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REP-WHOLE-NEET( EXCEPT,MODERN PHY,SEMICOND,COMMU.)-45-6-6-21-converted Flipbook PDF

REP-WHOLE-NEET( EXCEPT,MODERN PHY,SEMICOND,COMMU.)-45-6-6-21-converted


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FEGADE PHYSICS CLASSES Date : 06/05/2021 Time : 01:06:00 Marks : 264

TEST ID: 220 PHYSICS WHLE NEET(EXCEPT MODERN PHY,SEMICOND,COMM)

Single Correct Answer Type 1. A battery of π‘’π‘šπ‘“ E produces currents 𝐼1 and 𝐼2 when connected to external resistances 𝑅1 and 𝑅2 respectively. The internal resistance of the battery is 𝐼1 𝑅2 βˆ’ 𝐼2 𝑅1 𝐼1 𝑅2 + 𝐼2 𝑅1 a) b) 𝐼2 βˆ’ 𝐼1 𝐼1 βˆ’ 𝐼2 𝐼1 𝑅1 + 𝐼2 𝑅2 𝐼1 𝑅1 βˆ’ 𝐼2 𝑅2 c) d) 𝐼1 βˆ’ 𝐼2 𝐼2 βˆ’ 𝐼1 2. A current of 2A flows in an electric circuit as shown in figure. The potential difference(𝑉𝑅 βˆ’ 𝑉𝑆 ), in volts( 𝑉𝑅 βˆ’ 𝑉𝑆 are potentials at R and S respectively) is

melted into a disc of radius π‘Ÿ and thickness 𝑑. If its moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also equal to 𝐼 , then the value of π‘Ÿ is equal to

a)

7.

R

7Ξ© Q 2A

P

3Ξ©

3.

4.

5.

6.

7Ξ©

a) -4 b) +2 c) +4 d) -2 A body weighs 𝑀 newton at the surface of the earth. Its weight at a height equals to half the radius of the earth, will be 𝑀 4𝑀 2𝑀 8𝑀 a) b) c) d) 2 9 3 27 The ratio of the radii of the planets 𝑃1 and 𝑃2 is π‘Ž. The ratio of their acceleration due to gravity is 𝑏. The ratio of the escape velocities from them will be a) π‘Žπ‘ b) βˆšπ‘Žπ‘ c) βˆšπ‘Ž/𝑏 d) βˆšπ‘/π‘Ž Two discs of same thickness but of different radii are made of two different materials such that their masses are same. The densities of the materials are in the ratio 1:3. The moments of inertia of these discs about the respective axes passing through their centres and perpendicular to their planes will be in the ratio a) 1 : 3 b) 3 : 1 c) 1 : 9 d) 9 : 1 A solid sphere of radius 𝑅 has moment of inertia 𝐼 about its geometrical axis. If it is

8.

2

𝑅

b)

2

𝑅

c)

3

𝑅

d)

√3

𝑅 √15 √5 √15 √15 A person is sitting in a travelling train and facing the engine. He tosses up a coin and the coin falls behind him. It can be concluded that the train is a) Moving forward and gaining speed b) Moving forward and losing speed c) Moving forward with uniform speed d) Moving backward with uniform speed The pulley and strings shown in figure are smooth and of negligible mass. For the system to remain in equilibrium, the angle ΞΈ should be

a) 0Β° b) 30Β° c) 45Β° d) 60Β° A particle is projected with a velocity 𝑣 such that its range on the horizontal plane is twice the greatest height attained by it. The range of the projectile is (where 𝑔 is acceleration due to gravity) 4𝑣 2 4𝑣 2 𝑣2 4𝑔 a) b) 2 c) d) 5𝑔 𝑔 5𝑣 √5𝑔 10. A body moves along a circular path of radius 5 m. The coefficient of friction between the surface of path and the body is 0.5. The angular velocity, in radians/sec, with which 9.

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11.

12.

13.

14.

15.

