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PHYSICS

Years IIT-JEE CHAPTERWISE SOLVED PAPERS

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Preface

Whenever a student decides to prepare for any examination, her/his first and foremost curiosity

about the type of questions that he/she has to face. This becomes more important in the context of

competitive examinations where there is neck-to-neck race.

We feel great pleasure to present before you this book. We have made an attempt to provide

chapter wise questions asked in IIT-JEE / JEE Advanced from 1978 to 2016 along with solutions.

Solutions to the questions are not just sketch rather have been written in such a manner that the

students will be able to under the application of concept and can answer some other related

questions too.

We firmly believe that the book in this form will definitely help a genuine, hardworking student.

We have tried our best to keep errors out of this book. Comment and criticism from readers will be

highly appreciated and incorporated in the subsequent edition.

We wish to utilize the opportunity to place on record our special thanks to all team members of

Content Development for their efforts to make this wonderful book.

Career Point Ltd.

CONTENTS

Chapter Page No.

1. Unit, Dimension & Error 1‐16 ♦ Answers

♦ Solutions

9 10

2. Kinematics 17‐32 ♦ Answers

♦ Solutions

23 24

3. Laws of Motion & Friction 33‐54 ♦ Answers

♦ Solutions

41 42

4. Work, Power and Energy 55‐68 ♦ Answers

♦ Solutions

61 62

5 Conservation Law 69‐88 ♦ Answers

♦ Solutions

77 78

6. Rotational Motion 89‐130 ♦ Answers

♦ Solutions

106 108

7. Gravitation 131‐142 ♦ Answers

♦ Solutions

135 136

8. Simple Harmonic Motion 143‐162 ♦ Answers

♦ Solutions

151 152

9. Properties of Matter & Fluid Mechanics 163‐188

♦ Answers ♦ Solutions

173 174

10. Wave Motion 189‐222


201 203

11. Heat and Thermodynamics 223‐282


246 248

Cont....

12. Electrostatics 283‐332


303 305

13 Current Electricity 333‐360


345 346

14. Magnetic Effect of Current 361‐400


378 380

15. Electromagnetic Induction and Alternating Current 401‐434


415 417

16. Optics 435‐504


461 463

17. Modern Physics 505‐564


529 531

18. Model Test Papers 565‐598

♦ Practice Test‐1 [Paper‐1] ♦ Practice Test‐1 [Paper‐2] ♦ Practice Test‐2 [Paper‐1] ♦ Practice Test‐2 [Paper‐2]

565 573 581 589

1 Unit, Dimension & Error

Chapter

ONLY ONE CORRECT ANSWER 1. In the formula X = 3Y Z2, X and Z have

dimensions of capacitance and magnetic induction respectively. What are the dimensions of Y is MKSQ system? [1995, 2M]

(A) [M–3 L–1 T3 Q4] (B) [M–3 L–2 T4 Q4] (C) [M–2 L–2 T4 Q4] (D) [M–3 L–2 T4 Q]

2. The dimensions of 21

ε0E2 (ε0 : permittivity of

free space; E : electric field) is [2000, 2M]

(A) [MLT–1] (B) [ML2 T–2] (C) [MLT–2] (D) [ML2 T–1]

3. A quantity X is given by ε0 L tV

∆∆ , where ε0 is

the permittivity of free space, L is a length, ∆V is a potential difference and ∆t is a time interval. The dimensional formula for X is the same as that of [2001, 2M]

(A) resistance (B) charge (C) voltage (D) current 4. A cube has a side of length 1.2 × 10–2m.

Calculate its volume [2003, 2M] (A) 1.7 × 10–6 m3 (B) 1.73 × 10–6 m3 (C) 1.70 × 10–6 m3 (D) 1.732 × 10–6 m3

5. In the relation p = βα θ

αk

Z–e

p is pressure, Z is distance, k is Boltzmann constant and θ is the temperature. The dimensional formula of β will be - [2004, 2M]

(A) [M0 L2 T0] (B) [ML2 T] (C) [ML0 T–1] (D) [M0 L2 T–1]

6. A wire has a mass (0.3 ± 0.003) g, radius (0.5 ± 0.005) mm and length (6 ± 0.06) cm. The maximum percentage error in the measurement of its density is [2004, 2M]

(A) 1 (B) 2 (C) 3 (D) 4

7. Which of the following sets have different dimensions? [2005, 2M]

(A) Pressure, Young's modulus, Stress (B) Emf, Potential difference, Electric potential (C) Heat, Work done, Energy (D) Dipole moment, Electric flux, Electric field 8. The circular scale of a screw gauge has

50 divisions and pitch of 0.5 mm. Find the diameter of sphere. Main scale reading is 2.

