NEET Physics Class 11

Centre Of Mass Multiple Choice Question And Answers

Question 1. The center of mass of a body:

1. Lies always at the geometrical center
2. Lies always inside the body
3. Lies always outside the body
4. Lies within or outside the body

Answer: 4. Lies within or outside the body

Question 2. A body has its center of mass at the origin. The x-coordinates of the particles

1. Maybe all positive
2. Maybe all negative
3. It must be all non-negative
4. It may be positive for some particles and negative for other particles

Answer: 4. May be positive for some particles and negative in other particles

Centre of Mass MCQs for NEET Physics Class 11 with Answers

Question 3. All the particles of a body are situated at a distance R from the origin. The distance of the center of mass of the body from the origin is

1. = R
2. ≤ R
3. > R
4. ≥ R

Answer: 2. ≤ R

Question 4. Where will be the center of mass on combining two masses m and M (M > m):

1. Towards m
2. Towards M
3. In the middle of m and M
4. Anywhere

Answer: 2. Towards M

Question 5. Two homogenous spheres A and B of masses m and 2m having radii 2a and a respectively are placed in touch. The distance of the center of mass from the first sphere is :

1. a
2. 2a
3. 3a
4. None of these

Answer: 2. 2a

Question 6. In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1 Å = 10–10 m). The approximate location of the centre of mass of the molecule, distance from the hydrogen atom assuming the chlorine atom to be about 35.5 times as massive as hydrogen is

1. 1Å
2. 2.5 Å
3. 1.24 Å
4. 1.5 Å

Answer: 3. 1.24 Å

Question 7. The distance between the carbon atom and the oxygen atom in a carbon monoxide molecule is 1.1 Å. Given, that the mass of the carbon atom is 12 a.m.u. and the mass of the oxygen atom is 16 a.m.u., calculate the position of the center of mass of the carbon monoxide molecule

1. 6.3 Å from the carbon atom
2. 1 Å from the oxygen atom
3. 0.63 Å from the carbon atom
4. 0.12 Å from the oxygen atom

Answer: 3. 0.63 Å from the carbon atom

Question 8. Three identical metal balls each of radius r are placed touching each other on a horizontal surface such that an equilateral triangle is formed when the centers of the three balls are joined. The center of the mass of the system is located at

1. Horizontal surface
2. The center of one of the balls
3. Line joining centers of any two balls
4. Point of intersection of the medians

Answer: 4. Point of intersection of the medians

Question 9. Centre of mass is a point

1. Which is the geometric center of a body
2. From which distance of particles the same
3. Where the whole mass of the body is supposed to be concentrated
4. Which is the origin of the reference frame

Answer: 3. Where the whole mass of the body is supposed to be concentrated

Question 10. Choose the correct statement about the center of mass (CM) of a system of two particles

1. The CM lies on the line joining the two particles midway between them
2. The CM lies on the line joining them at a point whose distance from each particle is inversely proportional to the mass of that particle
3. The CM lies on the line joining them at a point whose distance from each particle is proportional to the square of the mass of that particle
4. The CM is on the line joining them at a point whose distance from each particle is proportional to the mass of that particle

Answer: 2. The CM lies on the line joining them at a point whose distance from each particle is inversely proportional to the mass of that particle

Question 11. The center of mass of a system of two particles divides the distance between them

1. In the inverse ratio of the square of masses of particles
2. In a direct ratio of the square of masses of particles
3. In inverse ratio of masses of particles
4. In direct ratio of masses of particles

Answer: 3. In inverse ratio of masses of particles

Question 12. A cricket bat is cut at the location of its center of mass as shown.

1. The two pieces will have the same mass
2. The bottom piece will have a larger mass
3. The handle piece will have a larger mass
4. The mass of the handle piece is double the mass of the bottom piece

Answer: 2. The bottom piece will have a larger mass

Question 13. The center of mass of the triangle shown in the figure has coordinates

1. \(x=\frac{h}{2}, y=\frac{b}{2}\)
2. \(\mathrm{x}=\frac{\mathrm{b}}{2}, \mathrm{y}=\frac{\mathrm{h}}{2}\)
3. \(x=\frac{b}{3}, y=\frac{h}{3}\)
4. \(x=\frac{h}{3}, y=\frac{b}{3}\)

Answer: 3. \(x=\frac{b}{3}, y=\frac{h}{3}\)

Question 14. Four bodies of equal mass start moving at the same speed as shown in the figure. In which of the following combination the center of mass will remain at the origin

1. c and b
2. a and d
3. a and c
4. b and c

Answer: 3. a and c

Question 15. Three identical spheres, each of mass 1 kg are kept as shown in the figure, touching each other, with their centers on a straight line. If their centers are marked P, Q, and R respectively, the distance of the center of mass of the system from P (origin) is

1. \(\frac{P Q+P R+Q R}{3}\)
2. \(\frac{P Q+P R}{3}\)
3. \(\frac{P Q+Q R}{3}\)
4. PR+QR

Answer: 2. \(\frac{P Q+P R}{3}\)

Question 16. A uniform square plate ABCD has a mass of 10 kg. If two point masses of 3 kg each are placed at the corners C and D as shown in the adjoining figure, then the center of mass shifts to the point which lies on –

1. OC
2. OD
3. OY
4. OX

Answer: 3. OY

Question 17. Three particles of masses 1 kg, 2 kg, and 3 kg are placed at the corners of an equilateral triangle of side 1.0 m as shown in Fig. The coordinates of the center of masses of the system are

1. \(\left(\frac{7}{12} m, \frac{\sqrt{3}}{4} m\right)\)
2. \(\left(\frac{1}{2} m, \frac{\sqrt{3}}{4} m\right)\)
3. \(\left(\frac{3}{12} m, \frac{1}{4} m\right)\)
4. \(\left(\frac{3}{12} m, \frac{\sqrt{3}}{4} m\right)\)

Answer: 3. \(\left(\frac{3}{12} m, \frac{1}{4} m\right)\)

Question 18. A non–uniform thin rod of length L is placed along the x-axis as such it is one of the ends at the origin. The linear mass density of the rod is λ = λ0x. The distance of the center of mass of the rod from the origin is:

1. L/2
2. 2L/3
3. L/4
4. L/5

Answer: 2. 2L/3

Question 19. The center of mass of the shaded portion of the disc is : (The mass is uniformly distributed in the shaded portion):

1. \(\frac{\mathrm{R}}{20}\) to the left of A
2. \(\frac{\mathrm{R}}{12}\) to the left of A
3. \(\frac{\mathrm{R}}{20}\) to the right of A
4. \(\frac{\mathrm{R}}{12}\) to the right of A

Answer: 1. \(\frac{\mathrm{R}}{20}\) to the left of A

Question 20. Four particles of masses m, 2m, 3m, and 4m are arranged at the corners of a parallelogram with each side equal to a, and one of the angles between two adjacent sides is 60º. The parallelogram lies in the x-y plane with mass m at the origin and 4m on the x-axis. The center of mass of the arrangement will be located at

1. \(\left(\frac{\sqrt{3}}{2} a, 0.95 a\right)\)
2. \(\left(0.95 a, \frac{\sqrt{3}}{4} a,\right)\)
3. \(\left(\frac{3 \mathrm{a}}{4}, \frac{\mathrm{a}}{2}\right)\)
4. \(\left(\frac{\mathrm{a}}{2}, \frac{3 \mathrm{a}}{4}\right)\)

Answer: 2. \(\left(0.95 a, \frac{\sqrt{3}}{4} a,\right)\)

Question 21. Masses 8, 2, 4, and 2 kg are placed at the corners A, B, C, and D respectively of a square ABCD of diagonal 80 cm. The distance of the center of mass from A will be

1. 20 cm
2. 30 cm
3. 40 cm
4. 60 cm

Answer: 2. 30 cm

Question 22. If the linear density of a rod of length 3m varies as λ = 2 + x, then the position of the center of gravity of the rod is

1. \(\frac{7}{3} m\)
2. \(\frac{12}{7} m\)
3. \(\frac{10}{7} m\)
4. \(\frac{9}{7} m\)

Answer: 2. \(\frac{12}{7} m\)

Question 23. A circular plate of uniform thickness has a diameter of 56 cm. A circular portion of diameter 42 cm is removed from one edge as shown in the figure. The center of mass of the remaining portion from the center of the plate will be:

1. 5 cm
2. 7 cm
3. 9 cm
4. 11 cm

Answer: 3. 9 cm

NEET Physics Chapter 3 Centre of Mass Multiple Choice Questions

Question 24. The coordinate of the center of mass of a system is shown in the figure:

1. \(\left(\frac{\mathrm{a}}{3}, 0\right)\)
2. \(\left(\frac{\mathrm{a}}{2}, \frac{\mathrm{a}}{2}\right)\)
3. \(\left(\frac{\mathrm{a}}{3}, \frac{\mathrm{a}}{3}\right)\)
4. \(\left(0, \frac{\mathrm{a}}{3}\right)\)

Answer: 3. \(\left(\frac{\mathrm{a}}{3}, \frac{\mathrm{a}}{3}\right)\)

Question 25. The coordinate of the center of mass of a system is shown in the figure:

1. \(\frac{\mathrm{a} \sqrt{3}}{2}, \frac{\mathrm{a}}{2}\)
2. \(\frac{a}{2}, \frac{a}{6} \sqrt{3}\)
3. \(\frac{a}{4}, \frac{a}{4} \sqrt{3}\)
4. \(\frac{\mathrm{a}}{2}, \frac{\mathrm{a}}{\sqrt{3}}\)

Answer: 2. \(\frac{a}{2}, \frac{a}{6} \sqrt{3}\)

Question 26. The center of masses of three particles of mass m₁= m₂= 1 kg at the corners of an equilateral triangle of side 1 m will be –

1. 0.50m, 0.43m
2. 0.43m, 0.50m
3. 0.25m, 0.25m
4. 0.22m, 0.22m

Answer: 1. 0.50m, 0.43m

Question 27. Two bodies of mass 1 kg and 3 kg have position vector \(\hat{i}+2 \hat{j}+\hat{k} \text { and }-3 \hat{i}-2 \hat{j}+\hat{k}\) respectively. The center of mass of this system has a position vector.

1. \(-2 \hat{i}+2 \hat{k}\)
2. \(-2 \hat{i}-\hat{j}+\hat{k}\)
3. \(2 \hat{i}-\hat{j}-2 \hat{k}\)
4. \(-\hat{i}+\hat{j}+\hat{k}\)

Answer: 2. \(-2 \hat{i}-\hat{j}+\hat{k}\)

Question 28. A circular disc of radius R is removed from a bigger circular disc of radius 2R such that the circumferences of the discs coincide. The center of mass of the new disc is αR from the center of the bigger disc. The value of α is:

1. 1/3
2. 1/2
3. 1/6
4. 1/4

Answer: 1. 1/3

Question 29. Two semicircular rings of linear mass densities λ and 2 λ and of radius ‘R’ each are joined to form a complete ring. The distance of the center of the mass of the complete ring from its geometrical center is:

1. \(\frac{3 R}{8 \pi}\)
2. \(\frac{2 R}{3 \pi}\)
3. \(\frac{3 R}{4 \pi}\)
4. None of these

Answer: 2. \(\frac{2 R}{3 \pi}\)

Question 30. A uniform metal disc of radius R is taken and out of it, a disc of diameter R is cut off from the end. The center of mass of the remaining part will be:

1. \(\frac{R}{4}\)from the centre
2. \(\frac{R}{3}\) from the centre
3. \(\frac{R}{5}\) from the centre
4. \(\frac{R}{6}\) from the centre

Answer: 4. \(\frac{R}{6}\) from the centre

Question 31. A uniform solid cone of height 40 cm is shown in the figure. The distance of the center of mass of the cone from point B (center of the base) is:

1. 20 cm
2. 10/3 cm
3. 20/3 cm
4. 10 cm

Answer: 4. 10 cm

Question 32. A bomb traveling in a parabolic path under the effect of gravity explodes in mid-air. The center of mass of fragments will:

1. Move vertically upwards and then downwards
2. Move vertically downwards
3. Move-in irregular path
4. Move-in the parabolic path which the unexploded bomb would have traveled.

Answer: 4. Move in the parabolic path that the unexploded bomb would have traveled.

Question 33. If a ball is thrown upwards from the surface of the earth and during upward motion :

1. The earth remains stationary while the ball moves upwards
2. The ball remains stationary while the earth moves downwards
3. The ball and earth both move toward each other
4. The ball and earth both move away from each other

Answer: 4. The ball and earth both move away from each other

Question 34. Internal forces can change :

1. The linear momentum but not the kinetic energy of the system.
2. The kinetic energy but not the linear momentum of the system.
3. Linear momentum as well as kinetic energy of the system.
4. Neither the linear momentum nor the kinetic energy of the system.

Answer: 2. The kinetic energy but not the linear momentum of the system.

Question 35. If the external forces acting on a system have zero resultant, the center of mass

1. Must not move
2. Must accelerate
3. May move
4. May accelerate

Answer: 3. May accelerate

Question 36. Two balls are thrown in the air. The acceleration of the center of mass of the two balls while in the air (neglect air resistance)

1. Depends on the direction of the motion of the balls
2. Depends on the masses of the two balls
3. Depends on the speeds of the two balls
4. Is equal to g

Answer: 4. Is equal to g

Question 37. Two particles of mass 1 kg and 0.5 kg are moving in the same direction with speeds of 2m/s and 6m/s respectively on a smooth horizontal surface. The speed of the center of mass of the system is:

1. \(\frac{10}{3} \mathrm{~m} / \mathrm{s}\)
2. \(\frac{10}{7} \mathrm{~m} / \mathrm{s}\)
3. \(\frac{11}{2} \mathrm{~m} / \mathrm{s}\)
4. \(\frac{12}{3} \mathrm{~m} / \mathrm{s}\)

Answer: 1. \(\frac{10}{3} \mathrm{~m} / \mathrm{s}\)

Question 38. The motion of the center of mass of a system of two particles is unaffected by their internal forces:

1. Irrespective of the actual direction of the internal forces
2. Only if they are along the line joining the particles
3. Only if they are at right angles to the line joining the particles
4. Only if they are obliquely inclined to the line joining the particles.