the body should move so that it does not leave the path is (g = 10msβˆ’2 ) a) 4 b) 3 c) 2 d) 1 A wooden ball of density D is immersed in water of density d to a depth h below the surface of water and then released. Upto what height will then ball jump out of water? 𝑑 𝑑 a) β„Ž b) ( βˆ’ 1) β„Ž 𝐷 𝐷 c) β„Ž d) Zero A metal plate of area 103 cm2 rests on a layer of oil 6 mmType equation here. thick. A tangential force of 10βˆ’2 N is applied on it to move it with a constant velocity of 6 cmsβˆ’1 . The coefficient of viscosity of the liquid is a) 0.1 poiseb) 0.5 poisec) 0.7 poised) 0.9 poise A frame made of a metallic wire enclosing a surface area 𝐴 is covered with a soap film. If the area of the frame of metallic wire is reduced by 50%, the energy of the soap film will be changed by a) 100% b) 75% c) 50% d) 25% A uniformly wound solenoidal coil of self inductance 1.8 Γ— 10βˆ’4 H and resistance 6 Ξ© is broken up into two identical coils. These identical coils are then connected in parallel across a 12 V battery of negligible resistance. The time constant of the current in the circuit and the steady state current through battery is a) 3 Γ— 10βˆ’5 s, 8 A b) 1.5 Γ— 10βˆ’5 s, 8 A c) 0.75 Γ— 10βˆ’4 s, 4 A d) 6 Γ— 10βˆ’5 s, 2 A In the circuit shown in figure neglecting source resistance, the voltmeter and ammeter readings will be respectively

a) 0 V, 3 A b) 150 V, 3 A c) 150 V, 6 A d) 0 V, 8 A 16. The dimensions of time constant are a) [M 0 L0 T 0 ] b) [M 0 L0 T] c) [MLT] d) None of these 𝛼𝑧 17. In the relation 𝑝 = Ξ± eβˆ’π‘˜ΞΈ , 𝑝 is the pressure, 𝑧 Ξ²

the distance, π‘˜ is Boltzmann constant and ΞΈ is the temperature, the dimensional formula of Ξ² will be

a) [M 0 L2 T 0 ] b) [ML2 T] 0 βˆ’1 c) [ML T ] d) [ML2 T βˆ’1 ] 18. A block of metal is heated to a temperature much higher than the room temperature and allowed to cool in a room free from air currents. Which of the following curves correctly represents the rate of cooling

a)

b)

c)

d)

19. Four rods of identical cross-sectional area and made 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 a) √2 b) 𝑇 √2 + 1 2 c) 0 d) None of these 20. Consider three concentric shells of metal A, B and C are having radii a, b and c respectively as shown in the figure (π‘Ž < 𝑏 < 𝑐). Their surface charge densities are 𝜎, βˆ’πœŽ and 𝜎 respectively. Calculate the electric potential on the surface of shell A

𝜎 𝜎 (π‘Ž βˆ’ 𝑏 + 𝑐) b) (π‘Ž βˆ’ 𝑏 βˆ’ 𝑐) πœ€0 πœ€0 𝜎 2 𝜎 c) (π‘Ž + 𝑏 2 + 𝑐 2 ) d) (π‘Ž + 𝑏 βˆ’ 𝑐) πœ€0 πœ€0 21. 1000 similar electrified rain drops merge together into one drop so that their total charge remains unchanged. How is the electric energy affected? a) 100 times b) 102 times c) 200 times d) 400 times a)

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22. Number of electric lines of force from 0.5 C if positive charge in a dielectric medium of constant 10 is a) 5.65 Γ— 109 b) 1.13 Γ— 1011 9 c) 9 Γ— 10 d) 8.85 Γ— 10βˆ’12 23. At a certain place, a magnet makes 30 oscillations per min. At another place where the magnetic field is double, its time period will be a) 4 s b) 2 s c) 1/2 s d) √2 s 24. The effect due to uniform magnetic field on a freely suspended magnetic needle is as follows a) Both torque and net force are present b) Torque is present but no net force c) Both torque and net force are absent d) Net force is present but not torque 25. By opening the door of a refrigerator inside a closed room, the room temperature a) Remains constant b) Decreases c) Increases d) None of these 26. Initial pressure and volume of a gas are 𝑃 and 𝑉 respectively. First it is expanded isothermally to volume 4𝑉 and then compressed adiabatically to volume 𝑉. The final pressure of gas will be (given 𝛾 = 3/2) a) 1𝑃 b) 2𝑃 c) 4𝑃 d) 8𝑃 27. A stress of 3.18 Γ— 108 π‘π‘šβˆ’2 is applied to a steel rod of length 1π‘š along its length. Its Young’s modulus is 2 Γ— 1011 π‘π‘šβˆ’2. Then the elongation produced in the rod in π‘šπ‘š is a) 3.18 b) 6.36 c) 5.18 d) 1.59 2 28. A steel wire of 1m long and 1π‘šπ‘š cross section area is hang from rigid end. When weight of 1π‘˜π‘” is hung from it then change in length will be (given π‘Œ = 2 Γ— 1011 𝑁/π‘š2) a) 0.5 π‘šπ‘š b) 0.25 π‘šπ‘š c) 0.05 π‘šπ‘š d) 5 π‘šπ‘š 29. How does time period of a pendulum very

32.

33.

34.

35.

36.