[2006, 3M]

: : : :: : : E

5 ′0

H N AB S

M

K

R

:: : :: : : E 25 ′′2

H N AB S

M

K

R

(A) 1.2 mm (B) 1.25 mm (C) 2.20 mm (D) 2.25 mm 9. A student performs an experiment to determine

the Young's modulus of a wire, exactly 2 m long, by Searle's method. In a particular reading, the student measures the extension in the length of the wire to be 0.8 mm with an uncertainty of ±0.05 mm at a load of exactly 1.0 kg. The student also measures the diameter of the wire to be 0.4 mm with an uncertainty of ±0.01 mm. Take g = 9.8 m/s2 (exact). The Young's modulus obtained from the reading is close to [2007, 3M]

(A) (2.0 ± 0.3) × 1011 N/m2 (B) (2.0 ± 0.2) × 1011 N/m2 (C) (2.0 ± 0.1) × 1011 N/m2 (D) (2.0 ± 0.5) × 1011 N/m2

39 YEARS TOPIC- WISE JEE Advanced Questions with Solutions

2

10. In the experiment to determine the speed of sound using a resonance column [2007, 3M]

(A) prongs of the tuning fork are kept in a vertical plane

(B) prongs of the tuning fork are kept in a horizontal plane

(C) in one of the two resonances observed, the length of the resonating air column is close to the wavelength of sound in air

(D) in one of the two resonances observed, the length of the resonating air column is close to half of the wavelength of sound in air

11. Students I, II and III perform an experiment for

measuring the acceleration due to gravity (g) using a simple pendulum.

They use different lengths of the pendulum and/or record time for different number of oscillations. The observations are shown table.

Least count for length = 0.1 cm, Least count for time = 0.1 s

Stud

ent Length

of the pendulum

(cm)

Number of oscillations

(n)

Total time for (n)

oscillations (s)

Time period

(s)

I 64.0 8 128.0 16.0

II 64.0 4 64.0 16.0

III 20.0 4 36.0 9.0

If EI, EII and EIII are the percentage errors in g,

i.e.,

×

∆ 100gg for students I, II and III,

respectively [2008, 3M] (A) EI = 0 (B) EI is minimum (C) EI = EII (D) EII is maximum 12. A vernier calipers has 1 mm marks on the main

scale. It has 20 equal divisions on the vernier scale which match with 16 main scale divisions. For this vernier calipers, the least count is [2010]

(A) 0.02 mm (B) 0.05 mm (C) 0.1 mm (D) 0.2 mm

13. The density of a solid ball is to be determined in an experiment. The diameter of the ball is measured with a screw gauge, whose pitch is 0.5 mm and there are 50 divisions on the circular scale. The reading on the main scale is 2.5 mm and that on the circular scale is 20 divisions. If the measured mass of the ball has a relative error of 2%, the relative percentage error in the density is [2011]

(A) 0.9% (B) 2.4% (C) 3.1% (D) 4.2% 14. In the determination of Young's modulus

π= 2ld

MLg4Y by using Searle's method, a wire

of length L = 2 m and diameter d = 0.5 mm is used. For a load M = 2.5 Kg, an extension l = 0.25 mm in the length of the wire is observed. Quantities d and l are measured using a screw gauge and a micrometer, respectively. They have the same pitch of 0.5 mm. The number of division on their circular scale is 100. The contribution to the maximum probable error of the Y measurement. [2012]

(A) due to the errors in the measurements of d and l are the same.

(B) due to the error in the measurement of d is twice that due to the error in the measurement of l

(C) due to the error in the measurement of l is twice that due to the error in the measurement of d

(D) due to the error in the measurement of d is four times that due to the error in the measurement of l

15. The diameter of a cylinder is measured using a

Vernier callipers with no zero error. It is found that the zero of the Vernier scale lies between 5.10 cm and 5.15 cm of the main scale. The Vernier scale has 50 divisions equivalent to 2.45 cm. The 24th division of the Vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is

[2013] (A) 5.112 cm (B) 5.124 cm (C) 5.136 cm (D) 5.148 cm

3 UNIT, DIMENSION & ERROR

LINKED COMPREHENSION TYPE

Passage # 1 (1 & 2) A dense collection of equal number of electrons

and positive ions is called neutral plasma. Certain solids containing fixed positive ions surrounded by free electrons can be treated as neutral plasma. Let N be the number density of free electrons, each of mass m. When the electrons are subjected to an electric field, they are displaced relatively away from the heavy positive ions. If the electric field becomes zero, the electrons begin to oscillate about the positive ions with a natural angular frequency ωp, which is called the plasma frequency. To sustain the oscillations, a time varying electric field needs to be applied that has an angular frequency ω, where a part of the energy is absorbed and a part of it is reflected. As ω approaches ωp, all the free electrons are set to resonance together and all the energy is reflected. This is the explanation of high reflectivity of metals. [2011]

1. Taking the electronic charge as e and the

permittivity as ε0, use dimensional analysis to determine the correct expression for ωp.