Answer: 1. Irrespective of the actual directions of the internal forces

Question 39. Two objects of masses 200 gm and 500 gm possess velocities \(10 \hat{i} \mathrm{~m} / \mathrm{s} \text { and } 3 \hat{i}+5 \hat{j} \mathrm{~m} / \mathrm{s}\) respectively. The velocity of their center of mass in m/s is :

1. \(5 \hat{i}-25 \hat{j}\)
2. \(\frac{5}{7} \hat{i}-25 \hat{j}\)
3. \(5 \hat{i}+\frac{25}{7} \hat{j}\)
4. \(25 \hat{i}-\frac{5}{7} \hat{j}\)

Answer: 3. \(5 \hat{i}+\frac{25}{7} \hat{j}\)

Question 40. 2 bodies of different masses of 2kg and 4kg are moving with velocities 20m/s and 10m/s towards each other due to mutual gravitational attraction. What is the velocity of their center of mass?

1. 5 m/s
2. 6 m/s
3. 8 m/s
4. Zero

Answer: 4. Zero

Question 41. Two spheres of masses 2M and M are initially at rest at a distance R apart. Due to the mutual force of attraction, they approach each other. When they are at separation R/2, the acceleration of the center of mass of spheres would be

1. 0 m / s²
2. g m / s²
3. 3g m / s²
4. 12g m / s²

Answer: 1. 0 m / s²

Question 42. Two bodies A and B have masses M and m respectively, where M > m and they are at a distance d apart. Equal force is applied to them so that they approach each other. The position where they hit each other is

1. Nearer to B
2. Nearer to A
3. At an equal distance from A and B
4. Cannot be decided

Answer: 2. Nearer to A

Question 43. Two particles whose masses are 10 kg and 30 kg and their position vectors are respectively would have the center of mass at –

1. \(-\frac{(\hat{\mathbf{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})}{2}\)
2. \(\frac{(\hat{i}+\hat{j}+\hat{k})}{2}\)
3. \(-\frac{(\hat{i}+\hat{j}+\hat{k})}{4}\)
4. \(\frac{(\hat{i}+\hat{j}+\hat{k})}{4}\)

Answer: 1. \(-\frac{(\hat{\mathbf{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})}{2}\)

Question 44. Two balls A and B of masses 100 gm and 250 gm respectively are connected by a stretched spring of negligible mass and placed on a smooth table. When the balls are released simultaneously the initial acceleration of B is 10 cm/sec² westward. What is the magnitude and direction of the initial acceleration of the ball A –

1. 25 cm/sec² Eastward
2. 25 cm/sec² Northward
3. 25 cm/sec² Westward
4. 25 cm/sec² Southward

Answer: 1. 25 cm/sec² Eastward

Question 45. A shell of mass m moving with velocity u suddenly breaks into 2 pieces. The part having mass m/4 remains stationary. The velocity of the other shell will be :

1. u
2. 2u
3. \(\frac{3}{4}\)u
4. \(\frac{4}{3}\)u

Answer: 4. \(\frac{4}{3}\)u

Question 46. A stone is projected with an initial velocity at some angle to the horizontal. A small piece separates from the stone before the stone reaches its maximum height. Then the piece will :

1. Fall to the ground
2. Fly horizontally initially and will then describe a parabolic path
3. Fly side by side with the parent stone along a parabolic path
4. The lag behind the parent stone increases the distance from it.

Answer: 3. Fly side by side with the parent stone along a parabolic path

Centre of Mass NEET Class 11 MCQs with Detailed Solutions

Question 47. Three particles with masses 10, 20 and 40gm are moving with velocities \(10 \hat{i}, 10 \hat{j}\) and \(10 \hat{k}_{}\) m/sec respectively. If due to some internal force, the first particle comes to rest and the velocity of the second becomes \((3 \hat{i}+4 \hat{j})\)+m/sec. then the velocity of the third particle after the interaction is-

1. \(\hat{i}+\hat{j}+5 \hat{k}\)
2. \(\hat{j}+10 \hat{k}\)
3. \(\hat{\mathrm{i}}+\hat{\mathrm{j}}+10 \hat{\mathrm{k}}\)
4. \(\hat{i}+3 \hat{j}+10 \hat{k}\)

Answer: 4. \(\hat{i}+3 \hat{j}+10 \hat{k}\)

Question 48. Two particles having mass ratio n: 1 are interconnected by a light inextensible string that passes over a smooth pulley. If the system is released, then the acceleration of the center of mass of the system is:

1. \((n-1)^2 g\)
2. \(\left(\frac{n+1}{n-1}\right)^2 g\)
3. \(\left(\frac{n-1}{n+1}\right)^2 g\)
4. \(\left(\frac{n+1}{n-1}\right) g\)

Answer: 3. \(\left(\frac{n-1}{n+1}\right)^2 g\)

Question 49. A uniform thin rod of mass M and Length L is standing vertically along the y-axis on a smooth horizontal surface, with its lower end at the origin (0,0). A slight disturbance at t = 0 causes the lower end to slip on the smooth surface along the positive x-axis, and the rod starts falling. The acceleration vector of the center of mass of the rod during its fall is :

⇒ [\(\vec{R}\) is reaction from surface]

1. \(\vec{a}_{C M}=\frac{M \vec{g}+\vec{R}}{M}\)
2. \(\vec{a}_{C M}=\frac{M \vec{g}-\vec{R}}{M}\)
3. \(\overrightarrow{\mathrm{a}}_{\mathrm{CM}}=\mathrm{Mg}-\overrightarrow{\mathrm{R}}\)
4. None of these

Answer: 1. [\(\vec{R}\) is reaction from surface]

Question 50. In a vertical plane inside a smooth hollow thin tube, a block of the same mass as that of the tube is released as shown in the figure. When it is slightly disturbed it moves towards the right. By the time the block reaches the right end of the tube then the displacement of the tube will be (where ‘R’ is the mean radius of a tube). Assume that the tube remains in the vertical plane.

1. \(\frac{2 R}{\pi}\)
2. \(\frac{4 R}{\pi}\)
3. \(\frac{R}{2}\)
4. R

Answer: 3. \(\frac{R}{2}\)

Question 51. A ball kept in a closed box moves in the box making collisions with the walls. The box is kept on a smooth surface. The centre of mass:

1. Of the box remains constant
2. Of the box plus the ball system remains constant
3. If the ball remains constant
4. Of the ball relative to the box remains constant

Answer: 2. Of the box plus the ball system remains constant

Question 52. A man of mass M stands at one end of a plank of length L which lies at rest on a frictionless surface. The man walks to the other end of the plank. If the mass of the plank is M/3, the distance that the plank moves relative to the ground is :

1. 3L/4
2. L/4
3. 4L/5
4. L/3

Answer: 1. 3L/4

Question 53. Two blocks A and B are connected by a massless string (shown in figure) A force of 30 N is applied to block B. The distance traveled by center of mass in 2s starting from rest is :

1. 1m
2. 2m
3. 3m
4. None of these

Answer: 2. 2m

Question 54. If the system is released, then the acceleration of the center of mass of the system :

1. \(\frac{g}{4}\)
2. \(\frac{g}{2}\)
3. g
4. 2g

Answer: 1. \(\frac{g}{4}\)

Question 55. Three particles of masses 1 kg, 2 kg and 3 kg are subjected to forces \((3 \hat{i}-2 \hat{j}+2 \hat{k}) N,(-\hat{i}+2 \hat{j}-\hat{k}) N\), and \((\hat{i}+\hat{j}+\hat{k}) N\)respectively. The magnitude of the acceleration of the CM of the system is:

1. \(\frac{\sqrt{11}}{6} \mathrm{~ms}^{-2}\)
2. \(\frac{\sqrt{14}}{6} \mathrm{~ms}^{-2} \)
3. \(\frac{11}{6} \mathrm{~ms}^{-2}\)
4. \(\frac{22}{6} \mathrm{~ms}^{-2}\)

Answer: 2. \(\frac{\sqrt{14}}{6} \mathrm{~ms}^{-2} \)

Question 56. Two bodies of mass 10 kg and 2 kg are moving with velocity \(2 \hat{i}-7 \hat{j}+3 \hat{k} \mathrm{~m} / \mathrm{s} \text { and }-10 \hat{i}+35 \hat{j}-3 \hat{k} \mathrm{~m} / \mathrm{s}\)respectively. The velocity of their centre of mass is :

1. \(2 \hat{i} \mathrm{~ms}\)
2. \(2 \hat{k} \text { ms }\)
3. \((2 \hat{j}+2 \hat{k}) \mathrm{ms}\)
4. \((2 \hat{i}+2 \hat{j}+2 \hat{k}) \mathrm{ms}\)

Answer: 2. \(2 \hat{k} \text { ms }\)

Question 57. Consider a system of two particles having masses m₁ and m₁. If the particle of mass m₁ is pushed towards the mass center of particles through a distance d, by what distance would the particle of mass m₂ move so as to keep the mass center of particles at the original position?

1. \(\frac{m_1}{m_1+m_2} d\)
2. \(\frac{m_1}{m_2} d\)
3. d
4. \(\frac{m_2}{m_1} d\)

Answer: 2. \(\frac{m_1}{m_2} d\)

Question 58. Two identical particles move towards each other with velocities 2v and v respectively. This velocity of the centre of mass is –

1. v
2. v/3
3. v/2
4. Zero

Answer: 3. v/2

Question 59. Two blocks of masses 10kg and 4kg are connected by a spring of negligible mass and are placed on a frictionless horizontal surface. An impulse gives a speed of 14 ms^-1 to the heavier block in the direction of the lighter block. Then, the velocity of the centre of mass is

1. 30 ms^-1
2. 20 ms^-1
3. 10 ms^-1
4. 5 ms^-1

Answer: 3. 10 ms^-1

Question 60. A uniform chain of length 2 m is kept on a table such that a length of 60 cm hangs freely from the edge of the table. The total mass of the chain is 4 kg. What is the work done in pulling the entire chain on the table?

1. 7.2 J
2. 3.6 J
3. 120 J
4. 1200 J

Answer: 2. 3.6 J

Question 61. A body A of mass M while falling vertically downwards under gravity breaks into two parts; a body B of mass \([\frac{1}{3}\)M and, a body C of mass \([\frac{2}{3}\)M. The centre of mass of bodies B and C taken together shifts compared to that of body A towards:

1. Depends on the height of the breaking
2. Does not shift
3. Shift towards body C
4. Shift towards body B

Answer: 2. Does not shift

Question 62. Consider a two-particle system with particles having masses m₁ and m₂. If the first particle is pushed towards the centre of mass through a distance d, by what distance should the second particle be moved, so as to keep the centre of mass at the same position?

1. d
2. \(\frac{m_2}{m_1} d\)
3. \(\frac{m_1}{m_1+m_2} d\)
4. \(\frac{m_1}{m_2} d\)

Answer: 4. \(\frac{m_1}{m_2} d\)

Question 63. A uniform sphere is placed on a smooth horizontal surface and a horizontal force F is applied on it at a distance h above the surface. The acceleration of the center

1. Is maximum when h = 0
2. Is maximum when h = R
3. Is maximum when h = 2R
4. Is independent of h

Answer: 4. Is independent of h

Question 64. A can of height h is filled with liquid of uniform density ρ. If the liquid is coming out from the bottom then a centre of mass of the ” can + water in the can “

1. First ascends and then descends
2. First descends and then ascends
3. Always decreases
4. None of these

Answer: 2. First descends and then ascends

Question 65. A man of mass M stands at one end of a plank of length L which lies at rest on a frictionless horizontal surface. The man walks to the other end of the plank. If the mass of the plank is M/3, the distance that the man moves relative to the ground is

1. 3 L/4
2. 4 L/5
3. L/4
4. None of these

Answer: 3. L/4

Question 66. When a block is placed on a wedge as shown in the figure, the block starts sliding down and the wedge also starts sliding on the ground. All surfaces are rough. The center of mass of the (wedge + block) system will

1. Leftward and downward.
2. Right ward and downward.
3. Leftward and upwards.
4. Only downward.

Answer: 2. Rightward and downward.