37.

with length? a) βˆšπ‘™

𝑙

b) √

2

c)

1 βˆšπ‘™

d) 2𝑙

38.

30. The displacement of a particle in SHM

various according to the relation π‘₯ = 4 (cosΟ€ 𝑑 + sinΟ€ 𝑑). The amplitude of the particle is a) -4 b) 4 c) 4√2 d) 8 31. Lenz’s law is expressed by the following formula (here 𝑒 = induced e.m.f., πœ™ = magnetic flux in one turn and 𝑁 = number of turns) 𝑑𝑁 π‘‘πœ™ a) 𝑒 = βˆ’πœ™ b) 𝑒 = βˆ’π‘ 𝑑𝑑 𝑑𝑑

𝑑 πœ™ π‘‘πœ™ d) 𝑒 = 𝑁 ( ) 𝑑𝑑 𝑁 𝑑𝑑 If a coil of metal wire is kept stationary in a non-uniform magnetic field, then a) An e.m.f. is induced in the coil b) A current is induced in the coil c) Neither e.m.f. nor current is induced d) Both e.m.f. and current is induced In a stationary wave all the particles a) On either side of a node vibrate in same phase b) In the region between two nodes vibrate in same phase c) In the region between two antinodes vibrate in same phase d) Of the medium vibrate in same phase A source of sound of frequency 500 Hz is moving towards an observer with velocity 30msβˆ’1.The speed of sound is 330ms βˆ’1 .The frequency heard by the observer will be a) 545 Hz b) 580 Hz c) 558.3 Hz d) 550 Hz The phase difference between two points separated by 0.8 m in a wave of frequency 120 Hz is0.5πœ‹. The wave velocity is a) 144msβˆ’1 b) 384msβˆ’1 c) 256msβˆ’1 d) 720msβˆ’1 A particle starting from rest falls from a certain height. Assuming that the value of acceleration due to gravity remains the same throughout motion, its displacement in three successive half second intervals are 𝑆1 , 𝑆2 , 𝑆3 . Then, a) 𝑆1 : 𝑆2 : 𝑆3 : 1: 5: 9 b) 𝑆1 : 𝑆2 : 𝑆3 : 1: 2: 3 c) 𝑆1 : 𝑆2 : 𝑆3 : 1: 1: 1 d) 𝑆1 : 𝑆2 : 𝑆3 : 1: 3: 5 A bus begins to move with an acceleration of 1π‘šπ‘  βˆ’2 . A man who is 48π‘š behind the bus starts running at 10 π‘šπ‘  βˆ’1 to catch the bus. The man will be able to catch the bus after a) 6𝑠 b) 5𝑠 c) 3𝑠 d) 8𝑠 A ball is allowed to fall from a height of 10 π‘š. If there is 40% loss of energy due to impact, then after one impact ball will go up to a) 10 π‘š b) 8 π‘š c) 4 π‘š d) 6 π‘š If a man increase his speed by 2 π‘š/𝑠, his K.E. is doubled, the original speed of the man is a) (1 + 2√2) π‘š/𝑠 b) 4 π‘š/𝑠 c) 𝑒 = βˆ’

39.

c) (2 + 2√2) π‘š/𝑠 d) (2 + √2) π‘š/𝑠 40. Two point charge βˆ’π‘ž and +π‘ž/2 are situated at the origin and at the point (π‘Ž, 0,0) respectively.

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The point along the 𝑋-axis where the electric field vanishes is π‘Ž a) π‘₯ = b) π‘₯ = √2π‘Ž √2 √2π‘Ž √2π‘Ž c) π‘₯ = d) π‘₯ = √2 βˆ’ 1 √2 + 1 41. A force of 2.25 N acts on a charge of 15 Γ— 10βˆ’4C. Calculate the intensity of electric field at that point a) 1500 NC βˆ’1 b) 150 NC βˆ’1 d) None of these c) 15000 NC βˆ’1 42. An electron enters a region where electrostatic field is 20𝑁/𝐢 and magnetic field is 5𝑇. If electron passes undeflected through the region, the velocity of electron will be a) 0.25π‘šπ‘  βˆ’1 b) 2π‘šπ‘  βˆ’1 c) 4π‘šπ‘  βˆ’1 d) 8π‘šπ‘  βˆ’1 43. In case of Hall effect for a strip having charge 𝑄 and area of cross-section 𝐴, the Lorentz force is

a) Directly proportional to 𝑄 b) Inversely proportional to 𝑄 c) Inversely proportional to 𝐴 d) Directly proportional to 𝐴 44. Inside a cylinder closed at both ends is a movable piston. On one side of the piston is a mass π‘š of a gas, and on the other side a mass 2 π‘š of the same gas. What fraction of the volume of the cylinder will be occupied by the larger mass of the gas when the piston is in equilibrium? The temperature is the same throughout. 2 1 1 1 a) b) c) d) 3 3 2 4 45. The average momentum of a molecule in an ideal gas depends on a) Temperature b) Volume c) Molecular mass d) None of these