(A) 0m

Neε

(B) Ne

m 0ε

(C) 0

2

mNe

ε (D) 2

0

Nemε

2. Estimate the wavelength at which plasma

reflection will occur for a metal having the density of electrons N = 4 × 1027 m–3. Take ε0 ≈ 10–11 and m ≈ 10–30, where these quantities are in proper SI units.

(A) 800 nm (B) 600 nm (C) 300 nm (D) 200 nm Passage # 2 (3 to 5) Phase space diagrams are useful tools in

analyzing all kinds of dynamical problems. They are especially useful in studying the changes in motion as initial position and momentum are changed. Here we consider

some simple dynamical systems in one-dimension. For such systems, phase space is a plane in which position is plotted along horizontal axis and momentum is plotted along vertical axis. The phase space diagram is x(t) vs p(t) curve in this plane. The arrow on the curve indicates the time flow. For example, the phase space diagram for a particle moving with constant velocity is a straight line as shown in the figure. We use the sign convention in which position or momentum upwards (or to right) is positive and downwards (or to left) is negative.

[2011] 3. The phase space diagram for a ball thrown

vertically up from ground is

(A)Position

Momentum

(B)Position

Momentum

(C) Position

Momentum

(D)Position

Momentum

4. The phase space diagram for simple harmonic

motion is a circle centered at the origin. In the figure, the two circles represent the same oscillator but for different initial conditions, and E1 and E2 are the total mechanical energies respectively. Then

Position

Momentum

a

2a E1 E2

E1

(A) E1 = 2 E2 (B) E1 = 2E2

(C) E1 = 4 E2 (D) E1 = 16 E2


4

5. Consider the spring-mass system, with the mass submerged in water, as shown in the figure. The phase space diagram for one cycle of this system is

(A)

Position

Momentum

(B)

Position

Momentum

(C)

Position

Momentum

(D)

Position

Momentum

ONE OR MORE THAN ONE CORRECT ANSWERS

1. L, C and R represent the physical quantities inductance, capacitance and resistance respectively. The combinations which have the dimensions of frequency are [1984, 2M]

(A) 1 / RC (B) R / L

(C) 1 / LC (D) C / L 2. The dimensions of the quantities in one

(or more) of the following pairs are the same. Identify the pair (s) [1986, 2M]

(A) torque and work (B) angular momentum and work (C) energy and Young's modulus (D) light year and wavelength 3. The pairs of physical quantities that have the

same dimensions is (are) [1995, 2M]

(A) Reynolds number and coefficient of friction (B) Curies and frequency of a light wave (C) Latent heat and gravitational potential (D) Planck's constant and torque

4. Let [ε0] denote the dimensional formula of the permittivity of the vacuum and [µ0] that of the permeability of the vacuum. If M = mass, L = length, T = time and I = electric current.

[1998, 2M] (A) [ε0] = [M–1 L–3 T2 I] (B) [ε0] = [M–1 L–3 T4 I2] (C) [µ0] = [ML T–2 I–2] (D) [µ 0] = [ML2 T–1 I]

5. The SI unit of the inductance, the henry can by written as [1998, 2M]

(A) weber/ampere (B) volt-second/ampere (C) joule/(ampere)2 (D) ohm-second

6. A student uses a simple pendulum of exactly 1 m length to determine g, the acceleration due to gravity. He uses a stop watch with the least count of 1 s for this and records 40 s for 20 oscillations. For this observation, which of the following statement(s) is/are true?

(A) Error ∆T in measuring T, the time period, is 0.05 s

(B) Error ∆T in measuring T, the time period, is 1 s

(C) Percentage error in the determination of g is 5%

(D) Percentage error in the determination of g is 2.5%

7. Planck's constant h, speed of light c and

gravitational constant G are used to form a unit of length L and a unit of mass M. Then the correct option(s) is(are) [2015]

(A) M ∝ c (B) M ∝ G

(C) L ∝ h (D) L ∝ G

8. In terms of potential difference V, electric current I, permittivity ε0, permeability µ0, and speed of light c, the dimensionally correct equation (s) is (are) [2015]

(A) µ0I2 = ε0V2 (B) ε0I = µ0V (C) I = ε0cV (D) µ0cI = ε0V


9. A length-scale (l) depends on the permittivity (ε) of a dielectric material, Boltzmann constant (kB), the absolute temperature (T), the number per unit volume (n) of certain charged particles, and the charge (q) carried by each of the particles. Which of the following expressions (s) for l is (are) dimensionally correct ? [2016]

(A) l =

ε Tknq

B

2 (B) l =

ε2

B

nqTk

(C) l =

ε Tknq

B3/2

2 (D) l=

ε Tknq

B3/1

2

10. In an experiment to determine the

acceleration due to gravity g, the formula used for the time period of a periodic motion

is T = 2πg5

)r–R(7 . The values of R and r

are measured to be (60 ± 1) mm and (10 ± 1) mm, respectively. In five successive measurements, the time period is found to be 0.52 s, 0.56 s, 0.57 s, 0.54 s and 0.59 s. The least count of the watch used for the measurement of time period is 0.01 s. Which of the following statement(s) is(are) true ?