Question 67. A 2 kg body and a 3 kg body are moving along the x-axis. At a particular instant the 2 kg body has a velocity of 3 ms^-1 and the 3 kg body has a velocity of 2 ms^-1. The velocity of the center of mass at that instant is

1. 5 ms^-1
2. 1 ms^-1
3. zero
4. None of these

Answer: 4. None of these

NEET Class 11 Physics Centre of Mass MCQs for Exam Preparation

Question 68. Two bodies of masses 2 kg and 4 kg are moving with velocities 2 m/s and 10m/s respectively along the same direction. Then the velocity of their centre of mass will be

1. 8.1 m/s
2. 7.3 m/s
3. 6.4 m/s
4. 5.3 m/s

Answer: 2. 7.3 m/s

Question 69. Two particles of masses m₁ and m₂ initially at rest start moving towards each other under their mutual force of attraction. The speed of the centre of mass at any time t, when they are at a distance r apart, is

1. Zero
2. \(\left(G \frac{m_1 m_2}{r^2} \cdot \frac{1}{m_1}\right) t\)
3. \(\left(G \frac{m_1 m_2}{r^2} \cdot \frac{1}{m_2}\right) t\)
4. \(\left(G \frac{m_1 m_2}{r^2} \cdot \frac{1}{m_1+m_2}\right) t\)

Answer: 1. Zero

Question 70. A body of mass 20 kg is moving with a velocity of 2ν and another body of mass 10 kg is moving with velocity V along the same direction. The velocity of their center of mass is

1. 5ν/3
2. 2ν/3
3. ν
4. Zero

Answer: 1. 5ν/3

Question 71. The two particles X and Y, initially at rest, start moving towards each other under mutual attraction. If at any instant the velocity of X is V and that of Y is 2V, the velocity of their centre of mass will be

1. Zero
2. V
3. 2V
4. V/2

Answer: 1. Zero

Question 72. Two particles A and B initially at rest move towards each other under a mutual force of attraction. The speed of the center of mass at the instant when the speed of A is v and the speed of B is 2v is :

1. v
2. Zero
3. 2 v
4. 3 v /2

Answer: 2. Zero

Question 73. If the KE of a body becomes four times its initial value, then the new momentum will be more than its initial momentum by;

1. 50%
2. 100%
3. 125%
4. 150%

Answer: 2. 100%

Question 74. A particle of mass 4m which is at rest explodes into three fragments. Two of the fragments, each of mass m are found to move with speed v each, in mutually perpendicular directions. The total energy released in the process of explosion is-

1. 3mv²/2
2. mv²
3. 4mv²
4. 2mv²

Answer: 1. 3mv²/2

Question 75. A bullet of mass m is being fired from a stationary gun of mass M. If the velocity of the bullet is v, the velocity of the gun is-

1. \(\frac{M v}{m+M}\)
2. \(\frac{\mathrm{mv}}{\mathrm{M}}\)
3. \(\frac{(M+m) v}{M}\)
4. \(\frac{\mathrm{M}+\mathrm{m}}{\mathrm{Mv}}\)

Answer: 2. \(\frac{\mathrm{mv}}{\mathrm{M}}\)

Question 76. A bomb explodes in the air in two equal fragments. If one of the fragments is moving vertically upwards with velocity v₀, then the other fragment is moving-

1. Vertically up with velocity v₀
2. Vertically downwards with velocity v₀
3. In any arbitrary direction
4. None of these

Answer: 2. Vertically downwards with velocity v₀

Question 77. Two particles with equal kinetic energies have masses in the ratio of 1: 2. Then linear momenta will be in the ratio-

1. 1
2. 4
3. 0.707
4. 2

Answer: 3. 0.707

Question 78. If a shell fired from a canon explodes in the air then-

1. Momentum decreases
2. Momentum increases
3. Kinetic energy increases
4. K.E. decreases

Answer: 3. Kinetic energy increases

Question 79. Three particles A, B and C of equal mass move with equal speeds v along the medians of an equilateral triangle as shown in the figure. They collide at the centroid G of the triangle. After collision A comes to rest, and B retraces its path with speed v. The velocity of C is-

1. \(\vec{v} \text {, direction } \vec{G} A\)
2. \(\overrightarrow{2} \mathrm{v} \text { and direction } \overrightarrow{\mathrm{G}} \mathrm{A}\)
3. \(2 \mathrm{v} \text {, direction } \overrightarrow{\mathrm{GB}}\)
4. \(\vec{v} \text {, and direction } \overrightarrow{B G}\)

Answer: 4. \(\vec{v} \text {, and direction } \overrightarrow{B G}\)

Question 80. Under the effect of mutual internal attractions-

1. The linear momentum of a system increases
2. The linear momentum of a system decreases
3. The linear momentum of the system is conserved
4. The angular momentum increases

Answer: 3. The linear momentum of the system is conserved

Question 81. A ball of mass 3 kg collides with a wall with a velocity of 10 m/sec at an angle of 30° and after collision reflects at the same angle with the same speed. The change in momentum of the ball in the MKS unit is-

1. 20
2. 30
3. 15
4. 45

Answer: 2. 30

Question 82. A particle is moving in X–Y plane under the action of a force \(\vec{F}\) such that at some instant ‘t’ the components of its linear momentum \(\vec{p}\) are px= 2 cos t and py= 2 sin t. At this instant the angle between \(\vec{F} \text { and } \vec{p}\) is

1. 90°
2. 0°
3. 180°
4. 30°

Answer: 1. 90°

Question 83. The kinetic energies of a lighter body and a heavier body are the same. Then the value of momentum is-

1. Higher for lighter body
2. Higher for heavier body
3. Same for both
4. Additional information is needed to reply to this question

Answer: 2. Higher for heavier body

Centre of Mass Multiple Choice Questions for NEET Class 11

Question 84. A bullet of mass m moving with a velocity v1strikes a suspended wooden block of mass M as shown in the figure and sticks to it. If the block rises to a height of h the initial velocity of the bullet is-

1. \(\frac{\mathrm{m}+\mathrm{M}}{\mathrm{m}} \sqrt{2 \mathrm{gh}}\)
2. \(\sqrt{2 g h}\)
3. \(\frac{M+m}{M} \sqrt{2 g h}\)
4. \(\frac{m}{M+m} \sqrt{2 g h}\)

Answer: 1. \(\frac{\mathrm{m}+\mathrm{M}}{\mathrm{m}} \sqrt{2 \mathrm{gh}}\)

Question 85. If the mass and kinetic energy of a particle are m and E respectively, then the value of its momentum is-

1. \(\sqrt{\mathrm{mE}}\)
2. \(\sqrt{2 m E}\)
3. \(\sqrt{2 E / m}\)
4. \(\sqrt{2 \mathrm{~m} / \mathrm{E}}\)

Answer: 2. \(\sqrt{2 m E}\)

Question 86. If a lighter body (mass M₁ and velocity V₁) and a heavier body (mass M₂ and velocity V₂) have the same kinetic energy, then-

1. M₂V₂< M₁V₁
2. M₂V₂= M₁V₁
3. M₂V₁= M₁V₂
4. M₂V₂> M₁V₁

Answer: 4. M₂V₂> M₁V₁

Question 87. A bomb of mass 12kg at rest explodes into two fragments of masses in the ratio 1: 3. The K.E. of the smaller fragment is 216 J. The momentum of heavier fragments is (in kg-m/sec) –

1. 36
2. 72
3. 108
4. Insufficient data

Answer: 1. 36

Question 88. A bomb is projected at 200m/s at an angle of 60° with horizontal. At the highest point, it explodes into three particles of equal masses. One goes vertically upward with a velocity of 100m/sec, second particle goes vertically downward with the same velocity as the first. Then what is the velocity of the third one-

1. 120 m/sec with 60° angle
2. 200 m/sec with 30° angle
3. 50 m/sec, in horizontal direction
4. 300 m/sec, in horizontal direction

Answer: 4. 300 m/sec, in horizontal direction

Question 89. The law of conservation of energy implies that the

1. Total mechanical energy is conserved
2. Total kinetic energy is conserved
3. Total potential energy is conserved
4. Sum of all kinds of energies is conserved

Answer: 4. Sum of all kinds of energies is conserved

Question 90. If the kinetic energy of a body becomes four times its initial value, then new momentum will-

1. Become twice its initial value
2. Become three times, its initial value
3. Become four times, its initial value
4. Remains constant

Answer: 1. Become twice its initial value

Question 91. A spacecraft of mass M is travelling in space with velocity v. It then breaks up into two parts such that the smaller part m comes to the rest, then the velocity of the remaining part is-

1. \(\frac{M v}{M-m}\)
2. \(\frac{M v}{M+m}\)
3. \(\frac{\mathrm{mv}}{\mathrm{M}-\mathrm{m}}\)
4. \(\frac{\mathrm{Mv}}{\mathrm{m}}\)

Answer: 1. \(\frac{M v}{M-m}\)

Question 92. A bomb at rest has a mass of 60 kg. It explodes and a fragment of 40 kg has a kinetic energy of 96 joules. Then the kinetic energy of the other fragment is-

1. 180 J
2. 190 J
3. 182 J
4. 192 J

Answer: 4. 192 J

Question 93. Consider the following two statements-

1. The linear momentum of a system of particles is zero
2. the kinetic energy of a system of particles is zero. Then

1. A does not imply B but B implies A
2. A implies B and B implies A
3. A does not imply B & B does not imply A
4. A implies B but B does not imply A

Answer: 1. A does not imply B but B implies A

Question 94. When a U238 nucleus originally at rest, decays emitting an alpha particle having a speed ‘u’, the recoil speed of the residual nucleus is-

1. \(\frac{4 u}{234}\)
2. \(-\frac{4 u}{238}\)
3. \(\frac{4 u}{238}\)
4. \(-\frac{4 u}{234}\)

Answer: 1. \(\frac{4 u}{234}\)

Question 95. A ball of mass 2 kg and another of mass 4 kg are dropped together from a 60-foot-tall building. After a fall of 30 feet each toward Earth, their respective kinetic energies will be in the ratio of-

1. \(\sqrt{2}\): 1
2. 1: 4
3. 1: 2
4. 1:\(\sqrt{2}\)

Answer: 3. 1: 2

Question 96. A moving body of mass m and velocity 3 km/hr collides with a body at rest and of mass 2m and then sticks to it. Now the combined mass starts to move, then the combined velocity will be:

1. 4 km/hr
2. 3 km/hr
3. 2 km/hr
4. 1 km/hr

Answer: 4. 1 km/hr

Question 97. A 50 g bullet moving with a velocity of 10 m/s strikes a block of mass 950 g at rest and gets embedded into it. The loss in kinetic energy will be

1. 100 %
2. 95 %
3. 5 %
4. 50 %

Answer: 2. 95 %

Question 98. A body at rest splits into three parts of mass m, m and 4m respectively. The two equal masses fly off perpendicular to each other and each with a speed of V. The speed of 4m will be

1. \(\frac{V}{2 \sqrt{2}}\)
2. \(\frac{\mathrm{V}}{\sqrt{2}}\)
3. \(\frac{\mathrm{V}}{2}\)
4. \(\sqrt{2} \mathrm{~V}\)

Answer: 1. \(\frac{V}{2 \sqrt{2}}\)

Question 99. A stationary body explodes into two fragments of masses m₁ and m₂. If the momentum of one fragment is p, the minimum energy of the explosion is

1. \(\frac{p^2}{2\left(m_1+m_2\right)}\)
2. \(\frac{p^2}{2 \sqrt{m_1 m_2}}\)
3. \(\frac{p^2\left(m_1+m_2\right)}{2 m_1 m_2}\)
4. \(\frac{p^2}{2\left(m_1-m_2\right)}\)

Answer: 3. \(\frac{p^2\left(m_1+m_2\right)}{2 m_1 m_2}\)

Question 100. A train of mass M is moving on a circular track of radius ‘ R ‘ with constant speed V. The length of the train is half of the perimeter of the track. The linear momentum of the train will be

1. 0
2. \(\frac{2 M V}{\pi}\)
3. MVR
4. MV

Answer: 2. \(\frac{2 M V}{\pi}\)

Question 101. Two bodies of masses m and 4m are moving with equal linear momentum. The ratio of their kinetic energies is:

1. 1: 4
2. 4: 1
3. 1: 1
4. 1: 2

Answer: 2. 4: 1

Question 102. If the momentum of a body increases by 20%, the percentage increase in its kinetic energy is equal to :

1. 44
2. 88
3. 66
4. 20

Answer: 1. 44

Question 103. A man is in a moving train, then the train :

1. His momentum must not be zero
2. His kinetic energy is zero
3. His kinetic energy is not zero
4. His kinetic energy may be zero

Answer: 4. His kinetic energy may be zero

Question 104. A bomb dropped from an airplane explodes in the air. It’s total :

1. Momentum decreases
2. Momentum increases
3. Kinetic energy increases
4. Kinetic energy decreases

Answer: 3. Kinetic energy increases

Question 105. Two blocks of masses m₁ and m₂ are connected by a massless spring and placed on a smooth surface. The spring initially stretched and released. Then :

1. The momentum of each particle remains constant separately
2. The momentums of each body are equal
3. The magnitude of momentums of each body are equal to each other
4. The mechanical energy of a system remains constant

1. a and b are correct
2. a, b and c are correct
3. c and d are correct
4. Only C is correct

Answer: 3. c and d are correct

Question 106. A bag of mass M hangs by a long thread and a bullet (mass m) comes horizontally with velocity v and gets caught in the bag. Then for the combined system (bag + bullet):

1. Momentum is mMv/(M + m)
2. KE is (1/2) Mv²
3. Momentum is mv
4. KE is m²v²/(M + m)

Answer: 3. Momentum is mv

Question 107. A man of mass ‘m’ climbs on a rope of length L suspended below a balloon of mass M. The balloon is stationary with respect to ground. If the man begins to climb up the rope at a speed v_rel (relative to rope). In what direction and with what speed (relative to ground) will the balloon move?

1. Downwards, \(\frac{\mathrm{mv}_{\mathrm{rel}}}{\mathrm{m}+\mathrm{M}}\)
2. Upwards, \(\frac{\mathrm{Mv}_{\mathrm{rell}}}{\mathrm{m}+\mathrm{M}}\)
3. Downwards, \(\frac{\mathrm{mV}_{\mathrm{rel}}}{\mathrm{M}}\)
4. Downwards, \(\frac{(M+m) v_{\text {rel }}}{M}\)

Answer: 1. Downwards, \(\frac{\mathrm{mv}_{\mathrm{rel}}}{\mathrm{m}+\mathrm{M}}\)

Question 108. In the figure shown the initial velocity of a boat (30 kg) + person (15 kg ) is 2 m/s. Find the velocity of the person w.r.t. boat so that the velocity of the boat will be 1 m/s in right (Neglect friction between boat and water)

1. 3 m/s towards right
2. 3 m/s towards left
3. 4 m/s towards right
4. 4 m/s towards left

Answer: 1. 3 m/s towards right

Question 109. 1 kg body explodes into three fragments. The ratio of their masses is 1: 1 : 3. The fragments of the same mass move perpendicular to each other at speeds of 30 m/s, while the heavier part remains in the initial direction. The speed of the heavier part is :

1. \(\frac{10}{\sqrt{2}} \mathrm{~m} / \mathrm{s}\)
2. \(10 \sqrt{2} \mathrm{~m} / \mathrm{s}\)
3. \(20 \sqrt{2} \mathrm{~m} / \mathrm{s}\)
4. \(30 \sqrt{2} \mathrm{~m} / \mathrm{s}\)

Answer: 2. \(10 \sqrt{2} \mathrm{~m} / \mathrm{s}\)

Question 110. A stationary particle explodes into two particles of masses m₁ and m₂ which move in opposite directions with velocities v₁ and v₂. The ratio of their kinetic energies E₁/E₂ is:-

1. m₂/m₁
2. m₁/m₂
3. 1
4. m₁v₂/m₂ v₁

Answer: 1. m₁/m₂

Question 111. A particle of mass m₁ is moving with a velocity v1and another particle of mass m₂ is moving with a velocity v₂. Both of them have the same momentum but their different kinetic energies are E₁ and E₂ respectively. If m₁> m₂ then :