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FEGADE PHYSICS CLASSES Date : 06/05/2021 TEST ID: 220 Time : 01:06:00 PHYSICS Marks : 264 1.ELECTRIC CHARGES AND FIELDS ,1.MEASUREMENTS,1.ROTATIONAL DYNAMICS ,10.ELECTROSTATICS ,10.MAGNETIC FIELD DUE TO ELECTRIC CURRENT,10.MECHANICAL PROPERTIES OF FLUIDS ,11.ELECTRIC CURRENT THROUGH CONDUCTORS ,11.MAGNETISM MATTERIALS,11.THERMAL PROPERTIES OF MATTER ,12.ELECTROMAGNETIC INDUCTION,12.MAGNETISM ,12.THERMODYNAMICS,13.AC CIRCUITS ,13.KINETIC THEORY ,14.OSCILLATIONS,15.WAVES ,2.ELECTROSTATIC POTENTIAL AND CAPACITANCE ,2.MECHANICAL PROPERTIES OF FLUIDS,2.UNITS AND MEASUREMENTS ,3.CURRENT ELECTRICITY ,3.KINETIC THEORY OF GASES,3.MOTION IN A PLANE,3.MOTION IN A STRAIGHT LINE ,4.LAWS OF MOTION,4.MOTION IN A PLANE ,4.MOVING CHARGES AND MAGNETISM ,4.THERMODYNAMICS,5.GRAVITATION,5.LAWS OF MOTION ,5.MAGNETISM AND MATTER ,5.OSCILLATIONS,6.ELECTROMAGNETIC INDUCTION ,6.MECHANICAL PROPERTIES OF SOLIDS,6.SUPERPOSITION OF WAVES ,6.WORK ENERGY AND POWER ,7.ALTERNATING CURRENT ,7.SYSTEM OF PARTICLES AND ROTATIONAL MOTION,7.THERMAL PROPERTIES OF MATTER,8.ELECTROSTATICS,8.GRAVITATION,8.SOUND ,9.CURRENT ELECTRICITY,9.MECHANICAL PROPERTIES OF SOLIDS

: ANSWER KEY : 1) 5) 9) 13) 17) 21) 25)

d b a c a a c

2) 6) 10) 14) 18) 22) 26)

c a d a b a b

3) 7) 11) 15) 19) 23) 27)

c a b d c d d

4) 8) 12) 16) 20) 24) 28)

b c a b a b c

29) 33) 37) 41) 45)

a b d a d

30) 34) 38) 42)

c d d c

31) 35) 39) 43)

b b c a

32) 36) 40) 44)

c d c a

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FEGADE PHYSICS CLASSES Date : 06/05/2021 TEST ID: 220 Time : 01:06:00 PHYSICS Marks : 264 1.ELECTRIC CHARGES AND FIELDS ,1.MEASUREMENTS,1.ROTATIONAL DYNAMICS ,10.ELECTROSTATICS ,10.MAGNETIC FIELD DUE TO ELECTRIC CURRENT,10.MECHANICAL PROPERTIES OF FLUIDS ,11.ELECTRIC CURRENT THROUGH CONDUCTORS ,11.MAGNETISM MATTERIALS,11.THERMAL PROPERTIES OF MATTER ,12.ELECTROMAGNETIC INDUCTION,12.MAGNETISM ,12.THERMODYNAMICS,13.AC CIRCUITS ,13.KINETIC THEORY ,14.OSCILLATIONS,15.WAVES ,2.ELECTROSTATIC POTENTIAL AND CAPACITANCE ,2.MECHANICAL PROPERTIES OF FLUIDS,2.UNITS AND MEASUREMENTS ,3.CURRENT ELECTRICITY ,3.KINETIC THEORY OF GASES,3.MOTION IN A PLANE,3.MOTION IN A STRAIGHT LINE ,4.LAWS OF MOTION,4.MOTION IN A PLANE ,4.MOVING CHARGES AND MAGNETISM ,4.THERMODYNAMICS,5.GRAVITATION,5.LAWS OF MOTION ,5.MAGNETISM AND MATTER ,5.OSCILLATIONS,6.ELECTROMAGNETIC INDUCTION ,6.MECHANICAL PROPERTIES OF SOLIDS,6.SUPERPOSITION OF WAVES ,6.WORK ENERGY AND POWER ,7.ALTERNATING CURRENT ,7.SYSTEM OF PARTICLES AND ROTATIONAL MOTION,7.THERMAL PROPERTIES OF MATTER,8.ELECTROSTATICS,8.GRAVITATION,8.SOUND ,9.CURRENT ELECTRICITY,9.MECHANICAL PROPERTIES OF SOLIDS

