, pr = 2 S/ c, where S is Poynting vector and c is speed of light. For a perfectly absorbing surface, pα = S/ c G

G

Note For the case of oblique incidence, where change in momentum at the perfectly reflecting surface is 2 p cos θ and the 2S corresponding radiation pressure is p = cosθ, where θ is the c angle of incidence.

Refractive Index ( n) The refractive index n of a material medium is given by ⎤ ⎡ μ 1 c με με ε × n= = = ⎢Q μ = μ r , ε = εr ⎥ 1 v μ 0 ε0 μ 0 ε0 ⎦ ⎣ 0 0 So, n = μ r εr where, μ r = relative permeability of the medium εr = relative permittivity of the medium.

Wave Impedance (Z ) G

G

G

The direction of poynting vector S of an electromagnetic wave at any point gives the wave’s direction of travel and the direction of energy transport at that point.

Momentum G

Electromagnetic waves also carry momentum. If a portion of electromagnetic wave of energy U propagating with Energy (U) speed c, then linear momentum, i.e. p = Speed (c)

The medium offers hindrance to the propagation of wave. Such hindrance is called wave impedance. It is given by Z=

Poynting’s Vector In electromagnetic waves, the rate of flow of energy crossing a unit area is described by a vector S, called as Poynting vector. Its unit is watt/m2 or Js −1 m−2 and it can be expressed as 1 S= (E × B) μ0

If the surface is a perfect reflector and incidence is normal, then the momentum is transported to the surface is twice, 2U then momentum becomes, p = c The average force exerted by electromagnetic waves on the p 2U surface will be F = = t tc

G

μ μr = ε εr

μ0 ε0

For vacuum or free space, Z =

μ0 = 376.6 Ω ε0

Electromagnetic Spectrum The orderly distribution of the electromagnetic waves in accordance with their wavelength or frequency into distinct groups having widely different properties is called electromagnetic spectrum as shown in the figure. Wavelength (m) 103 102 101

1

10–1 10–2 10–3 10–4 10–5 10–6 10–7 10–8 10–9 10–10 10–11 10–12

X-rays

Infrared Radio waves (AM, FM, TV)

Ultraviolet

Microwaves (radar, etc.)

γ-rays

Visible Light 7×10–7 Red

6×10–7 Yelloworange

5×10–7 Bluegreen

4×10–7 Violet

53

G

G

G

G

Radio waves are generated when charges are accelerating through conducting wires. Their wavelengths lie in the range of 104 m to about 0.1 m. It is used in radio and television communication system. Microwaves have wavelengths ranging from approximately 0.3 m to 10 −4 m and are also generated by electronic devices. They are used in radar systems and for studying the atomic and molecular properties of matter. Infrared waves have wavelengths ranging from 10 − 3 m to the longest wavelength of visible light, 7 × 10 − 7 m. These waves are produced by molecules and room temperature objects. It is used in physiotherapy, infrared photography and vibrational spectroscopy. Visible light is that component of the electromagnetic spectrum that can be detected by human eye. These extend from a wavelength of 7.8 × 10 − 7 m to 3.8 × 10 − 7 m and frequencies from 4 × 1014 Hz to 8 × 1014 Hz.

■

■

■

■

■ ■

■

Radiation γ-rays

Uses Gives information on nuclear structure, medical treatment etc

X-rays

Medical diagnosis and treatment study of crystal structure, industrial radiograph

UV-rays

Preserve food, sterilising the surgical instruments, detecting the invisible writings, finger prints etc

Visible light

To see objects

Infrared rays

To treat muscular strain for taking photography during the fog, haze, etc

Microwave and radio waves

In radar and telecommunication

G

G

Ultraviolet waves cover wavelengths ranging from approximately 4 × 10 − 7 m to 6 × 10 − 10 m. These are produced by the sun, special lamps like mercury lamp, hydrogen tube, etc., and very hot bodies. It has various uses such as in LASIK eye surgery, to kill germs in water purifiers as disinfectant in hospitals etc. As an UV light from the sun is absorbed by ozone (O3 ) molecules in the earth’s upper atmosphere in a layer called the stratosphere. This ozone shield converts (central) high energy UV radiations to infrared radiations. As a result absorption of UV radiation results in warming the stratosphere. X-rays have wavelengths in the range from approximately 10 − 8 m to 10 − 12 m. It is obtained by the deceleration of high energy electrons bombarding a metal target. It is used in the study of crystal structure and treatment of cancer.

54

I

The frequency of electromagnetic waves is its inherent characteristic. When an electromagnetic wave travels from one medium to another, its wavelength changes but frequency remains unchanged. All the types of electromagnetic waves travel with the same speed in free space. The amplitude of electric field in an electromagnetic waves of intensity I 2I is given by E 0 = ε 0c The current in L-C circuit during oscillatory discharge is given by dq d I = − = − (q 0 cosωt ) = q 0 sin ωt dt dt During the discharge of a capacitor, the current in the circuit increases. The ratio of the amplitudes of electric and magnetic fields is constant and it is equal to velocity of the electromagnetic waves in free space E i.e. 0 = c B0 The magnitudes and the amplitudes of the electric and magnetic components of the electromagnetic wave obey a wave equation : ⎛ δ2 δ 2 ⎞ ⎧E y (x , t )⎫ ⎜ 2 − μ 0ε 0 2 ⎟ ⎨ ⎬=0 δt ⎠ ⎩B z (x , t )⎭ ⎝ δx A standing electromagnetic waves do not propagate, but instead the electric and magnetic fields execute simple harmonic motion perpendicular to the direction of propagation, i.e. E y ( x , t ) = 2E 0 sin kx sinωt , B z ( x , t ) = 2B 0 cos kx cosωt .