1. E₁ < E₂
2. \(\frac{E_1}{E_2}=\frac{m_1}{m_2}\)
3. E₁ > E₂
4. E₁ = E₂

Answer: 1. E₁ < E₂

Question 112. A bomb of mass 30 kg at rest explodes into two pieces of masses 18 kg and 12 kg. The velocity of 18 kg mass is 6 ms^-1. The kinetic energy of the other mass is:

1. 256 J
2. 486 J
3. 524 J
4. 324 J

Answer: 2. 486 J

Question 113. A shell of mass 200 g is ejected from a gun of mass 4 kg by an explosion that generates 1.05 kJ of energy. The initial velocity of the shell is

1. 100 ms^-1
2. 80 ms^-1
3. 40 ms^-1
4. 120 ms^-1

Answer: 1. 100 ms^-1

NEET Physics Chapter 3 Centre of Mass MCQs: Key Concepts and Solutions

Question 114. A bomb of mass 3.0 kg explodes in the air into two pieces of masses 2.0 kg and 1.0 kg. The smaller mass goes at a speed of 80 m/s. The total energy imparted to the two fragments is :

1. 1.07 kJ
2. 2.14 kJ
3. 2.4 kJ
4. 4.8 kJ

Answer: 4. 4.8 kJ

Question 115. Consider the following two statements :

1. A. Linear momentum of a system of particles is zero
2. B. Kinetic energy of a system of particles is zero,

Then,

1. A does not imply B and B does not imply A
2. A implies B but B does not imply A
3. A does not imply B but B implies A
4. A implies B and B implies A

Answer: 3. A does not imply B but B implies A

Question 116. A bomb of mass 16 kg at rest explodes into two pieces of masses of 4 kg and 12 kg. The velocity of the 12 kg mass is 4 ms^-1. The kinetic energy of the other mass is :

1. 96 J
2. 144 J
3. 288 J
4. 192 J

Answer: 3. 288 J

Question 117. A block of mass 0.50 kg is moving with a speed of 2.00 ms^-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is :

1. 1.00 J
2. 0.67 J
3. 0.34 J
4. 0.16 J

Answer: 2. 0.67 J

Question 118. Which of the following is incorrect?

1. If the centre of mass of three particles is at rest, and it is known that two of them are moving along different non-parallel lines then the third particle must also be moving.
2. If the centre of mass remains at rest, then the net work done by the forces acting on the system must be zero.
3. If the centre of mass remains at rest then the net external force must be zero
4. None of this statement is incorrect

Answer: 2. If the centre of mass remains at rest, then the net work done by the forces acting on the system must be zero.

Question 119. A bomb at rest explodes in three segments of unequal masses. The most general description of the final state is that:

1. The fragments fly off in any arbitrary direction.
2. The fragments fly off in such a way that there directions lie in the same plane.
3. Two of the three must go opposite to each other.
4. Two of the three must fly off at right angles to each other.

Answer: 2. The fragments fly off in such a way that there directions lie in the same plane.

Question 120. Two particles A and B start moving due to their mutual interaction only. If at any time ‘t’, \(\overrightarrow{\mathrm{a}}_{\mathrm{A}} and \overrightarrow{\mathrm{a}}_{\mathrm{B}}\) respective accelerations, \(\vec{v}_A\) and \(\vec{v}_A\) are their respective velocities, and upto that time w_A and B w_B are the work done on A and B respectively by the mutual force, m_A and m_B are their masses respectively, then which of the following is always correct.

1. \(\vec{v}_A+\vec{v}_B=0\)
2. \(m_A \vec{v}_A+\vec{v}_B m_B=0\)
3. \(w_A+w_B=0\)
4. \(\vec{a}_A+\vec{a}_B=0\)

Answer: 2. \(m_A \vec{v}_A+\vec{v}_B m_B=0\)

Question 121. In the diagram shown, a block of mass M initially at rest on a frictionless horizontal surface is struck by a bullet of mass m moving with horizontal velocity v. What is the velocity of the bullet-block system after the bullet embeds itself in the block?

1. \(\left(\frac{M+m}{M}\right) v\)
2. \(\left(\frac{\mathrm{m}}{\mathrm{M}}\right) \mathrm{v}\)
3. \(\frac{\mathrm{mv}}{\mathrm{Nm}+\mathrm{M}}\)
4. \(\frac{\mathrm{Nm}+\mathrm{M}}{\mathrm{mv}}\)

Answer: 4. \(\frac{\mathrm{Nm}+\mathrm{M}}{\mathrm{mv}}\)

Question 122. A continuous stream of particles of mass m and velocity v, is emitted from a source at a rate of n per second. The particles travel along a straight line, collide with a body of mass M and are buried in this body. If the mass M was originally at rest, its velocity when it has received N particles will be:

1. \(\frac{\mathrm{mvn}}{\mathrm{Nm}+\mathrm{n}}\)
2. \(\frac{\mathrm{mvN}}{\mathrm{Nm}+\mathrm{M}}\)
3. \(\frac{\mathrm{mv}}{\mathrm{Nm}+\mathrm{M}}\)
4. \(\frac{\mathrm{Nm}+\mathrm{M}}{\mathrm{mv}}\)

Answer: 2. \(\frac{\mathrm{mvN}}{\mathrm{Nm}+\mathrm{M}}\)

Question 123. On doubling the speed of an object its-

1. K.E. is doubled
2. P.E. is doubled
3. Momentum is doubled
4. Acceleration is doubled

Answer: 3. Momentum is doubled

Question 124. A block moving in the air explodes in two parts then just after the explosion

1. The total momentum must be conserved
2. The total kinetic energy of the two parts must be the same as that of a block before the explosion.
3. The total momentum must change
4. The total kinetic energy must not be increased

Answer: 1. The total momentum must be conserved

Question 125. A particle of mass m moving with velocity v rebounds with the same speed after making an impact with a wall. The change in its momentum shall be

1. –2mv
2. mv
3. –mv
4. Zero

Answer: 1. –2mv

Question 126. A block of mass m slips down an inclined plane as shown in the figure and it presses a spring lying at the bottom. If the length of the spring h >> l and the spring constant is K the compression in the spring will be

1. \(\sqrt{\frac{\mathrm{mgh}}{\mathrm{k}}}\)
2. \(\sqrt{\frac{2 m g h}{k}}\)
3. \(\sqrt{\frac{g h}{m k}}\)
4. \(\sqrt{\frac{2 \mathrm{gh}}{\mathrm{mk}}}\)

Answer: 2. \(\sqrt{\frac{2 m g h}{k}}\)

Question 127. One end of a vertical ideal spring is attached to a rigid support and to the other end a weight of 200 gm is suspended. If this weight is doubled then the value of the spring constant is :

1. Halved
2. Unchanged
3. Doubled
4. Zero

Answer: 2. Unchanged

Question 128. A mass M is hanging from a spring. If on hanging an additional mass ‘m’ the string further gets stretched by x meters, then the spring constant is :

1. mg/x
2. (m + M)g/x
3. Mg/x
4. Mx/m

Answer: 1. mg/x

Question 129. After falling from a height h a mass m compresses a spring of force constant k. The compression produced in the spring is :

1. (mgh/k)^1/2
2. (2mgh/k)^1/2
3. (k/mgh)^1/2
4. (2k/mgh)^1/2

Answer: 2. (2mgh/k)^1/2

Question 130. With how much velocity a block of mass 2 kg should move on a frictionless surface so as to compress a spring with a force constant of 2 newton/meter by 4 meters:

1. 4 m/s
2. 16 m/s
3. 2 m/s
4. 8 m/s

Answer: 1. 4 m/s

Question 131. A body of mass 8kg moving with a velocity of 2 m/sec comes to the rest, after compressing a spring placed on a frictionless table. If the spring constant is 5000N/m then compression produced in the spring shall be :

1. 4 cm
2. 8 cm
3. 16 cm
4. 32 cm

Answer: 2. 8 cm

Question 132. A mass of 2 kg falls from a height of 0.4 m on a spring of force constant k = 1960 N/m. The maximum distance upto which the string can be compressed is :

1. 9 cm
2. 4.5 cm
3. 12.6 cm
4. 6.3 cm

Answer: 1. 9 cm

Question 133. An object is attached to a vertical spring and slowly lowered to its equilibrium position. This stretches the spring by 10 cm. If the same object is attached to the same vertical spring but permitted to fall instead, the maximum distance upto which the spring can be stretched is:

1. 5 cm
2. 10 cm
3. 20 cm
4. 40 cm

Answer: 3. 20 cm

Question 134. The figure shows the magnitude of change in momentum of the block when it comes to its initial position if the maximum compression of the spring is x0 will be:

1. \(2 \sqrt{k m} x_0\)
2. \(\sqrt{\mathrm{km}} x_0\)
3. Zero
4. None of these

Answer: 1. \(2 \sqrt{k m} x_0\)

Question 135. Two masses are connected by a spring as shown in the figure. One of the masses was given velocity v = 2 k, as shown in the figure where ‘k’ is the spring constant. The maximum extension in the spring will be

1. 2 m
2. m
3. \(\sqrt{2 m k}\)
4. \(\sqrt{3 \mathrm{mk}}\)

Answer: 3. \(\sqrt{2 m k}\)

Centre of Mass MCQ Practice Test with Answers for NEET Class 11

Question 136. A force of 50 dynes is acted on a body of mass 5gm which is at rest for an interval of 3 sec, then the impulse is-

1. 0.16 × 10^-3 N-S
2. 0.98 × 10^-3 N-S
3. 1.5 × 10^-3 N-S
4. 2.5 × 10^-3 N-S

Answer: 3. 1.5 × 10^-3 N-S

Question 137. The area of the F-t curve is A, where ‘F’ is the force on one mass due to the other. If one of the colliding bodies of mass M is at rest initially, its speed just after the collision is :

1. A/M
2. M/A
3. AM
4. \(\sqrt{\frac{2 A}{M}}\)

Answer: 1. A/M

Question 138. A body of mass 0.5 kg is projected under gravity with a speed of 98 m/s at an angle of 60° with the vertical. The change in momentum [in magnitude] of the body when it returns to the ground is

1. 24.5 N-s
2. 49.0 N-s
3. 98.0 N-s
4. 49 \(\sqrt{\frac{2 A}{M}}\)

Answer: 2. 49.0 N-s

Question 139. A body of mass ‘M’ collides against a wall with a velocity υ and retraces its path at the same speed. The change in momentum is (take the initial direction of velocity as positive) :

1. zero‘
2. 2Mυ
3. Mυ
4. –2Mυ

Answer: 4. –2Mυ

Question 140. If two balls, each of mass 0.06 kg, moving in opposite directions with a speed of 4m/s, collide and rebound with the same speed, then the impulse imparted to each ball due to the other (in kg-m/s) is :

1. 0.48
2. 0.53
3. 0.81
4. 0.92

Answer: 1. 0.48

Question 141. A ball of mass 50 gm is dropped from a height h = 10 m. It rebounds losing 75 percent of its kinetic energy. If it remains in contact with the ground for Δt = 0.01 sec., the impulse of the impact force is :

1. 1.3 N–s
2. 1.05 N-s
3. 1300 N–s
4. 105 N–s

Answer: 2. 1.05 N-s

Question 142. The given figure shows a plot of the time-dependent force F acting on a particle in motion along the x-axis. What is the total impulse delivered by this force to the particle from time t = 0 to t = 2second?

1. 0
2. 1 kg-m/s
3. 2 kg-m/s
4. 3 kg-m/s

Answer: 3. 2 kg-m/s

Question 143. A ball of mass 3 kg moving with a speed of 100 m/s, strikes a wall at an angle of 60º (as shown in the figure). The ball rebounds at the same speed and remains in contact with the ball for 0.2 seconds, the force exerted by the ball on the wall is :

1. 1500
2. 1500 N
3. 3003N
4. 300 N

Answer: 1. 1500

Question 144. In the figure given the position-time graph of a particle of mass 0.1 kg is shown. The impulse at t = 2 s is:

1. 0.2 kg m s^-1
2. – 0.2 kg m s^-1
3. 0.1 kg m s^-1
4. – 0.4 kg m s^-1

Answer: 2. – 0.2 kg m s^-1

Question 145. Two balls of the same mass are dropped from the same height onto the floor. The first ball bounces upwards from the floor elastically. The second ball sticks to the floor. The first applies an impulse to the floor of I₁ the second applies an impulse Ι₂(for the duration of collision). Then the relation between both impulses is,

1. Ι₂= 2Ι₁
2. \(\mathrm{I}_2=\frac{\mathrm{I}_1}{2}\)
3. Ι₂= 4Ι₁
4. \(\mathrm{I}_2=\frac{\mathrm{I}_1}{4}\)

Answer: 2. \(\mathrm{I}_2=\frac{\mathrm{I}_1}{2}\)

Question 146. A particle of mass m initially at rest, is acted upon by a variable force F for a brief interval of time T. It attains a velocity u after the force stops acting. F is shown in the graph as a function of time. The curve is a semicircle, find u.

1. \(\frac{\pi F_0^2}{2 m}\)
2. \(\frac{\pi T^2}{8 \mathrm{~m}}\)
3. \(\frac{\pi F_0 T}{4 m}\)
4. \(\frac{F_0 T}{2 m}\)

Answer: 3. \(\frac{\pi F_0 T}{4 m}\)

Question 147. A mass of 100g strikes the wall with a speed of 5m/s at an angle as shown in the figure and it rebounds with the same speed. If the contact time is 2 × 10^-3 sec., what is the force applied on the mass by the wall:

1. 250\(\sqrt{3}\) to right
2. 250 N to right
3. 250\(\sqrt{3}\)N to left
4. 250 N to left

Answer: 3. 250\(\sqrt{3}\)N to left

Question 148. Two particles of masses m₁ and m₂ in projectile motion have velocities \(\overrightarrow{\mathrm{u}}_1\) and \(\overrightarrow{\mathrm{u}}_2\) respectively at time t = 0. They collide at time t₀. Their velocities become \(\overrightarrow{\mathrm{v}}_1\) and \(\overrightarrow{\mathrm{v}}_2\)at time 2t₀ while still moving in the air. The value of

⇒ \(\left[\left(m_1 \vec{v}_1+m_2 \vec{v}_2\right)-\left(m_1 \vec{u}_1+m_2 \vec{u}_2\right)\right]\)is

1. Zero
2. (m₁+ m₂)gt₀
3. 2(m₁+ m₂)gt₀
4. \(\frac{1}{2}\)(m₁+ m₂)gt₀

Answer: 3. 2(m₁+ m₂)gt₀

Question 149. A body is moving towards a finite body which is initially at rest and collides with it. In the absence of any external impulsive force, it is not possible that

1. Both the bodies come to rest
2. Both bodies move after the collision
3. The moving body comes to rest and the stationary body starts moving
4. The stationary body remains stationary, the moving body does not change its velocity.