: HINTS AND SOLUTIONS : Single Correct Answer Type 1 (d) Let 𝐸 and π‘Ÿ be the emf and internal resistance of a battery respectively 3

𝑉𝑝 βˆ’ 𝑉𝑅 = 3𝑣 𝑉𝑝 βˆ’ 𝑉𝑠 = 7𝑉 From Eqs. (i) and (ii), we get 𝑉𝑅 βˆ’ 𝑉𝑠 = +4𝑉 (c) The value of acceleration due to gravity at a height β„Ž above the earth’s surface is given by

In the first case current flowing in the circuit 𝐸 𝐼1 = 𝑅1 π‘Ÿ Or 𝐸 = 𝐼1 (𝑅1 + π‘Ÿ)

𝑔′ =

2

In the second case current flowing in the circuit 𝐸 𝐼2 = 𝑅2 + π‘Ÿ Or 𝐸 = 𝐼2 (𝑅2 + π‘Ÿ) Equating equations (i) and (ii), we get 𝐼1 (𝑅1 + π‘Ÿ) = 𝐼2 (𝑅2 + π‘Ÿ) β‡’ 𝐼1 𝑅1 + 𝐼1 π‘Ÿ = 𝐼2 𝑅2 + 𝐼2 π‘Ÿ 𝐼1 𝑅1 βˆ’ 𝐼2 𝑅2 = (𝐼2 βˆ’ 𝐼1 )π‘Ÿ β‡’ (𝐼2 βˆ’ 𝐼1 )π‘Ÿ = 𝐼1 𝑅1 βˆ’ 𝐼2 𝑅2 𝐼1 𝑅1 βˆ’ 𝐼2 𝑅2 4 π‘Ÿ= 𝐼2 βˆ’ 𝐼1 (c) Current through each arm PRQ and PSQ=1A 5

𝑔 β„Ž 2

(1 + 𝑅)

where 𝑅 is radius of earth. 𝑅 When β„Ž= 2 𝑔 4𝑔 𝑔′ = = β„Ž 2 9 (1 + 𝑅) 4 4 Hence, weight 𝑀 β€² = π‘šπ‘”β€² = π‘šπ‘” = 𝑀. 9 9 (b) (𝑣𝑒 )𝑝1 √2g1 𝑅1 g1 𝑅1 = =√ Γ— = βˆšπ‘Žπ‘ (𝑣𝑒 )𝑝2 √2g1 𝑅2 g 2 𝑅2 (b) Page|6

1

1

𝑀

1 𝑀2

π‘‰π‘‘π‘Ž = 𝑉𝑑𝑔 βˆ’ 𝑉𝐷𝑔

M.I. of disc = 2 𝑀𝑅 2 = 2 𝑀 (πœ‹π‘‘πœŒ) = 2 πœ‹π‘‘πœŒ

π‘‘βˆ’π· )𝑔 𝐷

∴ π‘Ž=(

𝑀 𝑀 (𝐴𝑠 𝜌 = There fore R2 = ) 2 πœ‹π‘… 𝑑 πœ‹ π‘‘πœŒ If mass and thickness are same then, 𝐼 ∝

Velocity on reaching the surface, 𝑣 = √2π‘Žβ„Ž Further 𝑣 = √2π‘Žβ„Ž ∴ 2π‘Žβ„Ž = 2𝑔𝐻

1 𝜌

𝐼1 𝜌2 3 = = 𝐼2 𝜌1 1 (a) ∴

6

or 𝐻 =

8

9

Ι³=

We know 𝑅 = 4𝐻 cot πœƒ β‡’ cot πœƒ =

1 2

From triangle we can say that sin πœƒ =

2 , cos πœƒ √5

=

1 √5

∴ Range of projectile 𝑅 =

2𝑣 2 sin πœƒ cos πœƒ 𝑔

2𝑣 2 2 1 4𝑣 2 = Γ— Γ— = 𝑔 5𝑔 √5 √5

2



10 (d) Here, π‘Ÿ = 5 m, πœ‡ = 0.5, πœ” =? , g = 10msβˆ’2 π‘šπ‘Ÿπœ”2 = 𝐹 = πœ‡π‘… = πœ‡ π‘šg πœ‡g

0.5Γ—10 5

= 1 rad s βˆ’1

11 (b) Let the volume of the ball be V . Force on the hall due to upthrust =Vdg Net upward force = Vdg =VDg If upward acceleration is a, then