PHYSICS Superfast Light Pulses : A Measure of Response Time of Electrons to Light A team of researchers has found a way to measure the time. It takes for an electron in an atom to respond to a pulse of light. The team describes their use of a light field synthesizer to create pulses of light so fast that they were able to reveal the time, it took for electrons in an atom to respond when struck. As scientists have begun preparing for the day when photons will replace electrons in high speed computers, work is being done to better understand the link between the two. One important aspect of this is learning what happens when photons strike electrons that remain in their atom (rather than being knocked out of them), specifically, how long does it take them to respond.

13. The oscillating electric and magnetic field vectors of

20. The shortest wavelength of X-rays emitted from an

electromagnetic wave are oriented along

X-rays tube depends upon

(a) (b) (c) (d)

(a) nature of the gas in the tube (b) voltage applied to tube (c) current in the tube (d) nature of target of the tube

the same direction and in phase the same direction but have a phase difference of 90° mutually perpendicular directions and are in phase mutually perpendicular directions but has a phase difference of 90°

Hint In electric and magnetic waves are mutually perpendicular directions and they are in same phase.

14. The sun delivers 10 Wm 3

−2

of electromagnetic flux to the earth’s surface. The radiation force on the roof of dimensions 10 m × 20 m is

(a) 6.67 (b) 6.67 (c) 6.67 (d) 2.35

× 10 −6 × 10 −4 × 10 −3 × 10 −4

N N N N

Hint Radiation force =

1 μ 0 ε0 1 μ /ε

(b)

1 με

(d)

μ0 ε

(b) v s < v x < v m (d) v s = v x = v m

17. Electromagnetic waves travel in a medium at a speed

of 2.0 ×108 ms −1 . The relative permeability of the medium is 1.0. The relative permittivity is (b) 4. 25

(c) 8.62

(d) 0

23. If the wavelength of light is 4000 Å, then the number of wavelength in 1 mm length will be (d) 0

the given radiation sources is

microwaves respectively in vacuum, then (a) v s > v x > v m (c) v s > v x > v m

(b) 31 . × 10 −5 k$ T

24. The correct sequence of the increasing wavelength of

16. If v s , v x and v m are the speeds of γ-rays, X-rays and

(d) 2.25

18. An earth orbiting satellite has solar energy collecting

panel with total area 5 m2 . If solar radiations are perpendicular and completely absorbed, the average force associated with the radiation pressure is (Solar constant = 1.4 k Wm−2 )

(a) 2.33 × 10 −3 N (c) 2.33 × 10 −5 N

22. A plane electromagnetic wave of frequency 25 MHz

(a) 2.5 × 10 5 (b) 0.25 × 10 4 (c) 2.5 × 10 4

Hint Velocity of light in a medium, 1 1 c= = με μ 0ε 0 μ r ε r

(a) 9.25

(a) light and can easily deflect electrons (b) light and can absorb electrons (c) a heavy element with a high melting point (d) an element having high thermal conductivity

(a) 2.1 × 10 −8 k$ T (c) 5.0 × 10 −6 k$ T

Total power Velocity of light

in any medium is expressed as

(c)

production of X-rays because it is

travels in a free space along the X-direction. At the particular point in space and time, E = 63 $j V/ m . What is B at the point?

15. According to Maxwell’s equation, the velocity of light (a)

21. Molybdenum is used as a target element for the

(b) 2.33 × 10 −4 N (d) 2.33 × 10 −6 N

19. X-rays are produced by jumping of (a) electrons from lower to higher energy orbit of atom (b) electrons from higher to lower energy orbit of atom (c) protons from lower to higher energy orbit of nucleus (d) protons from higher to lower energy orbit of nucleus Hint X-rays are produced when there is vacancy for the electron on inner complete orbits of an atom and jump of electrons takes place from higher orbit to lower orbit of atom.

(a) radioactive sources, X-ray tube, crystal oscillator, sodium vapour lamp (b) radioactive sources, X-ray tube, sodium vapour lamp, crystal oscillator (c) X-ray tube, radioactive sources, crystal oscillator, sodium vapour lamp (d) X-ray tube, crystal oscillator, radioactive sourcess, sodium vapour lamp Hint Radioactive sources, X-ray tube, sodium vapour lamp, crystal oscillator.

25. A plane electromagnetic wave travels in vacuum along Z-direction. What can you say about the directions of its electric and magnetic field vectors ? If the frequency of the wave is 30 MHz, what is its wavelength? (a) 20 m

(b) 15 m

(c) 10 m

(d) 5 m

Hint Here, electromagnetic wave is travelling in Z-direction, then electric and magnetic fields are in XY-direction and are perpendicular to each other.

26. A TV tower has a height of 100 m. How much population is covered by the TV broadcast if the average population density around the tower is 1000 km−2 ? (Radius of the earth = 6.37 × 106 m) (a) 4 lakh (c) 40000

(b) 4 billion (d) 40 lakh

27. Ozone layer blocks the radiations of wavelength (a) less than 3 × 10 −7 m (c) more than 3 × 10 −7m

(b) equal to 3 × 10 −7m (d) All of these

Hint Ozone layer blocks the high energy radiations like ultraviolet (UV) (3 × 10 −7m).