Answer: 1. Both the bodies come to rest

Question 150. In head-on elastic collision of two bodies of equal masses, it is not possible :

1. The velocities are interchanged
2. The speeds are interchanged
3. The momenta are interchanged
4. The faster body speeds up and the slower body slows down

Answer: 4. The faster body speeds up and the slower body slows down

Question 151. A massive ball moving with speed v collides head-on with a tiny ball at rest having a mass very less than the mass of the first ball. If the collision is elastic, then immediately after the impact, the second ball will move with a speed approximately equal to:

1. v
2. 2v
3. v/2
4. ∞.

Answer: 2. 2v

Question 152. A ball of mass ‘m’, moving with uniform speed, collides elastically with another stationary ball. The incident ball will lose maximum kinetic energy when the mass of the stationary ball is

1. m
2. 2m
3. 4m
4. Infinity

Answer: 1. m

Question 153. In a collision between two solid spheres, the velocity of separation along the line of impact (assume no external forces act on the system of two spheres during impact) :

1. Cannot be greater than the velocity of the approach
2. Cannot be less than the velocity of the approach
3. Cannot be equal to the velocity of the approach
4. None of these

Answer: 1. Cannot be greater than the velocity of the approach

Question 154. In the figure shown block A collides head-on with another block B at rest. The mass of B is twice the mass of A. Block A stops after collision. The coefficient of restitution is :

1. 0.5
2. 1
3. 0.25
4. It is not possible

Answer: 1. 0.5

Question 155. A sphere of mass m moving with a constant velocity hits another stationary sphere of the same mass. If e is the coefficient of restitution, then the ratio of the speed of the first sphere to the speed of the second sphere after a collision will be :

1. \(\left(\frac{1-e}{1+e}\right)\)
2. \(\left(\frac{1+e}{1-e}\right)\)
3. \(\left(\frac{e+1}{e-1}\right)\)
4. \(\left(\frac{e-1}{e+1}\right)\)

Answer: 1. \(\left(\frac{1-e}{1+e}\right)\)

Question 156. A ball rebounds after colliding with the floor, then in case of inelastic collision-

1. The momentum of the ball before and after collision is the same
2. The mechanical energy of the ball is conserved
3. The total momentum of the earth-ball system is conserved
4. The total kinetic energy of the earth and ball is conserved

Answer: 3. The total momentum of the earth-ball system is conserved

Question 157. A ball is allowed to fall from a height of 8cm, if the ball is perfectly elastic, how much does it rise after rebound-

1. 8 cm
2. 1 cm
3. 0.5 cm
4. 0

Answer: 1. 8 cm

Question 158. A particle of mass m₁ moving with a velocity of 5m/s collides head-on with a stationary particle of mass m₂. After collision both the particle moves with a common velocity of 4m/s, then the value of m₁/m₂ is-

1. 4: 1
2. 2: 1
3. 1 : 8
4. 1 : 1

Answer: 1. 4: 1

Question 159. A body of mass m₁ collides head-on elastically with a stationary body of mass m₂. If velocities of m1 before and after the collision are v and –v/3 respectively then the value of m₁/m₂ is-

1. 1
2. 2
3. 0.5
4. 4

Answer: 3. 0.5

Question 160. A sphere of mass 0.1 kg is attached to a cord of 1m in length. Starting from the height of its point of suspension this sphere hits a block of the same mass at rest on a frictionless table. If the impact is elastic, then the kinetic energy of the block after the collision is-

1. 1 J
2. 10 J
3. 0.1 J
4. 0.5 J

Answer: 1. 1 J

Question 161. Two identical smooth spheres A and B are moving with the same velocity and collide with similar spheres C and D, then after elastics collision- (Consider dimensional collision)

1. D will move with greater speed
2. C and D will move with the same velocity v
3. C will stop and D will move with velocity v
4. All spheres A, B, C, and D will move with velocity v/2

Answer: 2. C and D will move with the same velocity v

Question 162. A ball is allowed to fall from a height of 1.0 m. If the value of the coefficient of restitution is 0.6, then after the impact ball will go up to

1. 0.16 m
2. 0.36 m
3. 0.40 m
4. 0.60 m

Answer: 2. 0.36 m

Question 163. A ball of mass m moving with velocity v collides elastically with another ball of identical mass coming from the opposite direction with velocity 2v. Their velocities after collisions are-

1. – v, 2v
2. – 2v, v
3. v, – 2v
4. 2v, – v

Answer: 2. – 2v, v

Question 164. A sphere of mass M moving with velocity u collides head-on elastically with a sphere of mass m at rest. After collision their respective velocities are V and v. The value of v is-

1. \(2 \mathrm{u} \frac{\mathrm{M}}{\mathrm{m}}\)
2. \(2 u \frac{\mathrm{m}}{\mathrm{m}}\)
3. \(\frac{2 u}{1+\mathrm{m} / \mathrm{M}}\)
4. \(\frac{2 \mathrm{u}}{1+\mathrm{M} / \mathrm{m}}\)

Answer: 3. \(\frac{2 u}{1+\mathrm{m} / \mathrm{M}}\)

Centre of Mass in NEET Physics Class 11 MCQs and Explanations

Question 165. A scooter of 40 kg mass moving with a velocity of 4 m/s collides with another scooter of 60 kg mass and moving with a velocity of 2 m/s. After the collision the two scooters stick to each other the loss in kinetic energy-

1. 392 J
2. 440 J
3. 48 J
4. 110 J

Answer: 3. 48 J

Question 166. Two spheres approaching each other collide elastically. Before collision the speed of A is 5m/s and that of B is 10m/s. Their masses are 1kg and 0.5kg. After collision velocities of A and B are respectively-

1. 5 m/s –10 m/s
2. 10 m/s, –5 m/s
3. –10 m/s, –5 m/s
4. –5 m/s, 10 m/s

Answer: 4. –5 m/s, 10 m/s

Question 167. After falling from a height of h and striking the ground twice, a ball rises up to the height [e = coefficient of restitution]

1. He
2. He²
3. He³
4. He⁴

Answer: 4. He⁴

Question 168. A metal ball of mass 2.0kg moving at 36km/hr collides with a stationary ball of mass 3.0kg. If after the collision both balls move together, the loss in kinetic energy will be

1. 40 J
2. 60 J
3. 100 J
4. 140 J

Answer: 2. 60 J

Question 169. A rubber ball is dropped from a height of 5m on a plane. On bouncing it rises to 1.8m. The ball loses its velocity on bouncing by a factor of-

1. 16/25
2. 2/5
3. 3/5
4. 9/25

Answer: 2. 2/5

Question 170. One sphere collides with another sphere of the same mass at rest inelastically. If the value of the coefficient of restitution is 1/2, the ratio of their speeds after collision shall be-

1. 1 : 2
2. 2 : 1
3. 1 : 3
4. 3 : 1

Answer: 3. 1 : 3

Question 171. A steel ball of radius 2cm is initially at rest on a horizontal frictionless surface. It is struck head-on by another steel ball of 4 cm radius travelling with a velocity of 81 cm/s. The velocities of two balls after collision are-

1. 72 cm/s and 56 cm/s
2. 144 cm/s and 56 cm/s
3. 144 cm/s and 63 cm/s
4. 63 cm/s and 72 cm/s

Answer: 3. 144 cm/s and 63 cm/s

Question 172. Which of the following statements is true for collisions-

1. Momentum is conserved in elastic collisions but not in inelastic collisions
2. Total kinetic energy is conserved in elastic collisions but momentum is not conserved
3. Total kinetic energy is not conserved in inelastic collisions but momentum is conserved
4. Total kinetic energy and momentum both are conserved in all types of collisions

Answer: 3. Total kinetic energy is not conserved in inelastic collisions but momentum is conserved

Question 173. For a two-particle collision, the following quantities are conserved in general-

1. Kinetic energy
2. Momentum
3. Both kinetic energy and momentum
4. Neither kinetic energy nor momentum

Answer: 2. Momentum

Question 174. A completely inelastic collision is one in which the two colliding particles-

1. Are separated after the collision.
2. Remain together after the collision.
3. Split into small fragments flying in all directions.
4. None of the above.

Answer: 2. Remain together after the collision.

Question 175. A bullet of mass m = 50 gm strikes a sandbag of mass M = 5 kg hanging from a fixed point, with a horizontal velocity \(\overrightarrow{\mathrm{v}}_{\mathrm{p}}\). If the bullet sticks to the sandbag then the ratio of the final and initial kinetic energy of the bullet is (approximately) :

1. 10^-2
2. 10^-3
3. 10^-6
4. 10^-4

Answer: 4. 10^-4

Question 176. There are a hundred identical sliders equally spaced on a frictionless track as shown in the figure. Initially, all the sliders are at rest. Slider 1 is pushed with velocity v towards slider 2. In a collision, the sliders stick together. The final velocity of the set of hundred stuck sliders will be :

1. \(\frac{v}{99}\)
2. \(\frac{v}{100}\)
3. Zero
4. v

Answer: 2. \(\frac{v}{100}\)

Question 177. The coefficient of restitution depends upon-

1. The masses of the colliding bodies
2. The direction of motion of the colliding bodies
3. The inclination between the colliding bodies
4. The materials of the colliding bodies

Answer: 4. The materials of the colliding bodies

Question 178. In an elastic collision of two particles, the following is conserved :

1. Momentum of each particle
2. The speed of each particle
3. The kinetic energy of each particle
4. The total kinetic energy of both the particles

Answer: 4. Total kinetic energy of both the particles

Question 179. A body of mass M₁ collides elastically with another mass M₂ at rest. There is maximum transfer of energy when :

1. M₁> M₂
2. M₁< M₂
3. M₁= M₂
4. Same for all values of M₁ and M₂

Answer: 3. M₁= M₂

Question 180. Two putty balls of equal mass moving with equal velocity in mutually perpendicular directions, stick together after collision. If the balls were initially moving with a velocity of \(45 \sqrt{2} \mathrm{~ms}^{-1}\) each, the velocity of their combined mass after a collision is :

1. \(45 \sqrt{2} \mathrm{~ms}^{-1}\)
2. 45 ms^-1
3. 90 ms^-1
4. \(22.5 \sqrt{2} \mathrm{~ms}^{-1}\)

Answer: 2. 45 ms^-1

Question 181. The coefficient of restitution e for a perfectly elastic collision is :

1. 1
2. 0
3. ∞
4. –1

Answer: 1. 1

Question 182. Two perfectly elastic particles P and Q of equal mass travelling along the joining them with velocities 15m/sec. and 10 m/sec. After the collision, their velocities respectively (in m/sec.) will be :

1. 0,25
2. 5,20
3. 10, 15
4. 20, 5

Answer: 3. 10, 15

Question 183. A particle of mass m moving with horizontal speed 6 m/sec. as shown in the figure. If m << M then for one-dimensional elastic collision, the speed of lighter particles after the collision will be :

1. 2 m/sec in original direction
2. 2 m/sec opposite to the original direction
3. 4 m/sec opposite to the original direction
4. 4 m/sec in original direction

Answer: 1. 2 m/sec in original direction

Question 184. A particle of mass m moving towards East with a velocity v collides with another particle of the same mass moving towards North with the same speed and adheres to it. The velocity of the combined particle is-

1. \(v / \sqrt{2}\) along North-East
2. \(v / \sqrt{2}\) along North-West
3. 2 \(\sqrt{2 v}\) along North-East
4. \(\sqrt{2 v}\) along North-West

Answer: 1. \(v / \sqrt{2}\) along North-East

Question 185. A particle of mass ‘m’ and velocity ‘ \(\) collides oblique elastically with a stationary particle of mass ‘m’. The angle between the velocity vectors of the two particles after the collision is :

1. 45°
2. 30°
3. 90°
4. None of these

Answer: 3. 90°

Question 186. An iron ball of mass 100gm moving at a speed of 10m/sec strikes a wall at an angle of 30° and reflects at the same angle. If the ball and wall remain in contact for 0.1 sec, the force exerted on the wall will be

1. 10 N
2. 100 N
3. 1.0 N
4. 0.1 N

Answer: 1. 10 N

Question 187. If a ball of mass 10 gm strikes perpendicular on a hard floor with a speed of 5 m/sec. and bounces with the same speed and remains in contact with the floor for 1 sec, then the force applied on the ball by the floor is-

1. 100 N
2. 10 N
3. 1.0 N
4. 0.1 N

Answer: 4. 0.1 N

Question 188. For inelastic collision between two spherical rigid bodies :

1. The total kinetic energy is conserved
2. The linear momentum is not conserved
3. The total mechanical energy is not conserved
4. The linear momentum is conserved

Answer: 4. The linear momentum is conserved

Question 189. A mass of 20 kg moving with a speed of 10 m/s collides with another stationary mass of 5 kg. As a result of the collision both masses stick together. The kinetic energy of the composite mass will be :

1. 600 J
2. 800 J
3. 1000 J
4. 1200 J

Answer: 2. 800 J

Question 190. When two bodies collide elastically, then :

1. The kinetic energy of the system alone is conserved
2. Only momentum is conserved
3. Both energy and momentum are conserved
4. Neither energy nor momentum is conserved

Answer: 3. Both energy and momentum are conserved

Question 191. A ball of mass ‘m’ moving with the velocity v collides head-on with another ball of mass m at rest, If the coefficient of restitution is e, then the ratio of the velocities of the first and second ball after the collision

1. \(\frac{1-e}{1+e}\)
2. \(\frac{1+e}{1-e}\)
3. \(\frac{1+e}{2}\)
4. \(\frac{1-e}{2}\)

Answer: 1. \(\frac{1-e}{1+e}\)

Question 192. When two bodies collide elastically, the force of interaction between them is :

1. Conservative
2. Non–conservative
3. Either conservative or non–conservative
4. Zero

Answer: 1. Conservative

Question 193. In an elastic collision in the absence of external force, which of the following is/are correct :

1. The linear momentum is not conserved
2. The potential energy is conserved in the collision
3. The final kinetic energy is less than the initial kinetic energy
4. The final kinetic energy is equal to the initial kinetic energy

Answer: 4. The final kinetic energy is equal to the initial kinetic energy

Question 194. A shell explodes in a region of negligible gravitational field, giving out n fragments of equal mass m. Then it’s total

1. Kinetic energy is smaller than that before the explosion
2. Kinetic energy is equal to the before the explosion
3. Momentum and kinetic energy depend on n
4. Momentum is equal to that before the explosion.