[2016] (A) The error in the measurement of r is 10% (B) The error in the measurement of T is

3.57% (C) The error in the measurement of T is 2% (D) The error in the determined value of g is

11%

MATRIX MATCH TYPE

1. Column I gives three physical quantities. Select the appropriate units for the choices given in Column II. Some of the physical quantities may have more than one choice. [1990, 3M]

Column I Column II Capacitance ohm-second Inductance coulomb2 – joule–1

Magnetic induction coulomb (volt)–1, newton (ampere metre)–1, volt-second (ampere)–1

2. Match the physical quantities given in Column I with dimensions expressed in terms of mass (M), length (L), time (T), and charge (Q) given in Column II and write the correct answer against the matched quantity in a tabular form in your answer book. [1993, 6M]

Column I Column II Angular momentum [M L2 T–2]

Latent heat [M L2 Q–2]

Torque [M L2 T–1]

Capacitance [M L3 T–1 Q–2]

Inductance [M–1 L–2 T2 Q2]

Resistivity [L2 T–2]

3. Some physical quantities are given in Column I

and some possible SI units in which these quantities may be expressed are given in Column II. Match the physical quantities in Column I with the units in Column II.

[2007, 6M] Column I Column II

(A) GMeMs G — universal gravitational constant, Me — mass of the earth, Ms — mass of the sun.

(p) (volt) (coulomb) (metre)

(B) MRT3

R — universal gas constant, T — absolute temperature, M — molar mass.

(q) (kilogram) (metre)3 (second)–2

(C) 22

2

BqF

F — force, q — charge, B — magnetic field.

(r) (metre)2 (second)–2

(D) e

e

RGM

G — universal gravitational constant, Me — mass of the earth, Re — radius of the earth.

(s) (farad) (volt)2 (kg)–1


6

4. Column II gives certain systems undergoing a process. Column I suggests changes in some of the parameters related to the system Match the statements in Column I to the appropriate process (es) from Column II. [2009]

Column I Column II (A) The energy of the

system is increased. (p) System : A capacitor,

initially uncharged. Process : It is connected

to a battery. (B) Mechanical energy

is provided to the system, which is converted into energy of random motion of its parts.

(q) System : A gas in anadiabatic container fitted with an adiabatic piston.

Process : The gas is compressed by pushing the piston.

(C) Internal energy of the system is converted into its mechanical energy.

(r) System : A gas in a rigid container.

Process : The gas gets cooled due to colder atmosphere surrounding it.

(s) System : A heavy nucleus, initially at rest.

Process : The nucleusfissions into two fragments of nearly equal masses and some neutrons are emitted.

(D) Mass of the system is decreased.

(t) System : A resistive wire loop.

Process : The loop is placed in a time varying magnetic field perpendicular to its plane.

5. Column II shows five systems in which two

objects are labelled as X and Y. Also in each case a point P is shown. Column I given some statements about X and/ or Y. Match these statements to the appropriate system(s) from Column II. [2009]

Column I Column II (A) The force exerted

by X on Y has a magnitude Mg.

(p) P

Y

X

Block Y of mass M left on a fixed inclined plane X, slides on it with a constant velocity.

(B) The gravitational potential energy of X is continuously increasing.

(q)

X

PZ

Y

Two ring magnets Y and Z, each of mass M, are kept in frictionless vertical plastic stand so that they repel each other. Y rests on the base X and Z hangs in air in equilibrium. P is the topmost point of the stand on the common axis of the two rings. The whole system is in a lift that is going up with a constant velocity.

(C) Mechanical energy of the system X + Y is continuously decreasing.

(r)

PX

Y

A pulley Y of mass m0

is fixed to a table through a clamp X. A block of mass M hangs from a string that goes over the pulley and is fixed at point P of the table. The whole system is kept in a lift that is going down with a constant velocity.


(s)

Y

P X

A sphere Y of mass M is put in a non-viscous liquid X kept in a container at rest. The sphere is released and it moves down in the liquid.

(D) The torque of the weight of Y about point P is zero.

(t)

Y

P X

A sphere Y of mass M is falling with its terminal velocity in a viscous liquid X kept in a container.