55

28. The radio wave (wavelength 21 cm) is emitted by hydrogen in interstellar space is due to the interaction called the hyperfine interaction in atomic hydrogen, the energy of the emitted wave is nearly

(a) 10 −17J

(c) 7 × 10 −8 J

(b) 1 J

(d) 10 −24 J

29. The electric field intensity produced by the radiations coming from 100 W bulb at a distance 3 m is E0 . The electric field intensity produced by the radiations coming from 50 W bulb at the same distance is

(a)

E 2

(b) 2 E

(c)

E 2

(d) 2 E

Hint The electric field intensity produced by the radiations coming from 50 W bulb at the same distance is 2 E.

30. Television signals reach us only through the ground waves. The range R related with the transmitter height h is in proportion to (c) h−1/ 2

(b) h1/ 2

(a) h

(d) h−1

Hint Range ( R ) = 2 hr , where r is the radius of the earth. So, R ∝ h1 / 2 .

Answers 1. (c) 11. (a) 21. (c)

2. (c) 12. (c) 22. (a)

3. (b) 13. (c) 23. (b)

4. (c) 14. (b) 24. (b)

5. (c) 15. (b) 25. (c)

MASTER

1. A radiation of 200 W is incident on a surface which is 60% reflecting and 40% absorbing. The total force on the surface is (b) 13 . × 10 −6 N (d) 103 . × 10 −7 N

(a) 107 . × 10 −6 N (c) 107 . × 10 −7 N

2. If c is the speed of electromagnetic waves in a vacuum, its speed in a medium of dielectric constant K and relative permeability μ is c μc

(b) v =

1 μK

(c) v =

K μc

(d) v = c μ K

3. A plane electromagnetic wave of intensity 10 Wm−2 2

strikes a small mirror of area 20 cm , held perpendicular to the approaching wave. The radiation force on the mirror will −11

−11

N N (b) 133 (a) 6.6 × 10 . × 10 (c) 133 (d) 6.6 × 10 −11 N . × 10 −10 N Hint Radiation force = momentum transferred per second by electromagnetic wave to the mirror.

4. If an electromagnetic wave is propagating in a medium μ with permittivity ε and permeability μ, then is ε

(a) intrinsic impedance of the medium (b) square of the refractive index of the medium (c) refractive index of the medium (d) energy density of the medium μ has the dimensions of resistance, hence it is called ε the intrinsic impedance of the medium.

Hint As,

5. The electromagnetic waves travel in a medium which has relative permeability by 1.3 and relative permittivity 2.14. Then, the speed of the electromagnetic wave in the medium will be

56

7. (b) 17. (d) 27. (b)

8. (c) 18. (c) 28. (d)

9. (a) 19. (b) 29. (d)

(a) 1.8 × 10 7 ms −1

(b) 1.8 × 10 8 ms −1

−1

(d) 5.7 × 10 9 ms −1

(c) 2.5 × 10 ms 6

Hint As, v =

Hint Here, Ftotal = Fre + Fab .

(a) v =

6. (b) 16. (d) 26. (d)

c μ ε0

=

3 × 10 8 1.3 × 2 .14

10. (c) 20. (b) 30. (b)

≈ 1.8 × 10 8 ms −1

6. The electric field (in Nms −1 ) in an electromagnetic

wave is given by E = 50 sin ω (t − x / c). The energy stored in a cylinder of cross-section 10 cm2 and the length 100 cm along the x-axis will be

(a) 5.5 ×10 −12 J (c) 2.2 ×10 −11 J

(b) 1.1 × 10 −11 J (d) 1.65 × 10 −11 J

7. The speed of electromagnetic wave in the medium of dielectric constant 2.25 and relative permittivity μ is

(a) 1 × 10 8 ms −1 (c) 4 × 10 ms 8

−1

(b) 2.5 × 10 8 ms −1

(d) 3 × 10 8 ms −1

Hint The speed of electromagnetic waves in a medium is 1 1 c v= = = μ 0 ε0 με μ 0 εr ε0 μ r

8. For an electromagnetic wave, the amplitude of electric

field is 10 V/ m. The frequency of wave is 5 × 1014 Hz. The wave is propagating along the Z-axis, then the average density of electric field is (a) 2.21 × 10 −10 Jm−3 (c) 81 . × 10 −12 Jm−3

(b) 3.25 × 10 −9 Jm−3 (d) 6.25 × 10 −3 Jm−3

9. A plane electromagnetic wave is incident on a material surface. The wave delivers momentum p and energy E is (a) p ≠ 0, E ≠ 0

(b) p = 0, E = 0

(c) p = 0, E ≠ 0

(d) p ≠ 0, E = 0

Hint When plane electromagnetic wave is incident on the material surface, the wave delivers some momentum and energy to the surface and hence, p ≠ 0, E ≠ 0.

10. In which one of the following regions of the electromagnetic spectrum will the vibrational motion of the molecules give rise to absorption? (a) Ultraviolet

(b) Microwaves

(c) Infrared

(d) Radio waves

Hint Molecular spectra due to vibrational motion lie in the microwave region of electromagnetic spectrum due to Kirchhoff’s law in spectroscopy the same will be absorbed.