10βˆ’2 6Γ—10βˆ’2 ) 6Γ—10βˆ’3

(103 Γ— 10βˆ’4 ) (

10βˆ’2 Γ— 6 Γ— 10βˆ’3 10βˆ’1 Γ— 6 Γ— 10βˆ’2 = 10βˆ’2 Nsm2 = 0.1 poise 13 (c) Surface energy = surface tension Γ— surface area 𝐸 = 𝑆 Γ— 2𝐴 New surface energy, 𝐸1 = 𝑆 Γ— 2(𝐴/2) = 𝑆 Γ— 𝐴 =

πΈβˆ’πΈ

1 % decrease in surface energy = Γ— 100 𝐸 2 𝑆𝐴 βˆ’ 𝑆𝐴 = Γ— 100 = 50% 2 𝑆𝐴 14 (a) 1 1 1 2 𝐿 = + = π‘œπ‘Ÿ 𝐿𝑃 = 𝐿𝑝 𝐿 𝐿 𝐿 2 Where L is inductance of each part

1.8 Γ—10βˆ’4 = 0.9 Γ— 2 𝐿 0.9Γ—10βˆ’4 𝐿𝑃 = = 2 2

= ∴

10βˆ’4 H = 0.45 Γ— 10βˆ’4H 6

Resistance of each part, π‘Ÿ = 2 = 3Ξ© Now,

1 π‘Ÿπ‘ƒ

1

1

2

=3+3=3

∴ π‘Ÿπ‘ƒ = 3/2Ξ© 𝐿

Time constant of circuit = π‘Ÿπ‘ƒ = 𝑃

0.45Γ—10βˆ’4 3/2

= 8A

15 (d) The voltage 𝑉𝐿 and 𝑉𝐢 are equal and opposite so, voltmeter reading will be zero. Also, 𝑅 = 30 Ξ©, 𝑋𝐿 = 𝑋𝐢 = 25 Ξ© So,

1

πœ” =βˆšπ‘Ÿ =√

𝑑 𝐷

= ( βˆ’ 1) β„Ž

𝐹 𝐴(𝑑𝑣/𝑑𝑦)

∴ ɳ=

(a) The coin falls behind him it means the velocity of train was increasing otherwise the coin fall directly into the hands of thrower (c) If 𝑇 is tension in each part of the string holding mass √2π‘š, then in equilibrium, 𝑇 cos ΞΈ + 𝑇 cos ΞΈ = √2 π‘šg 2 𝑇 cos ΞΈ = √2 π‘šg But 𝑇 = π‘šg; ∴ 2π‘šg cos ΞΈ = √2π‘šg 1 cos ΞΈ = √2 ΞΈ = 45Β° (a) 𝑅 = 2𝐻 Given

οƒ–5

π‘‘βˆ’π· )β„Ž 𝐷

=(

12 (a)

2 1 𝑀𝑅 2 = 2 π‘€π‘Ÿ 2 + π‘€π‘Ÿ 2 5 2 3 or 5 𝑀𝑅 2 = 2 π‘€π‘Ÿ 2 2 ∴ π‘Ÿ = 15 𝑅 √

7

π‘Žβ„Ž 8

𝑖=

𝑉

βˆšπ‘…2 +(𝑋𝐿 βˆ’π‘‹πΆ )2 𝑉 240 = 𝑅 = 30 = 8A

16 (b) Time constant in an 𝑅 βˆ’ 𝐢 circuit 𝜏 =π‘…βˆ’πΆ [𝜏] = [𝑅][𝐢] = [ML2 T βˆ’3 Aβˆ’2 ][M βˆ’1 Lβˆ’2 T 4 A2 ] = [M 0 L0 T] 17 (a)

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π‘Žπ‘§

In given equation, π‘˜ΞΈ should be dimensionless 𝛼= [𝛼] =

β‡’

π‘˜ΞΈ 𝑧 [ML2 T βˆ’2 K βˆ’1 Γ— K] = [MLT βˆ’2 ] [L] 𝑉𝐴 = π‘‰π‘Ž + 𝑉𝑏 + 𝑉𝑐

𝛼

And

𝑝=𝛽 [MLT βˆ’2 ]

𝛼 [Ξ²] = [ ] = = [M 0 L2 T 0 ] [MLβˆ’1 T βˆ’2 ] 𝑝

β‡’

18 (b) According to Newton’s law of cooling

𝑉𝐴 = Or 𝑉𝐴 =

πœƒ πœƒi

πœƒ0

Rate of cooling ∝ Temperature difference β‡’