Answer: 4. Momentum is equal to that before the explosion.

Question 195. During the head-on collision of two masses 1 kg and 2 kg the maximum energy of deformation is 1003J. If before the collision the masses are moving in the opposite direction, then their velocity of approach before the collision is :

1. 10 m/sec.
2. 5 m/sec.
3. 20 m/sec.
4. \(10 \sqrt{2}\) m/sec.

Answer: 1. 10 m/sec.

Question 196. A block A of mass m moving with a velocity ‘ v ‘ along a frictionless horizontal track and blocks of mass m/2 moving with 2 v collide with plank elastically. The final speed of the block A is :

1. \(\frac{5 v}{3}\)
2. v
3. \(\frac{2 v}{3}\)
4. None of these

Answer: 2. v

Question 197. A particle of mass m moves with velocity v₀= 20 m/sec towards a wall that is moving with velocity v = 5 m/sec. If the particle collides with the wall elastically, the speed of the particle just after the collision is :

1. 30 m/s
2. 20 m/s
3. 25 m/s
4. 22 m/s

Answer: 1. 30 m/s

Question 198. A super-ball is to bounce elastically back and forth between two rigid walls at a distance d from each other. Neglecting gravity and assuming the velocity of the super-ball to be v0horizontally, the average force being exerted by the super-ball on one wall is :

1. \(\frac{1}{2} \frac{m v_0^2}{d}\)
2. \(\frac{\mathrm{mv}_0^2}{\mathrm{~d}}\)
3. \(\frac{2 m v_0^2}{d}\)
4. \(\frac{4 m v_0^2}{d}\)

Answer: 2. \(\frac{\mathrm{mv}_0^2}{\mathrm{~d}}\)

Question 199. Which of the following relation(s) is/are always correct? [ p = linear momentum ]

1. Thrust = \(u_{\text {rel }} \frac{d m}{d t}\)
2. F = m \(\mathrm{m} \frac{\mathrm{dp}}{\mathrm{dt}}\)
3. F = m \(m \frac{d v}{d t}\)
4. F = m \(m \frac{d v}{d t}+v \frac{d m}{d t}\)

Answer: 4. F = m \(m \frac{d v}{d t}+v \frac{d m}{d t}\)

Question 200. Statement-1: In an elastic collision between two bodies, the relative speed of the bodies after a collision is equal to the relative speed before the collision. because Statement 2: In an elastic collision, the linear momentum of the system is conserved

1. Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
2. Statement-1 is True,Statement-2 is True;Statement-2 is NOT a correct explanation for Statement-1
3. Statement-1 is True, Statement-2 is False
4. Statement-1 is False, and Statement-2 is True.

Answer: 2. Statement-1 is True,Statement-2 is True;Statement-2 is NOT a correct explanation for Statement-1

Question 201. A ball hits a floor and rebounds after an inelastic collision. In this case

1. The momentum of the ball just after the collision is the same as that just before the collision
2. The mechanical energy of the ball remains the same during the collision
3. The total momentum of the ball and the earth is conserved
4. The total energy of the ball and the earth remains the same

Answer: 3. The total momentum of the ball and the earth is conserved

Question 202. Two balls having mass of 2 kg and 3 kg are approaching each other with velocities of 3 m/s and 2 m/s respectively on the horizontal frictionless surface. They undergo a head-on elastic collision. Find out the maximum potential energy of deformation.

1. Zero
2. 12.5 J
3. 15 J
4. None of these

Answer: 3. 15 J

Question 203. A particle ‘A’ of mass m collides head-on with another stationary particle ‘B’ of the same mass ‘m’. The kinetic energy lost by the colliding particle ‘A’ will be maximum if the coefficient of the restitution is

1. 1
2. 0
3. 0.5
4. None

Answer: 1. 1

Question 204. Two particles A and B of masses 10 kg and 38 kg respectively are moving along the same straight line with velocities of 15 m/s and 3 m/s respectively in the same direction. After the elastic collision, the velocities of A and B are v_A and v_B in the direction of initial motion. Then :

1. v_A= 20, v_B= 8
2. v_A= − 4, v_B= 8
3. v_A= 16, v_B= 28
4. v_A= − 5, v_B= 10

Answer: 2. v_A= − 4, v_B= 8

NEET Physics Chapter 3 Centre of Mass: MCQs for Revision and Practice

Question 205. Two small spheres of equal mass, heading towards each other with equal speeds, undergo a head-on collision (no external force acts on the system of two spheres). Then which of the following statements is correct?

1. Their final velocities must be zero.
2. Their final velocities may be zero.
3. Each must have a final velocity equal to the other’s initial velocity.
4. Their velocities must be reduced in magnitude

Answer: 2. Their final velocities may be zero.

Question 206. In a perfectly inelastic direct collision, maximum transfer of energy takes place if-

1. m₁>> m₂
2. m₁<< m₂
3. m₁= m₂
4. m₂= 0

Answer: 3. m₁= m₂

Question 207. Which of the following statements is true for collisions-

1. Momentum is conserved in elastic collisions but not in inelastic collisions.
2. Total K.E. is conserved in elastic collisions but momentum is not conserved.
3. Total K.E. is not conserved in inelastic collisions but momentum is conserved.
4. Total K.E. and momentum both are conserved in all types of collisions.

Answer: 3. Total K.E. is not conserved in inelastic collisions but momentum is conserved.

Question 208. A body falls on a surface with a coefficient of restitution of 0.6 from a height of 1m. Then the body rebounds to a height of :

1. 0.6 m
2. 0.4 m
3. 1m
4. 0.36 m

Answer: 4. 0.36 m

Question 209. If the force on a rocket which is ejecting gases with a relative velocity of 300 m/s, is 210 N. Then the rate of combustion of the fuel will be :

1. 10.7 kg/sec
2. 0.07 kg/sec
3. 1.4 kg/sec
4. 0.7 kg/sec

Answer: 4. 0.7 kg/sec

Question 210. A belt is moving horizontally with a speed of 2m/s and sand is falling on it at the rate of 150 gm/sec. The additional force required to keep the speed of the belt is-

1. 0.015 N
2. 0.30 N
3. 3N
4. 300 N

Answer: 2. 0.30 N

Question 211. A rocket with a lift-off mass of 3.5 × 104 kg is blasted upwards with an initial acceleration of 10 m/s². The initial thrust of the blast is-

1. 14.0 × 105 N
2. 1.76 × 105 N
3. 3.5 × 105 N
4. 7.0 × 105 N

Answer: 4. 7.0 × 105 N

Question 212. Fuel is consumed at the rate of 100 kg/sec. in a rocket. The exhaust gases are ejected at a speed of 4.5 × 104 m/s. What is the thrust experience by the rocket-

1. 3 × 106 N
2. 4.5 × 106 N
3. 6 × 106 N
4. 9 × 106 N

Answer: 2. 4.5 × 106 N

Question 213. A rocket of initial mass 6000 kg. ejects mass at a constant rate of 16 kg/sec. with a constant relative speed of 11 km/sec. What is the acceleration of the rocket, a minute after the blast-(Consider acceleration due to gravity g = 10msec^-2)

1. 28.3 m/sec²
2. 42 m/sec²
3. 34.9 m/sec²
4. 24.92 m/sec²

Answer: 4. 24.92 m/sec²

Question 214. A 6000 kg rocket is set for vertical firing. If the exhaust speed is 1000 m/sec. How much gas must be ejected each second to supply the thrust needed to give the rocket an initial upward acceleration of 20 m/sec²- (consider g = 9.8 msec^-2 acceleration due to gravity )

1. 92.4 kg/sec
2. 178.8 kg/sec
3. 143.2 kg/sec
4. 47.2 kg/sec

Answer: 2. 178.8 kg/sec

Question 215. The rocket works on the principle of conservation of-

1. Energy
2. Angular momentum
3. Momentum
4. Mass

Answer: 3. Momentum

Question 216. A rocket with a lift-off mass of 3.5 × 104 kg is blasted upwards with an initial acceleration of 10 m/s². Then the initial thrust of the blast is :

1. 3.5 × 105 N
2. 7.0 × 105 N
3. 14.0 × 105 N
4. 1.75 × 105 N

Answer: 2. 7.0 × 105 N

Question 217. A balloon having a mass ‘ m ‘ is filled with gas and is held in the hands of a boy. Then suddenly it gets released and gas starts coming out of it at a constant rate. The velocities of the ejected gases are also constant at 2 m/s with respect to the balloon. Find out the velocity of the balloon when the mass of gas is reduced to half.

1. l n 2
2. 2 ln 4
3. 2 ln 2
4. None of these

Answer: 3. 2 in 2

Question 218. In the figure shown a hole of radius 2 cm is made in a semicircular disc of radius 6 π cm at a distance 8 cm from the centre C of the disc. The distance of the center of mass of this system from point C is:

1. 4 cm
2. 8 cm
3. 6 cm
4. 12 cm

Answer: 2. 8 cm

Question 219. A semicircular portion of radius ‘r’ is cut from a uniform rectangular plate as shown in the figure. The distance of the center of mass ‘C’ of the remaining plate, from point ‘O’ is :

1. \(\frac{2 r}{(3-\pi)}\)
2. \(\frac{3 r}{2(4-\pi)}\)
3. \(\frac{2 r}{(4+\pi)}\)
4. \(\frac{2 r}{3(4-\pi)}\)

Answer: 4. \(\frac{2 r}{3(4-\pi)}\)

Question 220. Look at the drawing given in the figure which has been drawn with ink of uniform line-thickness. The mass of ink used to draw each of the two inner circles, and each of the two line segments is m. The mass of the ink used to draw the outer circle is 6m. The coordinates of the centers of the different parts are outer circle (0, 0), left inner circle (–a, a), right inner circle (a, a), vertical line (0, 0), and horizontal line (0, –a). The y-coordinate of the center of mass of the ink in this drawing is

1. \(\frac{a}{10}\)
2. \(\frac{3 r}{2(4-\pi)}\)
3. \(\frac{2 r}{(4+\pi)}\)
4. \(\frac{2 r}{3(4-\pi)}\)

Answer: 1. \(\frac{a}{10}\)

Question 221. Two spherical bodies of mass M and 5M and radii R and 2R respectively are released in free space with initial separation between their centres equal to 12R. If they attract each other due to gravitational force only, then the distance covered by the smaller body just before the collision is :

1. 2.5R
2. 4.5 R
3. 7.5R
4. 1.5 R
5. Answer: 3. 7.5R

Question 222. A radioactive nucleus initially at rest decays by emitting an electron and neutrino at right angles to one another. The momentum of the electron is 3.2 × 10^-23 kg-m/sec. and that of the neutrino is 6.4 × 10^-23 kg-m/sec. The direction of the recoiling nucleus with that of the electron motion is-

1. tan^-1(0.5)
2. tan^-1
3. π – tan^-1
4. 2π+ tan^-1(2)

Answer: 3. π – tan^-1

Question 223. A particle of mass 1 kg is thrown vertically upwards with a speed 100 m/s. After 5s it explodes into two parts. One part of mass 400g comes back with a speed of 25 m/s, what is the speed of the other part just after the explosion?

1. 100 m/s upwards
2. 600 m/s upwards
3. 100 m/s downward
4. 300 m/s upward

Answer: 1. 100 m/s upwards

Question 224. A mass ‘m’ moves with a velocity ‘v’ and collides inelastically with another identical mass at rest. After collision the 1st mass moves with velocity \(\frac{v}{\sqrt{3}}\) in a direction perpendicular to the initial direction of motion. Find the speed of the 2nd mass after collision:

1. v
2. \(\sqrt{3 v}\)
3. \(\frac{2}{\sqrt{3}}\)
4. \(\frac{v}{\sqrt{3}}\)

Answer: 3. \(\frac{2}{\sqrt{3}}\)

Question 225. A mass of 10 gm, moving horizontally with a velocity of 100 cm/sec, strikes the bob of a pendulum and sticks to it. The mass of the bob is also 10 gm. The maximum height to which the system can be raised is (g = 10 m/sec²)

1. Zero
2. 5 cm
3. 2.5 cm
4. 1.25 cm

Answer: 4. 1.25 cm

Question 226. A solid iron ball A of radius r collides head-on with another stationary solid iron ball B of radius 2r. The ratio of their speeds just after the collision (e = 0.5) is :

1. 3
2. 4
3. 2
4. 1

Answer: 3. 2

NEET Physics Chapter 3 Centre of Mass: MCQs for Revision and Practice

Question 227. Two balls, having linear momenta \(\overrightarrow{\mathrm{p}}_1=p \hat{\mathrm{i}} \text { and } \overrightarrow{\mathrm{p}}_2=-p \hat{\mathrm{i}} \text {, }\) undergo a collision in free space. There is no external force acting on the balls. Let \(\overrightarrow{\mathrm{p}}_1^{\prime} \text { and } \overrightarrow{\mathrm{p}}_2^{\prime}\) be their final momenta. The following option(s) is(are) Not Allowed for any non-zero value of p, a₁, a₂, b₁, b₂, c₁and c₂.