11. The amplitude of an electromagnetic wave in vacuum is doubled with no other changes made to the wave. As a result of this doubling of the amplitude, which of the following statement is true? (a) The speed of wave propagation changes only (b) The frequency of the wave changes only (c) The wavelength of the wave changes only (d) None of the above Hint Velocity of an electromagnetic wave, 1 c= = 3 × 10 8 m / s is independent of amplitude, frequency μ 0 ε0 and wavelength of an electromagnetic wave.

12. The mean electric energy density between the plates of

a charged capacitor is (here, q = charge on the capacitor and A = area of the capacitor plate) (a)

q2 2 ε0 A 2

(b)

q 2 ε0 A 2

(c)

q2 2 ε0 A

(d) zero

Hint Electric energy stored in a charge capacitor U=

1 q2 1 q2 1 q = = × Ad 2 c 2 ε 0 A / d 2 ε 0 A2

13. If μ 0 , ε0 are the absolute permeability and permittivity

respectively of space μ r , εr are the relative permeability and permittivity respectively of the media and μ , ε are the absolute permeability and permittivity of medium respectively, then the refractive index of the medium is (a)

με μ 0 ε0

(b)

μ r εr μ 0ε0

(c) μ r ε r

(d)

μ ε μ 0εr

14. For a plane of electromagnetic wave, the electric field

oscillates sinusoidally at a frequency of 2 × 1010 Hz and the amplitude 48 Vm−1 . The amplitude of oscillating magnetic field will be (a) 16 × 10 −8 Wbm−2 (c) 18 × 10

−5

Wbm

−2

(b) 12 × 10 −8 Wbm−2

(d) 2.0 × 10 −6 Wbm−2

15. Radio waves received by a radio telescope from distant aparts may have a wavelength of about 0.20 m. If the speed of the wave is 3 ×108 ms −1 , then frequency of the wave will be (a) 1.5 × 10 9 Hz (c) 1.5 × 10 3 Hz

(b) 15 . × 10 8 Hz (d) 135 Hz

Answers 1. (a) 6. (b) 11. (d)

2. (a) 7. (a) 12. (a)

3. (c) 8. (a) 13. (c)

4. (a) 9. (a) 14. (a)

5. (b) 10. (b) 15. (b)

57

Formulae at a Glance Electromagnetic Waves G

G

G

df E The displacement current is given by the relation, ID = e 0 dt df E where, e 0 = absolute permittivity of space, = rate of change of dt electric flux e df ù é Ampere’s Maxwell’s law ò B × dl = m 0 ( I + ID ) = m 0 ê1 + 0 E ú dt û ë Maxwell's equations are as follows

ò

E × dS =

(ii)

ò

E × dS = 0 (Gauss’s law in magnetostatics)

S

-d (iii) ò E × dl = S dt induction) (iv) G

G

ò

S

G

G

q (Gauss’s law in electrostatics) e0

(i)

S

G

ò B × dS S

E × dl ± m 0 I + m 0 e0

d dt

G

(Faraday’s law of electromagnetic

Ray Optics G

ò E × dS (Ampere-Maxwell’s law)

Velocity of electromagnetic waves in free space is given by 1 c= = 3 ´ 10 8 m/s m 0 e0 The amplitudes of electric and magnetic fields in free space in electromagnetic waves are related by, E 0 = cB0

G

G

Alternating Current G

G

G

G

G

G

G

G

Alternating current vary as sine function of time is given by I = I0 sin wt and alternating voltage, V = V0 sin wt . where, I0 and V0 are the maximum or peak values of current and voltage, wis the angular frequency, n is the frequency and T is the time period of given AC. T Average or mean value of an AC over a half cycle, i.e. t = 0 to . 2 2I Iav = 0 = 0.637 I0 Similarly, Vav = 0.637 V0 \ p I V The rms value of an AC is defined as, ( Irms ) = 0 and Vrms = 0 2 2 1 Reciprocal of reactance is known as susceptance, i.e S = X 1 Reciprocal of impedance is known as admittance. Thus, Y = . Its unit Z is Siemen (S). V V In pure resistive circuit, V = V0 sin wt , where current, I = or Irms = rms R R and current, voltage are in same phase, i.e. current is given by p I = I0 sin wt , average power, Pav = Vrms Irms cos = 0. 2 V In inductive reactance, X L: = 2p nL = wL, current, I = , current lags XL p behind the voltage by , i.e. I = I0 sin( wt - p / 2 ) for V = V0 sin wt . 2 1 As pure capacitive circuit, capacitive reactance, XC = , current wC V p and current leads the voltage by flowing, I = , i.e. XC 2 I = I0 sin ( wt + p / 2 ) for V = V0 sin wt .