1 4πœ‹π‘Ž2 𝜎 1 4πœ‹π‘ 2 (βˆ’πœŽ) . + . 4πœ‹πœ€0 π‘Ž 4πœ‹πœ€0 𝑏 1 4πœ‹π‘ 2 𝜎 + . 4πœ‹πœ€0 𝑐

(∡ π‘ž = 4πœ‹π‘Ÿ 2 𝜎)

t

π‘‘πœƒ βˆ’ 𝑑𝑑

1 π‘žπ‘Ž 1 π‘žπ‘ 1 π‘žπ‘ . + + + . 4πœ‹πœ€0 π‘Ž 4πœ‹πœ€0 𝑏 4πœ‹πœ€0 𝑐

∝ (πœƒ βˆ’ πœƒ0 ) β‡’

π‘‘πœƒ βˆ’ 𝑑𝑑

= 𝛼(πœƒ βˆ’ πœƒ0 ) [𝛼 =

or 𝑉𝐴 =

𝜎 (π‘Ž βˆ’ 𝑏 + 𝑐) πœ€0

constant]

21 (a) Let q be charge on each small drop of radius r. If R π‘‘πœƒ β‡’ ∫ = βˆ’π›Ό ∫ 𝑑𝑑 β‡’ πœƒ is radius of big drop, then (πœƒ βˆ’ πœƒ0 ) πœƒ1 0 4 3 4 πœ‹π‘… = 1000 Γ— πœ‹π‘Ÿ 3 = πœƒ0 + (πœƒ1 βˆ’ πœƒ0 )𝑒 βˆ’π›Όπ‘‘ 3 3 This relation tells us that, temperature of the body ∴ 𝑅 = 10 π‘Ÿ varies exponentially with time from πœƒ1 to πœƒ0 Cβ€˜=10 C Hence graph (b) is correct π‘ž2 Initial energy 𝐸1 = 1000 Γ— 2𝐢 β€² 19 (c) (1000π‘ž)2 Final energy 𝐸2 = 𝐢 β€² 𝑄 πΎπ΄Ξ”πœƒ π‘šπΏ 𝐾(πœ‹π‘Ÿ 2 )Ξ”πœƒ = = = 𝐸2 1000 Γ— 𝐢 𝑑 𝑙 𝑑 𝑙 = π‘š πΎπ‘Ÿ 2 𝐸1 𝐢′ β‡’ Rate of melting of ice ( 𝑑 ) ∝ 𝑙 1 1 = 1000 Γ— = 100 Since for second rod 𝐾 becomes 4 π‘‘β„Ž π‘Ÿ becomes 10 double and length becomes half, so 22 (a) π‘š π‘š π‘ž 0.5 rate of melting will to twice 𝑖. 𝑒. ( 𝑑 ) = 2 ( 𝑑 ) = ϕ𝐸 = = = 5.65 Γ— 109 2 1 𝐾Ρ0 10 Γ— 8.85 Γ— 1012 2 Γ— 0.1 = 0.2𝑔/𝑠 23 (d) Temperature difference between 𝐢 and 𝐷 is zero 𝐼 1 C As 𝑇 = 2πœ‹βˆšπ‘€π΅ ∴ 𝑇 β‡’ 2 time. πœƒ

1

√

R

R

Initial time period = 60/30 = 2 s

B

A

T R

R D

20 (a) The electric potential on the surface of shell 𝐴

∴ New 𝑇 =

2 √2

= √2 s

24 (b) When a magnetic needle is placed in a uniform magnetic field, equal and opposite forces act on the poles of the needle which give rise to a torque, but not net force. 25 (c) Page|8

26

27

28

29

While working refrigerator reject heat from its inside into the room continuously to keep it cool inside. Now, if the door of the refrigerator is open the heat rejected will be more than that in the previous case. So, the room temperature in this case will be more than the temperature when the door of the refrigerator is closed. Hence, room temperature will increase gradually. (b) In isothermal process 𝑃1 𝑉1 = 𝑃2 𝑉2 𝑃 β‡’ 𝑃𝑉 = 𝑃2 Γ— 4𝑉 ∴ 𝑃2 = 4 In adiabatic process 𝑃 𝛾 𝛾 𝑃2 𝑉2 = 𝑃3 𝑉3 β‡’ Γ— (4𝑉)1.5 = 𝑃3 𝑉 1.5 β‡’ 𝑃3 = 2𝑃 4 (d) 𝐹/𝐴 π‘Œ= βˆ†π‘™/𝑙 Given, 𝐹/𝐴 = stress = 3.18 Γ— 108 π‘π‘šβˆ’2 𝑙 = 1π‘š, π‘Œ = 2 Γ— 1011 π‘π‘šβˆ’2 𝑙𝐹/𝐴 1 Γ— 3.18 Γ— 108 βˆ†π‘™ = = = 1.59 Γ— 10βˆ’3 π‘š π‘Œ 2 Γ— 1011 = 1.59π‘šπ‘š (c) 𝑀𝑔𝐿 1 Γ— 10 Γ— 1 𝑙= = = 0.05 π‘šπ‘š π‘Œπ΄ 2 Γ— 1011 Γ— 10βˆ’6 (a) 𝑙