1. \(\overrightarrow{\mathrm{p}}_1^{\prime}=\mathrm{a}_1 \hat{\mathrm{i}}+\mathrm{b}_1 \hat{\mathrm{j}}+\mathrm{c}_1 \hat{\mathrm{k}}, \overrightarrow{\mathrm{p}}_2^{\prime}=\mathrm{a}_2 \hat{\mathrm{i}}+\mathrm{b}_2 \hat{\mathrm{j}}\)
2. \(\overrightarrow{\mathrm{p}}_1^{\prime}=\mathrm{c}_1 \hat{\mathrm{k}}, \overrightarrow{\mathrm{p}}_2^{\prime}=c_2 \hat{\mathrm{k}}\)
3. \(\overrightarrow{\mathrm{p}}_1^{\prime}=\mathrm{a}_1 \hat{\mathrm{i}}+\mathrm{b}_1 \hat{\mathrm{j}}+\mathrm{c}_1 \hat{\mathrm{k}}, \overrightarrow{\mathrm{p}}_2^{\prime}=\mathrm{a}_2 \hat{\mathrm{i}}+\mathrm{b}_2 \hat{\mathrm{j}}-\mathrm{c}_1 \hat{\mathrm{k}}\)
4. \(\overrightarrow{\mathrm{p}}_1^{\prime}=\mathrm{a}_1 \hat{\mathrm{i}}+\mathrm{b}_1 \hat{\mathrm{j}}, \overrightarrow{\mathrm{p}}_2^{\prime}=\mathrm{c}_2 \hat{\mathrm{k}}\)

Answer: 1 and 4

Question 228. An explosion blows a rock into three parts. Two parts go off at right angles of each other. These two are 1 kg first part moving with a velocity of 12 ms–1 and 2 kg second part moving with a velocity of 8 ms–1. If the third part files off with a velocity of 4 ms–1, its mass would be :

1. 5 kg
2. 7 kg
3. 17 kg
4. 3 kg

Answer: 1. 5 kg

Question 229. A ball moving with a velocity 2 m/s collides head-on with another stationary ball of double the mass. If the coefficient of restitution is 0.5, then their velocities (in m/s) after the collision will be

1. 0, 1
2. 1, 1
3. 1, 0.5
4. 0, 2

Answer: 1. 0, 1

Question 230. Two particles which are initially at rest, move towards each other under the action of their internal attraction. If their speeds are v and 2v at any instant, then the speed of the center of mass of the system will be

1. 2v
2. Zero
3. 1.5 v
4. v

Answer: 2. Zero

Question 231. A mass of m moving horizontally (along the x-axis) with velocity v collides and sticks to the mass of 3m moving vertically upward (along the y-axis) with velocity 2v. The final velocity of the combination is :

1. \(\frac{1}{4} v \hat{i}+\frac{3}{2} v \hat{j}\)
2. \(\frac{1}{3} v \hat{i}+\frac{2}{3} v \hat{j}\)
3. \(\frac{2}{3} v \hat{i}+\frac{1}{3} v \hat{j}\)
4. \(\frac{3}{2} v \hat{i}+\frac{1}{4} v \hat{j}\)

Answer: 1. \(\frac{1}{4} v \hat{i}+\frac{3}{2} v \hat{j}\)

Question 232. Two persons of masses 55 kg and 65 kg respectively, are at the opposite ends of a boat. The length of the boat is 3.0 m and weighs 100 kg. The 55 kg man walks up to the 65 kg man and sits with him. If the boat is in still water the center of mass of the system shifts by :

1. 3.0 m
2. 2.3 m
3. Zero
4. 0.75 m

Answer: 3. Zero

Question 233. Two spheres A and B of masses m₁ and m₂ respectively collide. A is at rest initially and B is moving with velocity v along the x-axis. After collision B has a velocity \(\frac{v}{2}\) in a direction perpendicular to the original direction. The mass A moves after collision in the direction.

1. Same as that of B
2. Opposite to that of B
3. θ = tan^-1(1/2) to the x-axis
4. θ = tan^-1(–1/2) to the x-axis

Answer: 4. θ = tan^-1(–1/2) to the x-axis

Question 234. Three masses are placed on the x-axis: 300 g at origin, 500g at x = 40 cm and 400g at x = 70 cm. The distance of the centre of mass from the origin is :

1. 40 cm
2. 45 cm
3. 50 cm
4. 30 cm

Answer: 1. 40 cm

Question 235. An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to each other. The first part of mass 1kg moves with a speed of 12 ms^-1 and the second part of mass 2 kg moves with 8 ms^-1 speed. If the third part flies off with 4 ms^-1 speed, then its mass is :

1. 5 kg
2. 7 kg
3. 17 kg
4. 3 kg

Answer: 1. 5 kg

Question 236. A body of mass (4m) is lying in the x-y plane at rest. It suddenly explodes into three pieces. Two pieces, each of mass (m) move perpendicular to each other with equal speeds (υ). The total kinetic energy generated due to the explosion is :

1. mv
2. \(\frac{3}{2} m v^2\)
3. 2mv²
4. 4mv²

Answer: 2. \(\frac{3}{2} m v^2\)

Question 237. Two particles of masses m₁, m₂ move with initial velocities u₁ and u₂. On collision, one of the particles gets excited to a higher level, after absorbing energy ε If the final velocities of particles be v₁ and v₂ then we must have:

1. \(\frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2=\frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 v_2^2-\varepsilon\)
2. \(\frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2-\varepsilon=\frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 v_2^2\)
3. \(\frac{1}{2} m_1^2 u_1^2+\frac{1}{2} m_2^2 u_2^2+\varepsilon=\frac{1}{2} m_1^2 v_1^2+\frac{1}{2} m_2^2 v_2^2\)
4. \(m_1^2 u_1+m_2^2 u_2-\varepsilon=m_1^2 v_1+m_2^2 v_2\)

Answer: 2. \(\frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2-\varepsilon=\frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 v_2^2\)

Question 238. Two spherical bodies of mass M and 5 M and radii R and 2R are released in free space with initial separation between their centres equal to 12 R. If they attract each other due to gravitational force only, then the distance covered by the smaller body before a collision is:

1. 4.5 R
2. 7.5 R
3. 1.5 R
4. 2.5 R

Answer: 2. 7.5 R

Question 239. A rigid ball of mass m strikes a rigid wall at 60º and gets reflected without loss of speed as shown in the figure below. The value of impulse imparted by the wall in the ball will be

1. \(\frac{\mathrm{mV}}{3}\)
2. mV
3. 2mV
4. \(\frac{\mathrm{mV}}{2}\)

Answer: 2. mV

Question 240. A bullet of mass 10 g moving horizontally with a velocity of 400 ms–1 strike of a wooden block of mass 2 kg which is suspended by a light inextensible string of length 5 m. As a result, the center of gravity of the block is found to rise a vertical distance of 10 cm. The speed of the bullet after it emerges horizontally from the block will be

1. 160 ms^-1
2. 100 ms^-1
3. 80 ms^-1
4. 120 m^-1

Answer: 4. 120 m^-1

Question 241. Two identical balls A and B having velocities of 0.5 m/s and –0.3 m/s respectively collide elastically in one dimension. The velocities of B and A after the collision respectively will be

1. 0.3 m/s and 0.5 m/s
2. – 0.5 m/s and 0.3 m/s
3. 0.5 m/s and – 0.3 m/s
4. – 0.3 m/s and 0.5 m/s

Answer: 3. 0.5 m/s and – 0.3 m/s

Question 242. A moving block having a mass of m collides with another stationary block having a mass of 4m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of the coefficient of restitution (e) will be

1. 0.5
2. 0.4
3. 0.8
4. 0.25

Answer: 4. 0.25

Question 243. Body A of mass 4 m moving with speed u collides with another body B of mass 2 m at rest the collision is head-on and elastic in nature. After the collision, the fraction of energy lost by colliding body A is :

1. \(\frac{5}{9}\)
2. \(\frac{1}{9}\)
3. \(\frac{8}{9}\)
4. \(\frac{4}{9}\)

Answer: 3. \(\frac{8}{9}\)

Question 244. A particle of mass 5 m at rest suddenly breaks on its own into three fragments. Two fragments of mass m each move along a mutually perpendicular direction with speed v each. The energy released during the process is

1. \(\frac{3}{5} m v^2\)
2. \(\frac{5}{3} m v^2\)
3. \(\frac{3}{2} m v^2\)
4. \(\frac{4}{3} m v^2\)

Answer: 4. \(\frac{4}{3} m v^2\)

Question 245. Two small particles of equal masses start moving in opposite directions from point A in a horizontal circular orbit. Their tangential velocities are v and 2v, respectively, as shown in the figure. Between collisions, the particles move at constant speeds. After making how many elastic collisions, other than that at A, these two particles will again reach point A?

1. 4
2. 3
3. 2
4. 1

Answer: 3. 2

NEET Physics Chapter 3 Centre of Mass: MCQs for Revision and Practice

Question 246. Statement-1: Two particles moving in the same direction do not lose all their energy in a completely inelastic collision.

Statement 2: The principle of conservation of momentum holds true for all kinds of collisions.

1. Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.
2. Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1
3. Statement-1 is false, Statement-2 is true.
4. Statement-1 is true, Statement-2 is false.

Answer: 1. Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.

Question 247. A ball of mass 0.2 kg rests on a vertical post of height 5 m. A bullet of mass 0.01 kg, traveling with a velocity of V m/s in a horizontal direction, hits the center of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of 20 m and the bullet at a distance of 100 m from the foot of the post. The initial velocity V of the bullet is

1. 250 m/s
2. 250\(\sqrt{2}\)m/s
3. 400 m/s
4. 500 m/s

Answer: 4. 500 m/s

Question 248. This question has Statement 1 and Statement 2. Of the four choices given after the Statements, choose the one that best describes the two Statements.

Statement – 1: A point particle of mass m moving with speed υ collides with stationary point particle of mass M. If the maximum energy loss possible is given as\(f\left(\frac{1}{2} m v^2\right) \text { then } f=\left(\frac{m}{M+m}\right)\)

Statement 2: Maximum energy loss occurs when the particles get stuck together as a result of the collision.

1. Statement -1 is true, Statement -2 is true, and Statement -2 is the correct explanation of Statement -1.
2. Statement -1 is true, Statement – 2 is true, Statment-2 is not the correct explanation of Statement – 1.
3. Statement -1 is true, Statement 2 is false.
4. Statement -1 is false, and Statement – 2 is true.

Answer: 4. Statement -1 is false, Statement – 2 is true.

Question 249. A particle of mass m is projected from the ground with an initial speed u0at an angle α with the horizontal. At the highest point of its trajectory, it makes a completely inelastic collision with another identical particle, which was thrown vertically upward from the ground with the same initial speed u0. The angle that the composite system makes with the horizontal immediately after the collision is :

1. \(\frac{\pi}{4}\)
2. \(\frac{\pi}{4}+\mathrm{a}\)
3. \(\frac{\pi}{4}-\mathrm{a}\)
4. \(\frac{\pi}{4}\)

Answer: 1. \(\frac{\pi}{4}\)

Question 250. A tennis ball is dropped on a horizontal smooth surface. It bounces back to its original position after hitting the surface. The force on the ball during the collision is proportional to the length of compression of the ball. Which one of the following sketches describes the variation of its kinetic energy K with time t most appropriately? The figures are only illustrative and not to the scale.

Answer: 2.

Question 251. A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving in the y direction with speed v. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to :

1. 44%
2. 50%
3. 56%
4. 62%

Answer: 3. 56%

Question 252. The distance of the centre of mass of a solid uniform cone from its vertex is z0. If the radius of its base is R and its height is h then z0is equal to

1. \(\frac{h^2}{4 R}\)
2. \(\frac{3 h}{4}\)
3. \(\frac{5 h}{8}\)
4. \(\frac{3 h^2}{8 R}\)

Answer: 2. \(\frac{3 h}{4}\)

Question 253. It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is pd, while for its similar collision with carbon nucleus at rest, fractional loss of energy is pc. The values of pd and pc are respectively:

1. (0, 0)
2. (0, 1)
3. (.89, .28)
4. (.28, .89)

Answer: 3. (.89, .28)

Question 254. The mass of a hydrogen molecule is 3.32 × 10^-27 kg. If 10²³ hydrogen molecules strike, per second, a fixed wall of area 2cm² at an angle of 45° to the normal, and rebound elastically with a speed of 103 m/s, then the pressure on the wall is nearly :

1. 2.35 × 10² N/m²
2. 4.70 × 10² N/m²
3. 2.35 × 10³ N/m²
4. 4.70 × 10³ N/m²

Answer: 3. 2.35 × 10³ N/m²

Question 255. In a collinear collision, a particle with an initial speed v₀ strikes a stationary particle of the same mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after collision, is :

1. \(\frac{v_0}{2}\)
2. \(\frac{v_0}{\sqrt{2}}\)
3. \(\frac{v_0}{4}\)
4. \(\sqrt{2} v_0\)

Answer: 4. \(\sqrt{2} v_0\)

Question 256. Three blocks A, B and C are lying on a smooth horizontal surface, as shown in the figure. A and B have equal masses, m which C has mass M. Block A is given an initial speed of v towards B due to which it collides with B perfectly inelastically. The combined mass collides with C, also perfectly inelastically of the initial kinetic energy is also perfectly inelastically lost in the whole process. What is the value of M/m?