G

G

G

Power factor, cos f =

G

Average power over a full cycle of AC, Pav = Vrms Irms cos f =

1 V0 I0 cos f 2

1 1 1 2 = + = f u v R where, f is focal length, u is distance of the object, R is radius of curvature and v is distance of the image from the pole of mirror. h +v Linear magnification, m = 2 = h1 - u where, h2 is size of image and h1 is size of the object. sin i (As light goes from rarer to denser medium) and Snell’s law, m = sin r sin r (When light goes from denser to rarer medium) m= sin i Mirror formula,

Lens maker’s formula, for both convex and concave lenses is é1 1 1 ù = (u - 1) ê ú f R R ë 1 2 û where, R1 and R2 are radii of curvature of the two surfaces of the lens and m is refractive index of material of lens. 1 Power of lens, i.e. P = when f = 1m, P = 1 dioptre f If m is refractive index of material prism, then from Snell’s law, sin i sin( A + dm ) / 2 = sin r sin A / 2

m=

G

G

G

G

G

G

R True power = Z Apparent power

G

1 , i.e. the natural LC frequency of the circuit is equal to applied frequency, then the circuit is said to be in resonance. At resonance, current in the circuit is maximum and impedance is minimum and Zmin = R. E At resonance, I0 = 0 and VL = I0 X L R wL 1 VC = I0 XC , i.e. VL = ´ E0 = ´ E 0 = QE 0 R wRC wL 1 or is termed as quality factor circuit. where, Q = R wRC It determines the sharpness of resonance. Higher the value of Q, sharper is the resonance. In series L-C-R circuit, if X L = XC = w 0 =

G

This formula is called prism formula, where A is angle of prism and dm is net deviation angle. Angular dispersion Dispersive power of prism, w = Mean deviation é dù Magnifying power of a simple microscope, m = ê1 + ú, where d is fû ë least distance of distinct vision from the eye and f is focal length of an eye. In normal adjustment, when final image is at infinity, m = d / f . v é dù Magnifying power of a compound microscope, m = 0 ê1 + ú -u 0 ë fe û where, v 0 and u 0 are distance of image and object from optical centre of objective lens, fe is focal length of eyelens. The magnifying power of astronomical telescope is given by f f m= 0 = 0 - fe |fe | When final image is at the least distance of distinct vision from the eye, -f é f ù the magnifying power is given by m = 0 ê1 + e ú fe ë d û where, f0 is focal length of objective lens. f (R / 2 ) Magnifying power of a reflecting type telescope is m = 0 = fe fe

77

1. In a vernier callipers, N divisions of vernier scale coincide with (N −1) divisions of main scale (in which length of one division is 1mm. The least count of the instrument should be (a) N

(b) N − 1

(c)

1 10N

(d)

1 N−1

2. The dimensional formula of self-inductance is (a) [MLT −2A −2 ]

(b) [ML2T −1A −2 ]

(c) [ML2T −2A −2 ]

(d) [ML2T −2A −1 ]

3. What will be the ratio of the distance moved by a freely falling body from rest in 4th and 5th seconds of journey? (a) 4 : 5

(b) 7 : 9

(c) 16 : 25

(d) 1 : 1

4. A person swims in a river aiming to reach exactly opposite point on the bank of a river. His speed of swimming is 0.5 m/s at an angle 120° with the direction of flow of ρ water. The speed of water in stream is (a) 1.0 m/s (c) 0.25 m/s

(b) 0.5 m/s (d) 0.43 m/s

5. The maximum range of a gun of horizontal terrain is 16 km. If g = 10 m/s2 , then muzzle velocity of a shell must be (a) 160 m/s

(b) 200 2 m/s (c) 400 m/s

(d) 800 m/s

6. An electric fan has blades of length 30 cm measured from the axis of rotation. If the fan is rotating at 120 rev/min, the acceleration of a point on the tip of the blade is

78

(a) 1600 m/s 2

(b) 47.4 m/s 2

(c) 23.7 m/s 2

(d) 50.50 m/s 2

7. A 600 kg rocket is set for a vertical firing. If the exhaust speed is 1000 m/s, the mass of the gas ejected per second to supply the thrust needed to overcome the weight of rocket is

(a) 117.6 kgs −1

(b) 58.6 kgs −1 (c) 6 kgs −1

(d) 76.4 kgs −1

8. Consider a car moving along a straight horizontal road with a speed of 72 km/h. If the coefficient of static friction between the tyres and the road is 0.5, the shortest distance in which the car can be stopped is (Take g = 10 m/s2 ) (a) 30 m

(b) 40 m

(c) 72 m

(d) 20 m

9. Two identical balls A and B moving with velocities + 0.5 m/s and − 0.3 m/s respectively collide head on elastically. The velocity of the balls A and B after collision will be respectively, (a) + 0.5 m/s and + 0.3 m/s (c) + 0.3 m/s and + 0.5 m/s

(b) − 0.3 m/s and + 0.5 m/s (d) − 0.5 m/s and + 0.3 m/s

10. A metal ball of mass 2 kg moving with a velocity of 36 km/h has a head on collision with a stationary ball of mass 3 kg. If after the collision, the two balls move together, the loss in kinetic energy due to collision is (a) 140 J

(b) 100 J

(c) 60 J

(d) 40 J

11. A circular disc is to be made using iron and aluminium. To keep its moment of inertia maximum about a geometrical axis, it should be prepared that (a) (b) (c) (d)

aluminium is at the interior and iron surrounds it iron is at the interior and aluminium surrounds it aluminium and iron layers are in alternate order sheet of iron is used at both external surfaces and aluminium sheet as inner material

AIPMT PREP UP 12. The moment of inertia of a body about a given axis is 1.2 kg- m2 . Initially, the body is at rest. In order to produce a rotational kinetic energy of 1500 J, an angular acceleration of 25 rad / s2 must be applied about the axis for a duration of (a) 4 s

(b) 2 s

(c) 8 s

(d) 10 s

13. Two satellites of earth, S1 and S2 , are moving in the

21. A bullet of mass 2 g is having a charge of 2 μC. Through, what potential difference must it be accelerated, starting from rest to acquire a speed of 10 m/s? (a) 5 kV

(b) 50 kV

(c) 5 V

(d) 50 V

22. An electric dipole of moment p is lying along a uniform electric field E. The work done in rotating the dipole by 90° is

same orbit. The mass of S1 is four times that of mass of S2 . Which one of the following statements is true?