6. Match List I with List II and select the correct

answer using the codes given below the lists : [2013] List I List II (P) Boltzmann constant (1) ML2T–1 (Q) Coefficient of viscosity (2) ML–1T–1 (R) Planck constant (3) MLT–3K–1 (S) Thermal conductivity (4) ML2T–2K–1 Codes : P Q R S (A) 3 1 2 4 (B) 3 2 1 4 (C) 4 2 1 3 (D) 4 1 2 3

FILL IN THE BLANKS

1. Planck's constant has dimensions …….. [1985, 2M] 2. In the formula X = 3Y Z2, X and Z have

dimensions of capacitance and magnetic induction respectively. The dimensions of Y in MKSQ system are …….. [1988, 2M]

3. The dimensions of electrical conductivity is …….. [1997,1M]

4. The equation of state of a real gas is given by

+ 2V

ap (V – b) = RT

where p, V and T are pressure, volume and temperature respectively and R is the universal gas constant. The dimensions of the constant a in the above equation is ….. [1997, 2M]

ANALYTICAL & DESCRIPTIVE QUESTIONS

1. Give the MKS units for each of the following quantities [1980, 3M]

(a) Young's modulus, (b) Magnetic induction, (c) Power of a lens. 2. A gas bubble, from an explosion under water,

oscillates with a period T proportional to pa db Ec, where p is the static pressure, d is the density of water and E is the total energy of the explosion. Find the values of a, b and c.

[1981, 3M] 3. Write the dimensions of the following in terms

of mass, time, length and charge [1982, 2M] (a) Magnetic flux, (b) Rigidity modulus. 4. N divisions on the main scale of a vernier

callipers coincide with (N + 1) divisions on the vernier scale. If each division on the main scale is of a units, determine the least count of instrument. [2003, 2M]

5. In a Searle's experiment, the diameter of the

wire as measured by a screw gauge of least count 0.001 cm is 0.050 cm. The length, measured by a scale of least count 0.1 cm, is 110.0 cm. When a weight of 50 N is suspended from the wire, the extension is measured to be 0.125 cm by a micrometer of least count 0.001 cm. Find the maximum error in the measurement of Young's modulus of the material of the wire from these data.

[2004, 2M]


8

6. The pitch of a screw gauge is 1 mm and there are 100 divisions on the circular scale. While measuring the diameter of a wire, the linear scale reads 1 mm and 47th division on the circular scale coincides with the reference line. The length of the wire is 5.6 cm. Find the curved surface area (in cm2) of the wire in appropriate number of significant figures.

[2004, 2M] 7. The edge of a cube is measured using a vernier

calliper. (9 divisions of the main scale is equal to 10 divisions of vernier scale and 1 main scale division is 1 mm). The main scale division reading is 10 and 1 division of vernier scale was found to be coinciding with the main scale. The mass of the cube is 2.736 g. Calculate the density in g/cm3 upto correct significant figures. [2005, 2M]

8. To find the distance d over which a signal can be seen clearly in foggy conditions, a railways engineer uses dimensional analysis and assumes that the distance depends on the mass density ρ of the fog, intensity (power/area) S of the light from the signal and its frequency ƒ. The engineer finds that d is proportional to S1/n. The value of n is.

[2013] 9. The energy of a system as a function of time t is

given as E(t) = A2 exp(–αt), where α = 0.2 s–1. The measurement of A has an error of 1.25 %. If the error in the measurement of time is 1.50 %, the percentage error in the value of E(t) at t = 5 s is [2015]


ANSWERS

Only One Correct Answer

1. (B) 2. None of the four choices 3. (D) 4. (A) 5. (A) 6. (D)

7. (D) 8.(A) 9.(B) 10. (A) 11. (B) 12. (D) 13. (C)

14. (A) 15. (B)

Linked Comprehension Type

1. (C) 2. (B) 3. (D) 4. (C) 5. (B)

One or More than One Correct Answers

1. (A,B,C) 2. (A,D) 3. (A,B,C) 4. (B,C) 5. (A,B,C,D) 6.(A,C)

7. (A,C,D) 8. (A,C) 9. (B,D) 10. (A,B,D)

Matrix Match Type

1. See the solution 2. See the solution 3. A → p,q ; B→ r, s; C→ r,s; D→ r, s

4. A→ p,q,s,t; B→ q; C→ s; D → s 5. A → p, t; B → q,s,t; C → p,r,t; D → q 6. (C)

Fill in the Blanks

1. [ML2T–1] 2. [M–3L–2T4Q4] 3. [M–1L–3T3A2] 4. [ML5T–2]

Analytical & Descriptive Questions

1. (a) N/m2 (b) Tesla (c) m–1 2. a = 65– , b =

21 , c =

31 3. (a) [ML2T–1Q–1]; (b)[ML–1T–2]

4. 1N

a+

5. 1.09 × 1010 N / m2 6. 2.6 cm2 7. 2.66 g/cm3 8. 3 9. 4


10

SOLUTIONS

Only One Correct Answer

1. [Y] =

2ZX

=

2)inductionMagnetic(

cetanCapaci

=

2–2–2

222–1–

TQMTQLM = [M–3L–2T4Q4]

∴ Correct answer is (B).