Time period of pendulum 𝑇 = 2πœ‹βˆšg ∴

𝑇 ∝ βˆšπ‘™

30 (c)

and

𝑅 sin 𝛿 = 4 𝑅 cos 𝛿 = 4 𝑅 = 4√2

32 (c) E.m.f. or current induces only when flux linked with the coil changes 33 (b) In stationary wave all the particles in one particular segment (𝑖. 𝑒., between two nodes) vibrates in the same phase 34 (d) 𝑒×𝑣 𝑣′ = 𝑒 βˆ’ 𝑒𝑠 =

330 Γ— 500 = 550 Hz 330 βˆ’ 30

35 (b) Here , 𝑛 = 120 𝐻𝑧, π‘₯ = 0.8π‘š, Ο• = 0.5πœ‹.

From Ο• =

2πœ‹ π‘₯; πœ† πœ†

=

2πœ‹π‘₯ Ο•

=

2πœ‹Γ—0.8 0.5πœ‹

= 3.2 m

𝑣 = π‘›πœ† = 120 Γ— 3.2 = 384 msβˆ’1 36 (d) 1

9

(𝑆 β€² ∝ 𝑑 2 . Now, 𝑆1β€² : 𝑆2β€² : 𝑆3β€² ∷ 4 : 1: 4 or 1:4:9 For successive intervals, 𝑆1 : 𝑆2 : 𝑆3 ∷ 1: (4 βˆ’ 1): (9 βˆ’ 4) or 𝑆1 : 𝑆2 : 𝑆3 ∷ 1: 3: 5 37 (d) Let the man will be able to catch the bus after 𝑑 𝑠 Then 1 10𝑑 = 48 + Γ— 1 Γ— 𝑑 2 2 2 𝑑 βˆ’ 20𝑑 + 96 = 0 (𝑑 βˆ’ 12)(𝑑 βˆ’ 8) = 0 𝑑 = 8𝑠 and 𝑑 = 12𝑠 Thus the man will be able to catch the bus after 8𝑠 38 (d) π‘šπ‘”β„Ž = initial potential energy π‘šπ‘”β„Žβ€² = final potential energy after rebound As 40% energy lost during impact ∴ π‘šπ‘”β„Žβ€² = 60% of π‘šπ‘”β„Ž 60 60 β‡’ β„Žβ€² = Γ—β„Ž = Γ— 10 = 6π‘š 100 100 39 (c) 1

Initial kinetic energy 𝐸 = 2 π‘šπ‘£ 2 …(i) 1

Final kinetic energy 2𝐸 = 2 π‘š(𝑣 + 2)2 …(ii) By solving equation (i) and (ii) we get 𝑣 = (2 + 2√2) π‘š/𝑠 40 (c)

Suppose the field vanishes at distance π‘₯, we have π‘˜π‘ž π‘₯2

π‘˜π‘ž/2

= (π‘₯βˆ’π‘Ž)2 or 2(π‘₯ βˆ’ π‘Ž)2 = π‘₯ 2 or √2(π‘₯ βˆ’ π‘Ž) = π‘₯ √2π‘Ž ) √2βˆ’1

or (√2 βˆ’ 1)π‘₯ = √2π‘Ž or π‘₯ = ( 41 (a) 𝐸=

𝐹 2.25 = = 1500NC βˆ’1 π‘ž 15 Γ— 10βˆ’4

42 (c) 𝑣=

𝐸 20 = = 4 π‘š/𝑠 𝐡 5

44 (a) When the piston is in equilibrium, the pressure is same on both the sides of the piston. It is given Page|9

that temperature and weight of gas on the two sides of piston not change. From ideal gas equation, 𝑝𝑉 = 𝑛 𝑅𝑇, we have 𝑉 ∝ mass of the gas.

𝑉

π‘š

𝑉1 π‘š + 1 = π‘š1 + 1 𝑉2 2 𝑉1 +𝑉2 π‘š1 +π‘š2 Or = 𝑉2 π‘š2 𝑉2 π‘š2 2π‘š 2 Or = = = 𝑉1 +𝑉2 π‘š1 +π‘š2 π‘š+2π‘š 3

So, 𝑉1 = π‘š1 or 2

2

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