(a) 2 pE

(a) The time period of S1 is four times that of S 2 (b) The potential energies of earth and satellite in the two cases are equal (c) S1 and S 2 are moving with the same speed (d) The kinetic energies of the two satellites are equal

(c) 2 pE

pE 2 (d) pE (b)

23. A student measures the terminal potential difference

14. A roller coaster is designed such that riders experience

(V ) of a cell (of emf ε and internal resistance r) as a function of the current (I) flowing through it. The slope and intercept of the graph between V and I, then respectively equal to

weightlessness, as they go round the top of the hill whose radius of curvature is 20 m. The speed of the car at the top of the hill is in between

24. The mean free path of electrons in a metal is 4 × 10 −8 m.

(a) 14 m/s and 15 m/s (c) 16 m/s and 17 m/s

(b) 15 m/s and 16 m/s (d) 13 m/s and 14 m/s

15. If λ m denotes the wavelength at which the radiative emission from a black body at a temperature T K is maximum, then (a) λ m ∝ T 4

(b) λ m is independent of T

(c) λ m ∝ T

(d) λ m ∝ T −1 −2 −1

7 cal - cm s . At a temperature of 727° C, the rate of heat radiated in the same units will be (b) 50

(c) 112

(d) 80

17. Two springs of spring constants k1 and k2 are joined in series. The effective spring constant of the combination is given by (a) k1k 2

(b)

( k1 + k 2 ) 2

(c) k1 + k 2

(d)

k1k 2 ( k1 + k 2 )

18. The potential energy of a long spring when stretched by 2 cm is U. If the spring is stretched by 8 cm, the potential energy stored in it, is (a) 4 U

(b) 8 U

(c) 16 U

(d) U / 4

19. Two vibrating tunning forks produce progressive

waves given by the y1 = 4 sin500 πt and y2 = 2 sin506 πt . Number of beats produced per minute are (a) 360 (c) 30

(b) 180 (d) 60

20. A transverse wave propagating along x-axis is represented by π y(x,t ) = 8.0 sin⎛⎜ 0.5 πx − 4 πt − ⎞⎟ ⎝ 4⎠ where, x is in metre and t is in second. The speed of the wave is (a) 4 π m/s π (c) m/s 4

(b) 0.5 π m/s (d) 8 m/s

(b) −r and ε (d) −ε and r

The electric field which can give an average 2 eV energy to an electron in the metal will be (in unit of Vm−1 ) (a) 8 × 10 7 (c) 8 × 10 −11

(b) 5 × 10 −11 (d) 5 × 10 7

25. Two bulbs 25 W, 220 V and 100 W, 220 V are given. Which has higher resistance?

16. A black body at 227° C radiates heat at the rate of (a) 60

(a) ε and r (c) r and −ε

(a) 25 W bulb (b) 100 W bulb (c) Both bulbs will have equal resistance (d) Resistance of bulbs cannot be compared

26. Two 220 V, 100 W bulbs are connected first in series and then in parallel. Each time the combination is connected to a 220 V AC supply line. The power drawn by the combination in each case respectively will be (a) 200 W, 150 W (c) 50 W, 100 W

(b) 180 W, 200 W (d) 50 W, 200 W

27. To convert a galvanometer into a voltmeter, one should connect (a) high resistance in series with galvanometer (b) low resistance in series with galvanometer (c) high resistance in parallel with galvanometer (d) low resistance in parallel with galvanometer

28. A galvanometer of resistance 50 Ω is connected to a battery of 3V alongwith a resistance of 2950 Ω in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be (a) 5050 Ω (c) 6050 Ω

(b) 5550 Ω (d) 4450 Ω

29. Above Curie temperature, (a) a ferromagnetic substance becomes paramagnetic (b) a paramagnetic substance becomes diamagnetic (c) a diamagnetic substance becomes paramagnetic (d) a paramagnetic substance becomes ferromagnetic

79

AIPMT PREP UP 30. A bar magnet having a magnetic moment of

2 × 104 JT −1 is free to rotate in a horizontal plane. A horizontal magnetic field B = 6 × 10 −4 T exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction 60° from the field is (a) 0.6 J

(b) 12 J

(c) 6 J

(d) 2 J

31. In an inductor of self-inductance L = 2 mH, current changes with time according to relation i = t 2 e− t . At what time emf is zero? (a) 4 s

(b) 3 s

(c) 2 s

(d) 1 s

32. The total charge induced in a conducting loop when it is moved in magnetic field depends on

flowing through the circuit is

D1

30 Ω

(b) 5.6 × 10 −10 m

(a) 2.5 m (c) 4 × 10

−10

(d) 4.6 × 10 −10 m

m

38. Monochromatic light of frequency 6.0 × 1014 Hz is

produced by a laser. The power emitted is 2 × 10 −3 W. The number of photons emitted on the average by the source per second is (a) 5 × 1015