2. (None of the four choices) 21

ε0 E2 is the

expression of energy density (Energy per unit volume)

ε 2

0E21 =

3

2–2

LTML = [ML–1 T–2]

3. C = Vq

∆∆

or ε0 LA =

Vq

∆∆

or ε0 = )V.(A

L)q(∆

∆

X = ε0 L tV

∆∆ =

)V(AL)q(

∆∆ L

tV

∆∆

but [A] = [L2]

∴ X = tq

∆∆ = current

4. V = l3 = (1.2 × 10–2 m)3 = 1.728 × 10–6 m3 Q Length (l) has two significant figures, the

volume (V) will also have two significant figures. Therefore, the correct answer is

V = 1.7 × 10–6 m3

5.

θαk

Z = [M0L0T0]

[α] =

θ

Zk

Further [p] =

βα

∴ [β] =

αp

=

θZpk

Dimensions of kθ are that to energy. Hence,

[β] =

2–1–

2–2

TLMLTML = [M0L2T0]

Therefore, the correct option is (A).

6. Density ρ = Lr

m2π

∴ ρρ∆ × 100 =

∆

+∆

+∆

LL

rr2

mm × 100

After substituting the values, we get the maximum percentage error in density = 4%

Hence, the correct option is (D). 7. Dipole moment = (charge) × (distance) Electric flux = (electric field) × (area) Hence, the correct option is (D). 8. Least count (LC)

= scalecircularondivisionsofNumber

Pitch

= 50

5.0 = 0.01 mm

Now, Diameter of ball = (2 × 0.5 mm) + (25 – 5) (0.01) = 1.2 mm

9. Y = lA

FL = l2d

FL4π

= )108.0()104.0(

)2)(8.90.1)(4(3–23– ××π

×

= 2.0 × 1011 N / m2

Further YY∆ = 2

∆

dd +

∆ll

∴ ∆Y =

∆

+

∆

ll

dd2 Y

=

+×

8.005.0

4.001.02 × 2.0 × 1011

= 0.225 × 1011 N/m2 = 0.2 × 1011 N/m2 (By rounding off) or (Y + ∆Y) = (2 + 0.2) × 1011 N/m2 ∴ Correct option is (B).


10. Length of air column in resonance is odd integer

multiple of 4λ .

∴ Correct option is (A).

11. T = 2πgl

or nt = 2π

gl ∴g =

2

22

t)n)(4( lπ

% error in g = gg∆ × 100 =

∆

+∆

tt2

ll × 100

EI =

×

+128

1.0264

1.0 × 100 = 0.3125%

EII =

×

+64

1.0264

1.0 × 100 = 0.46875%

EIII =

×

+36

1.0220

1.0 × 100 = 1.055%

Hence EI is minimum. ∴ Correct option is (B). 12. Least count of vernier calipers LC = 1 MSD – 1 VSD

=scalevernierondivisionsofNumber

scalemainondivisionSmallest

20 divisions of vernier scale = 16 divisions of main scale

∴ 1 VSD = 2016 mm = 0.8 mm

∴ LC = 1 MSD – 1 VSD = 1 mm – 0.8 mm ∴ = 0.2 mm ∴ Correct option is (D).

13. Least count of screw gauge = 50

5.0

= 0.01 mm = ∆r

Diameter r = 2.5 mm + 20 × 50

5.0 = 2.70 mm

rr∆ =

70.201.0

or rr∆ × 100 =

7.21

Now density d = Vm =

3

2r

34

m

π

Here, r is the diameter.

∴dd∆ × 100 =

∆

+∆

rr3

mm × 100

= mm∆ × 100 + 3 ×

∆

rr × 100

= 2% + 3 × 7.2

1 = 3.11%

14. 2dMLg4Ylπ

= & % ymax

= %M + %L + %l + 2%d Least count of both instrument,

∆l = ∆d = 3105100

5.0 −×=

%l = %225.0

1051003

=×

=×∆ −

l

l

%11005.0

105100ddd%

3=×

×=×

∆=

−

here we see that, contribution of l, = 2% contribution of d = 2% d = 2 × 1 = 2% hence both terms l and d contribute equally.

15. Here 1 MSD = 0.05 cm

1 VSD = 5045.2 = 0.049 cm

Least count = 1 MSD – 1VSD = 0.001 cm diameter = 5.10 + (0.001) × 24 diameter = 5.124 cm Linked Comprehension Type

1. N = Number of electrons per unit volume ∴ [N] = [L–3], [e] = [q] = [It] = [AT] [ε0] = [M–1L–1T4A2] Substituting the dimensions we can see that,

ε0

2

mNe = [T–1]

Angular frequency has also the dimension [T–1] ∴ Correct option is (C).


12

2. ω = 2πf = λπc2

∴λ = ωπc2 =

02 m/Ne

c2

ε

π

Substituting the values, we get

λ ~– 600 nm. ∴ Correct option is (B). 3. Momentum is first positive but decreasing.