(b) 5 × 1016

(c) 5 × 1017

(d) 5 × 1014

39. The energy of ground electronic state of hydrogen atom is −13.6 eV. The energy of first excited state will be (b) −27.2 eV

(c) −6.8 eV

(d) −3.4 eV

40. The ionization energy of the electron in the hydrogen

33. If internal resistance of cell is negligible, then current D2

lattice planes of diffraction grating is 2.8 × 10 −10 m, then the maximum wavelength of X-rays (in metre) is

(a) −54.4 eV

(a) the rate of change of magnetic flux (b) initial magnetic flux only (c) the total change in magnetic flux (d) final magnetic flux only

30 Ω

37. In X-rays diffraction experiment, distance between atomic

atom in its ground state is 13.6 eV. The atoms are excited to higher energy levels to emit radiations of 6 wavelengths. Maximum wavelength of emitted radiation corresponds to transition between (a) n = 3 to n = 2 states (c) n = 2 to n = 1 states

(b) n = 3 to n = 1 states (d) n = 4 to n = 3 states

41. The relationship between disintegration constant (λ) and the half-time (T) will be (a) λ =

+ –

5 (b) A 50

4 (c) A 50

2 (d) A 50

34. A coil of inductance 8.4 mH and resistance 6 Ω are connected to a 12 V battery. The current in the coil is 1.0 A at approximately time (a) 500 s

(b) 25 s

(c) 35 s

(d) 1 ms

is situated at a distance of 1 km from two objects. The minimum distance between these two objects, which can be resolved by the telescope, when the mean wavelength of light is 5000 Å is of the order of (b) 5 m

(c) 5 mm

constants 5λ and λ, respectively. If initially they have the same number of nuclei, then the ratio of the 1 number of nuclei of N1 to that of N2 will be after a time e (a) λ

(b)

1 λ 2

(c)

(d) 5 cm

(a) OR gate

(b) XOR gate

(c) AND gate (d) NAND gate

junction diode, which of the following is correct?

(a)

80

5 d 4

3L/4

(b)

4 d 5

Diode

(a) In forward biasing, the voltage across R is V V (b) In reverse biasing, the voltage across R is V (c) In forward biasing, the voltage across R is 2 V (d) In reverse biasing, the voltage across R is 2 V

A B

(c) d

R

45. Following diagram performs the logic function of

2d

H/2

e λ

B

44. For a given circuit of ideal p- n

L

(d)

Y

Cross-sectional area A/5 is immersed such that it floats with its axis vertical at the liquid-liquid interface with length L/4 in the denser liquid as shown in the figure. The lower density liquid is open to atmosphere having pressure p0 . Then, density of solid is given by

d

1 4λ

A

36. A homogeneous solid cylinder of length L (L ν 2 , therefore stopping potential will be more for the radiation of frequency ν1 . The properties of nuclear forces are (a) exchange forces (b) spin dependent (c) charge independent (d) do not obey inverse square law As we know, V = E − Ir E −V r= = 5Ω I

12. (i) In figure I, both diodes are forward bias V 2 = = 0.1 A R 20 (ii) No current flows in the circuit. So,

I

I=

Intensity −10

II

J or We know from radioactive equation N = N 0 e − λt . The activity of a radioactive substance can also be expressed in terms of its half-life 0.693 N I = − λN = − T1 / 2

Saturation current

20. (i) As short radiowaves are reflected

21.

24.

(b) We know that P = I V cos φ, where cos φ is the power factor, to supply given power at a given voltage (c) Saurabh → Social awareness, adaptation etc. Uncle → Knowledge, adaptation, concern of others etc. (i) A field in which work done does not depend on the path followed. (ii) Capacitance of a spherical capacitance depends on

(a) radius (b) medium between the plates (iii) E =

V 12 = = 4 × 10 3 V/m d 3 × 10 −3

(iv) Q

V

or +q

F–q = – qE V

19. E = mc = 1.5 × 10

O Collector plate potential Retarding potential

VCE

K2

IC

2

–V01 –V02

–

IE

–Vo

ν 1 > ν2

23. (a) To minimise power loss due to heat.

Rh2 +

n-p-n + VC V2 –

E

14. (i) As majority charge carriers in doped

(Short circuit current)

The significance of the graph is that a solar cell does not draw current but supplies same to the load. Using formula, δ 30 + 1, δ = A(μ − 1), μ = + 1 = A 60 μ = 1.5 or If the intensity of light remains same in all the orientations of the polariser, then the light is unpolarised. But if the intensity of light varies from maximum to minimum then the light is a polarised light beam.

–IB

VCC Rh1 +

Photo current

Y Y=A+B

B

IC – mA +

C

d

2.

13. A thin and lightly doped base region

2 i = A + δm Space wave or Line of Sight (LOS) propagation. It is an OR gate and symbol is

Photo current

1. The relationship between i, A and δ m is

back by ionosphere. (ii) These are not reflected by the ionosphere therefore to reflect back the TV signals to desire location on Earth, then satellites are used. It is based on the Wheatstone bridge principle for the present situation. R 40 2 2X or = = R= X 60 3 3 When the resistors are doubled and interchanged, we have 2X L 2X×3 L or = = 2 R 100 − L 2 × 2 X 100 − L 3 L ⇒ = 2 100 − L Solving, L = 60 cm. There will be no change in the balance point on interchanging the battery and the galvanometer.

F+q = qE

θ d sin θ

–q

(i) Fnet = zero ∴ |F+q | = |F−q | and opposite direction (ii) Torque, τ =p ×E (iii) Work done, W = ∫ τdθ = 2 pE ⇒ 2 pE

25. (i) α-particle possess minimum kinetic

energy. (ii) Electron has maximum kinetic energy. or μ NIR 2 (iii) For coil 1, B1 = 0 2 3 / 2 2 (2 R ) For coil 2, B2 =

μ 0 NIR 2 2 (2 R 2 )3 / 2

Both are in same direction, so resultant magnetic field is μ NIR 2 B = 0 2 3/ 2 (2 R )

26. (i) Total power,

P = − 4 + 2 = − 2D 1 P= f 1 1 ⇒ f= = × 100 = − 50 cm P −2 (ii) Focal length increases, i.e. fr > fv (iii) Using lens maker formula, ⎛1 1 1⎞ = ( n − 1)⎜ − ⎟ f ⎝ R1 R2 ⎠

Just Solve & Send

Quizzer (No.16) 1. An astronaut marooned on the surface of an asteroid of radius r and mean density equal to the earth, find that he can escape by jumping. What is the minimum value of r? (a) r 2 > Rh (c) r 2 < Rh

(b) r > Rh (d) None of these

2. Free neutrons have a decay constant of -3

-1

1.10 ´ 10 s . If the de-Broglie wavelength of the neutrons in a parallel beam is 1 nm, determine the distance from the source where the beam intensity has dropped to half its starting value. (a) 2.5 ´ 10 5 m (c) 2.5 ´ 10 4 m

(b) 3.5 ´ 10 4 m (d) 15 . ´ 10 6 m

3. A telescope is used to observe at a distance of 10 km, two objects which are 0.12 m apart and illuminated by light of wavelength 600 nm. Estimate the diameter of the objective lens of the telescope, if it can just resolve the two objects. (a) 5 cm

(b) 4 cm

(c) 6 cm

(d) 8 cm

4. Consider a gas at temperature of 2500 K whose atoms can occupy only two energy levels separated by 1.5 eV energy gap. Ratio of number of atoms in the higher energy level to the lowest energy level is (a) 1

(b) 8.2 ´ 10 -4 (c) 9.64 ´ 10 -4 (d) 0

5. A singly charged ions He are accelerated in a +

cyclotron, so that their maximum orbital radius r = 60 cm. The frequency of a cyclotron oscillator is equal to n = 10 MHz, the effective accelerating voltage across the dee is V = 50 kV. Neglecting the gap between the dees. Find the total time of acceleration of the ion and approximate distance covered by the ion in the process of its acceleration. (a) 15 ms, 0.74 km (c) 16 ms, 0.74 km

(b) 20 ms, 1.25 km (d) 22 ms, 3.24 km

6. A 2.24 m high cylinder whose base area is 1 dm

2

contains 4 g of helium gas at a temperature of 0°C and pressure of 10 N /cm2 . An 80 kg piston is then dropped into the cylinder. Find the maximum

speed of the piston, if it moves without friction. There is no heat transfer between the gas, the cylinder and the piston, because of the rapidity of the process. (Take g = 10 m/s2 ) C V = 3150 Jkg -1K -1 , C p = 5250 Jkg -1K -1 (a) 2.7 m/s (c) 4.5 m/s

(b) 3.4 m/s (d) 5.9 m/s

80 kg

2.24 m

1 dm2

7. A metallic ring of cross-section 2.5 m , mean radius 2

40 cm and relative permeability 1500 is wound uniformly with 3000 turns of wire. If a current of 1.6 A passes through the wire, find the magnetic field B and magnetisation in the ring. (a) 2.9 ´ 10 6 Am-1 (c) 3.2 ´ 10 6 Am-1

(b) 4.2 ´ 10 4 Am-1 (d) 9.2 ´ 10 6 Am-1

8. A superconductor used at room temperature for an energy storage device. What is the maximum energy stored in a solenoid with the dimensions of a torch battery of length 0.05 m and diameter 0.03 m, if the maximum magnetic field sustainable by the superconductor is 15 T? (a) 3.2 kJ

(b) 4.89 kJ

(c) 9.2 kJ

(d) 10.1 kJ

9. The particle of rest mass m moving along the X-axis

m 2 moving along the axis with velocity -v. If the two particles coalesce, find the rest mass of the resulting particle.

with velocity v collides with a particle of rest mass

m 2 7m (c) 2

(a)

3m 2 9m (d) 2 (b)

10. The spherical nucleus has a total charge q (uniformly distributed) and radius R. Find the electric field at any point inside the nucleus at a distance r from the centre. Hence, find the potential difference between the centre of the nucleus and its surface. (a) 4 MV (c) 1 MV

(b) 2 MV (d) 0.1 MV

95

MAGAZINE QUIZ

KNOWLEDGE Coefficient Quizzer (No. 16)

Winner of Knowledge Coefficient Quizzer (No. 15) (February Issue) Ashish Singh (Haridwar)

Physics Spectrum Arihant Media Promoters c/o Arihant Prakashan Kalindi, T.P. Nagar, Meerut (UP) - 250002

fjohtu cum dSz 'k dkslZ

JEE Main 2016

96

MARCH 2016