Displacement (or say position) is initially zero. It will first increase. At highest point, momentum is zero and displacement is maximum. After that momentum is downwards (negative) and increasing but displacement is decreasing. Only (D) option satisfies these conditions.

4. E = 21 mω2 A2

or E ∝ A2,

1

2

EE

= 2

1

2

AA

=

2

a2a

or E1 = 4E2 ∴ Correct option is (C). 5. In all the given four figures, at mean position

the position coordinate is zero. At the same time mass is starting from the

extreme position in all four cases. In figures (C) and (D), extreme position is more than the initial extreme position. But due to viscosity opposite should be the case.

Hence, the answer should be either option (A) or option (B) . Correct answer is (B), because mass starts from positive extreme position (from uppermost position). Then, it will move downwards or, momentum should be negative.

One or More than One Correct Answers

2. (A) Torque and work both have the dimensions [ML2T–2].

(D) Light year and wavelength both have the dimensions of length ie, [L].

∴ Answer is (A) and (D).

3. Reynold's number and coefficient of friction are dimensionless quantities.

Curie is the number of atoms decaying per unit time and frequency is the number of oscillations per unit time.

Latent heat and gravitational potential both have the same dimension corresponding to energy per unit mass.

4. F = 04

1πε

⋅2

21

rqq

[ε0] = ]r][F[]q][q[

221

= ]L][MLT[

]IT[22–

2

= [M–1 L–3 T4I2] Speed of light,

c = 00

1µε

∴ [µ0] = 2

0 ]c][[1

ε

=21–243–1– ]LT][ITLM[

1

= [MLT–2I–2]

5. (A) L = iφ or henry =

ampereweber

(B) e = – L

dtdi

∴L = –)dt/di(

e

or henry = ampere

ondsecvolt −

(C) U = 21 Li2 ∴ L =

2iU2

or henry = 2)ampere(joule

(D) U = 21 Li2 = i2 Rt

∴ L = Rt or henry = ohm-second


6. T = 20

s40 = 2 s.

Further, t = nT = 20T or ∆t = 20 ∆T

∴ tt∆ =

TT∆

or ∆T = tT

⋅ ∆t

=

402 (1)

= 0.05 s Further,

T = 2πgl or T ∝ g–1/2

∴ TT∆ × 100 = –

21 ×

gg∆ × 100

or % error in determination of g is

gg∆ × 100 = – 200 ×

TT∆

= – 2

05.0200 ×

= – 5% ∴ Correct options are (A) and (C).

7. E = λhc

E = r

mGm– 21

λhc =

rGM 2

unit of λ and r is L

So M = 2/1

Ghc

…(1)

M ∝ c M ∝ G1 Ans. (A)

E = mc2

E = λhc

λhc = mc2

λ = mh

∴ L = Mh

L = 2/1)hc(h G1/2 from (1)

∴ L ∝ G L ∝ h 8. W = qV ML2T–2 = ATV V = M1L2T–3A–1

I ⇒ A

F = 04

1πε

2

2

rq

ε0 = 2

2

Frq

ε0 = 22

22

LMLTTA

×−

ε0 = M–1L–3T4A2

c = LT–1 Similarly µ0I2 = ε0V2 I = ε0cV A = M0L0T0A1 9. l α εa kb Tc nd qe

(A) l = θθ

×1–2–213–421–

223–

TLMLTAMTAL

l = 2Ll =

Ll

(B) l = 2B

nqTkε

= 213–

1–2–213–421–

TALTLM)LTAM( θθ

= 2L = L

(C) l = θθ 1–2–212–3–421–

22

TLMLLTAMTA

(D) l = θθ+ 1–2–211–3–421–

22

TLMLLTAMTA

= 2L = L


14

10. Time period

T = 2πg5

)r–R(7

Time period I 0.52 sec II 0.56 sec III 0.57 sec IV 0.54 sec V 0.59 sec

Av ⟨ T ⟩ = 5

59.054.057.056.052.0 ++++

⟨ T ⟩ = 578.2

⟨ T ⟩ = 0.556 sec

Error = 0.03 0.01 0.02 0.01 0.03

absolute error = 0.1/5 = 0.02

T2g (R – r)–1 = const

T

T2∆ + gg∆ –

r–R)r–R(∆ = 0

T

T2∆ + gg∆ –

)r–R(1 (∆R – ∆r) = 0

gg∆ =

+

×502

10057.32

% error =

+

×502

10057.32 × 100 ≈ 11%

Matrix Match Type

1. t ≡ RL ∴ L ≡ tR ≡ ohm-second

U ≡ C2

q 2

∴ C ≡ Uq 2

≡ coulomb2 / joule

q ≡ CV ∴ C ≡ Vq ≡ coulomb/volt

L ≡ dt/die–

∴ L ≡ ( )

)di(dte ≡ volt-second/ampere

F = ilB

∴ B ≡ liF ≡ Newton / ampere-metre

I II Capacitance coulomb-volt coulomb2 joule–1 Inductance ohm-sec, volt second ampere–1

Magnetic induction

newton (ampere-metre)–1

2. The correct table is as under

I II Angular momentum [M L2T–1] Latent heat [L2T–2]

Torque [M L2T–2]

Capacitance [M–1 L–2T2Q2]

Inductance [M L2Q–2]

Resistivity [M L3T–1 Q–2]

6. Boltzmann constant

Q E = 23 KT

ML2T–2 = k[K] k = [ML2T–2K–1] Coff. of viscocity F = 6πηRV

η = 1–

2

LTLMLT

×

−

η = [ML–1T–1] Planck constant E = hν ML2T–2 = h.T–1 h = [ML2T–1] Thermal conductivity (K) K = [MLT–3K–1] Fill in the Blanks

1. E = hν

∴ h = νE

or [h] = ][]E[

ν =

]T[]TML[

1–

2–2 = [ML2T–1]

2. [X] = [C] = [M–1L–2T2Q2] [Z] = [B] = [M T–1 Q–1]

∴ [Y] = 21–1–

222–1–

]QMT[]QTLM[

= [M–3L–2T4Q4]


3. Electrical conductivity σ = EJ

= q/FA/i =

FAqi

= FA

)i)(it( = FA

ti 2

σ = ]L][MLT[

]T][A[22–

2

= [M–1 L–3 T3 A2]

4.

2Va = [p]

∴[a] = [pV2] = [ML–1T–2] [L6] = [ML5T–2]

Analytical & Descriptive Question

2. [T] = [padbEc] = [ML–1T–2]a [ML–3]b [ML2T–2]c Equating the powers both sides, we have a + b + c = 0 … (i) – a – 3b + 2c = 0 … (ii) – 2a – 2c = 1 … (iii) Solving these three equations, we have

a = – 65 , b =

21 and c =

31

4. (N + 1) divisions on the vernier scale = N divisions on main scale ∴ 1 division on vernier scale

= 1N

N+

divisions on main scale

Each division on the main scale is of a units.

∴ 1 division on vernier scale =

+ 1NN a

units = a′(say) Least count = 1 main scale division – 1 vernier scale division

= a – a′ = a –

+ 1NN a

= 1N

a+

5. Young's modulus of elasticity is given by

Y = strainstress

= L/A/F

l =

AFLl

=

π4

FL2dl

Substituting the values, we get

Y = 24–3– )100.5()1025.1(41.150

××π××××

= 2.24 × 1011 N/m2 Now,

YY∆ =

LL∆ +

ll∆ + 2

dd∆

=

110

1.0 +

125.0001.0 + 2

05.0001.0

= 0.0489 ∆Y = (0.0489)Y = (0.0489) × (2.24 × 1011) N/m2 = 1.09 × 1010 N/m2

6. Least count of screw gauge = 100mm1

= 0.01 mm Diameter of wire = (1 + 47 × 0.01) mm = 1.47 mm

Curved surface area (in cm2) = (2π)

2d (L)

or S = πdL = (π) (1.47 × 10–1) (5.6) cm2 = 2.5848 cm2 Rounding off to two significant digits S = 2.6 cm2

7. 1 MSD = 1 mm 9 MSD = 10 VSD ∴ Least count, LC = 1 MSD – 1 VSD

= 1 mm – 109 mm

= 101 mm

Measure reading of edge = MSR + VSR (LC)


16

= 10 + 1 × 101

= 10.1 mm Volume of cube V = (1.01)3 cm3 = 1.03 cm3 [After rounding off upto 3 significant digits, as

edge length measured upto 3 significant digits]

∴ Density of cube = 03.1736.2

= 2.6563 g/cm3 = 2.66 g/cm3 (After rounding off to 3 significant digits) 8. d = ρa Sb ƒc

d′ = (ML–3)a b

2

1211

LLTTLM

−−

(T–1)c

M0L1T0 = Ma + bL–3aT–3b – c

a + b = 0; – 3a = 1, a = –31 ; – 3b – c = 0

b = – a

⇒ b = 31

⇒ n = 3

9. E = A2e–αt E = A2e–0.2t ln E = 2lnA – 0.2t

E

dE = 2 A

dA – 0.2 dt

E

dE = 2 A

dA – 0.2 dt

t

dt = 1.5 %

∴ dt = 1.5 × 5 % = 1.5 × 5 = 7.5 = 2 × 1.25 + 0.2 × 7.5 = 2.5 + 1.5 = 4 %

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PHYSICS - Career Pointcareerpoint.ac.in/dlp/books/39-Years-Sample/Physics 39 (1).pdf · A vernier calipers has 1 mm marks on the main scale. It has 20 equal divisions on the vernier