1. 2
2. 4
3. 5
4. 3

Answer: 2. 4

Question 257. A piece of wood of mass 0.03 kg is dropped from the top of a 100-height building. At the same time, a bullet of mass 0.02 kg is fired vertically upwards, with a velocity of 100 ms^-1 from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is : (g = 10 m/s^-2)

1. 10 m
2. 20 m
3. 30 m
4. 40 m

Answer: 4. 40 m

NEET Physics Class 11 Chapter 3 Centre of Mass MCQs and Answers

Question 258. A simple pendulum, made of a string of length  and a bob of mass m, is released from a small angle θ0. It strikes a block of mass M, kept on a horizontal surface at its lowest point of oscillations, elastically. It bounces back and goes up to an angle θ1. Then M is given by :

1. \(\mathrm{m}\left(\frac{\theta_0+\theta_1}{\theta_0-\theta_1}\right)\)
2. \(\frac{\mathrm{m}}{2}\left(\frac{\theta_0-\theta_1}{\theta_0+\theta_1}\right)\)
3. \(\frac{\mathrm{m}}{2}\left(\frac{\theta_0+\theta_1}{\theta_0-\theta_1}\right)\)
4. \(\mathrm{m}\left(\frac{\theta_0-\theta_1}{\theta_0+\theta_1}\right)\)

Answer: 4. \(\mathrm{m}\left(\frac{\theta_0-\theta_1}{\theta_0+\theta_1}\right)\)

Question 259. The position vector of the center of mass\(\overrightarrow{\mathrm{r}}[latex]cm of an asymmetric uniform bar of the negligible area of cross-section as shown in the figure is:

1. [latex]\vec{r}_{\mathrm{cm}}=\frac{3}{8} L \hat{x}+\frac{11}{8} L \hat{y}\)
2. \(\vec{r}_{c m}=\frac{11}{8} L \hat{x}+\frac{3}{8} L \hat{y}\)
3. \(\vec{r}_{\mathrm{cm}}=\frac{13}{8} L \hat{x}+\frac{5}{8} L \hat{y}\)
4. \(\vec{r}_{\mathrm{cm}}=\frac{5}{8} L \hat{x}+\frac{13}{8} L \hat{y}\)

Answer: 3. \(\vec{r}_{\mathrm{cm}}=\frac{13}{8} L \hat{x}+\frac{5}{8} L \hat{y}\)

Elasticity And Viscosity Solids

The materials having a definite shape and volume are known as solids. All solids have the property of elasticity by virtue of which solids behave as incompressible substances and exhibit rigidity and mechanical strength. Solids are classified into two categories namely Crystalline solids and amorphous solids (or glassy solids).

Crystalline Solids: A solid in which atoms or molecules are arranged in a regular three-dimensional pattern is known as a crystalline solid shown in figure (1) For example quartz, mica, sugar, copper sulphate, sodium chloride, potassium iodide, cesium chloride, carbon, etc.

Amorphous Solids: A solid in which atoms or molecules are not arranged in a regular manner is known as an amorphous solid shown in figure (2) For example: talc powder, glass, rubber, plastics, etc.

Elasticity and Viscosity Notes for NEET Physics Class 11

Unit Cell And Crystal Lattice

Unit cell is the building block of a crystal. It is defined as the smallest pattern of atoms in a lattice, the repetition of which in three dimensions forms a crystal lattice.

Crystal lattice: It is defined as a regular arrangement of a large number of points in space, each point representing the position of an atom or a group of atoms in a crystal. The crystal lattice is shown in the figure.

Elasticity And Plasticity

The property of a material body by virtue of which it regains its original configuration (i.e. shape and size) when the external deforming force is removed is called elasticity.

• The property of the material body by virtue of which it does not regain its original configuration when the external force is removed is called plasticity.

Deforming Force: An external force applied to a body that changes its size or shape or both is called a deforming force.

Perfectly Elastic Body: A body is said to be perfectly elastic if it completely regains its original form when the deforming force is removed. Since no material can regain completely its original form the concept of a perfectly elastic body is only an ideal concept. A quartz fiber is the nearest approach to the perfectly elastic body.

Perfectly Plastic Body: A body is said to be perfectly plastic if it does not regain its original form even slightly when the deforming force is removed.

• Since every material partially regains its original form on the removal of the deforming force, the concept of a perfectly plastic body is only an ideal concept. Paraffin wax and wet clay are the nearest approaches to perfectly plastic bodies.

Cause of Elasticity: In a solid, atoms and molecules are arranged in such a way that each atom/molecule is acted upon by the forces due to the neighboring atom/molecules. These forces are known as intermolecular forces.

• When no deforming force is applied to the body, each atom/molecule of the solid (i.e. body) is in its equilibrium position and the intermolecular forces between the molecules of the solid are zero.
• On applying the deforming force on the body, the molecules either come closer or go far apart from each other. As a result of this, the atoms/molecules are atoms displaced from their equilibrium position.
• In other words, equilibrium intermolecular forces get disturbed or changed, and restoring forces are developed on the molecules. When the deforming force is removed, these restoring forces bring the molecules of the solid to their respective equilibrium positions and hence the solid (or the body) regains its original form.

Stress

When deforming force is applied to the body then the upto a certain limit equal restoring force in the opposite direction is developed inside the body. The restoring forces per unit area developed in the body is called stress.

stress = \(=\frac{\text { restoring force }}{\text { Area of the body }}=\frac{F}{A}\)

The unit of stress is N/m² or Nm^-2. There are three types of stress

1. Longitudinal or Normal stress: When an object is one dimensional or linear then force acting per unit area is called longitudinal stress. It is of two types :

1. Compressive stress
2. Tensile stress

Longitudinal or Normal stress Examples:

Consider a block of solid as shown in the figure. Let a force F be applied to the face which has area A. Resolve \(\vec{F}\) into two components:

F_n= F sin θ is called normal force and F_t= F cos θ called tangential force.

∴ Normal (tensile) stress = \(\frac{F_{\mathrm{n}}}{\mathrm{A}}=\frac{\mathrm{F} \sin \theta}{\mathrm{A}}\)

2. Tangential or shear stress: It is defined as the restoring force acting per unit area tangential to the surface of the body.

Tangential (shear) stress = \(\frac{F_t}{A}=\frac{F \cos \theta}{A}\)

The effect of stress is to produce distortion or a change in size, volume, and shape
(i.e. configuration of the body).

3. Bulk stress or All-around stress or Pressure: When force F is acting all along the surface normal (ΔA) to the area, then force acting per unit area is known as pressure. The effect of pressure is to produce volume change. The shape of the body may or may not change depending upon the homogeneity of the body.

Question1. Find out longitudinal stress and tangential stress on a fixed block

NEET Physics Chapter 4 Elasticity and Viscosity Study Notes

Answer:

Longitudinal or normal stress ⇒ \(\sigma_{\mathrm{n}}=\frac{100 \sin 30^{\circ}}{5 \times 2}=5 \mathrm{~N} / \mathrm{m}^2\)

Tangential stress ⇒ \(\sigma_{\mathrm{t}}=\frac{100 \cos 30^{\circ}}{5 \times 2}=5 \sqrt{3} \mathrm{~N} / \mathrm{m}^2\)

Question 2. Find out Bulk stress on the spherical object of radius\(\frac{10}{\pi}\) cm if the area and mass of the piston are 50 cm² and 50 kg respectively for a cylinder filled with gas. 

Answer:

⇒ \(p_{\text {gas }}=\frac{m g}{A}+p_a=\frac{50 \times 10}{50 \times 10^{-4}}+1 \times 10^5\)

⇒ \(2 \times 10^5 \mathrm{~N} / \mathrm{m}^2\)

Bulk stress = p_gas= 2 × 10⁵ N/m²

Strain

The ratio of the change in configuration (i.e. shape, length, or volume) to the original configuration of the body is called strain

i.e. Strain,\(\epsilon=\frac{\text { change in configuration }}{\text { original configuration }}\)

Types Of Strain: There are three types of strain

Longitudinal Strain: This type of strain is produced when the deforming force causes a change in the length of the body. It is defined as the ratio of the change in length to the original length of the body.

Consider A Wire Of Length L: When the wire is stretched by a force F, then let the change in length of the wire be ΔL as shown in the Figure.

∴ Longitudinal strain, \(\epsilon_{\ell}=\frac{\text { change in length }}{\text { original length }}\)

or Longitudinal strain = \(\frac{\Delta \mathrm{L}}{\mathrm{L}}\)

Volume Strain: This type of strain is produced when the deforming force produces a change in the volume of the body as shown in the Figure. It is defined as the ratio of the change in volume to the original volume of the body.

If the upper face is displaced through x keeping the lower face fixed at a distance of l then

If ΔV = change in volume

V = original volume

Δ tangentialdisplacement

⇒ \(\epsilon_{\mathrm{v}}=\text { volume strain }\)

⇒ \(\frac{\Delta \mathrm{V}}{\mathrm{V}}=\frac{\text { tangential displacement }}{\text { transverse distance }}\)

Shear Strain: This type of strain is produced when the deforming force causes a change in the shape of the body. It is defined as the angle (θ) through which a face originally perpendicular to the fixed face is turned as shown in the Figure.

⇒ \(\tan \phi \text { or } \phi=\frac{\mathrm{x}}{\ell}\)

Hooke’s Law And Modulus Of Elasticity

According to this law, within the elastic limit, stress is proportional to the strain.

i.e. stress ∝ strain

or stress = constant × strain or \(\frac{\text { stress }}{\text { strain }}=\text { Modulus of Elasticity. }\)

This Constant Is Called The Modulus Of Elasticity.

Thus, the modulus of elasticity is defined as the ratio of the stress to the strain.

The modulus of elasticity depends on the nature of the material of the body and is independent of its dimensions (i.e. length, volume, etc.).

Unit: The Sl unit of modulus of elasticity is Nm^-2 or Pascal (Pa).

Types Of Modulus Of Elasticity

Corresponding to the three types of strain there are three types of modulus of elasticity.

1. Young’s modulus of elasticity (Y)
2. Bulk modulus of elasticity (B)
3. Modulus of rigidity (η).

Young’s modulus of elasticity

It is defined as the ratio of the normal stress to the longitudinal strain.

i.e. Young’s modulus (Y) = \(\frac{\text { Longitudinal stress }}{\text { Longitudinal strain }}\)

Normal stress = F/A,

Longitudinal strain = ΔL/L

Y = \(Y=\frac{F / A}{\Delta L / L}=\frac{F L}{A \Delta L}\)

Question 1. Find out the shift in points B, C, and D in the compound wire shown in Figure.

Answer:

⇒ \(\Delta L_B=\Delta L_{A B}=\frac{F L}{A Y}=\frac{M g L}{A Y}\)

⇒ \(\frac{10 \times 10 \times 0.1}{10^{-7} \times 2.5 \times 10^{10}}=4 \times 10^{-3} \mathrm{~m}=4 \mathrm{~mm}\)

⇒ \(\Delta \mathrm{L}_{\mathrm{c}}=\Delta \mathrm{L}_{\mathrm{B}}+\Delta \mathrm{L}_{\mathrm{BC}}=4 \times 10^{-3}+\frac{100 \times 0.2}{10^{-7} \times 4 \times 10^{10}}\)

⇒ \(4 \times 10^{-3}+5 \times 10^{-3}=9 \mathrm{~mm}\)

⇒ \(\Delta \mathrm{L}_{\mathrm{D}}=\Delta \mathrm{L}_{\mathrm{c}}+\Delta \mathrm{L}_{\mathrm{CD}}=9 \times 10^{-3}+\frac{100 \times 0.15}{10^{-7} \times 1 \times 10^{10}}\)

⇒ \(9 \times 10^{-3}+15 \times 10^{-3}=24 \mathrm{~mm}\)

Elongation Of Rod Under Its Self Weight

Let rod has a self-weight ‘W’, area of cross-section ‘A” and length ‘L’. Considering on element of length dx at a distance ‘x’ from the bottom. then \(\mathrm{T}=\frac{\mathrm{W}}{\mathrm{L}} \mathrm{x}\) is down word force on dx elongation in ‘dx’ element = \(\frac{\text { T.dx }}{\mathrm{AY}}\)

Total elongation s = \(\int_0^L \frac{T d x}{A Y}=\int_0^L \frac{W x d x}{L A Y}=\frac{W L}{2 A Y}\)

Note: One can do this directly by considering total weight at C.M. and using effective length L/2.

Class 11 NEET Physics Elasticity and Viscosity Notes

Question 1. Find out the elongation in the block shown in Figure. If mass, area of cross-section, and Young modulus of a block are m, A, and Y respectively.

Answer:

Acceleration, a = \(\frac{T d x}{A Y}\) then T = m′a

where ⇒ m′ = \(\frac{m}{L} x\)

T = \(\frac{m}{L} \times \frac{F}{m}=\frac{F x}{L}\)

Elongation in element ‘dx’ = \(\frac{T d x}{A Y}\) total elongation, δ = \(\int_0^L \frac{T d x}{A Y}=\int_0^L \frac{F x d x}{A L Y}=\frac{F L}{2 A Y}\)

Note: Try this problem, if friction is given between block and surface (µ = friction coefficient), and Case :

F < µmg

F > µmg

In both cases answer will be \(\frac{F L}{2 A Y}\)

Bulk Modulus:

It is defined as the ratio of the normal stress to the volume strain i.e. B = \(\frac{\text { Pressure }}{\text { Volume strain }}\)

The stress is the normal force applied per unit area and is equal to the pressure applied (p). p

⇒ \(B=\frac{p}{\frac{-\Delta V}{V}}=-\frac{p V}{\Delta V}\)

A negative sign shows that an increase in pressure (p) causes a decrease in volume (ΔV).

Compressibility: The reciprocal of bulk modulus of elasticity is called compressibility K. The Unit of compressibility in Sl is N^-1 m² or pascal^-1(Pa^-1).

The bulk modulus of solids is about fifty times that of liquids, and for gases it is 10–8 times of solids.

B_Solids > B_liquids > B_gases

Isothermal modulus of elasticity of gas B = P (pressure of gas)

Adiabatic modulus of elasticity of gas B = γ × P where γ = \(\) = ratio of specific beats.

Modulus of Rigidity:

It is defined as the ratio of the tangential stress to the shear strain. Let us consider a cube whose lower face is fixed and a tangential force F acts on the upper face whose area is A.

∴ Tangential stress = F/A.

Let the vertical sides of the cube shift through an angle θ, called shear strain

∴ The modulus of rigidity is given by

η = \(\frac{\text { Tangential stress }}{\text { Shear strain }}\)

or \(\eta=\frac{F / A}{\phi}=\frac{F}{A \phi}\)

Question 2. A rubber cube of side 5 cm has one side fixed while a tangential force equal to 1800 N is applied to the opposite face to find the shearing strain and the lateral displacement of the strained face. The modulus of rigidity for rubber is 2.4 × 10⁶ N/m².
Answer:

L = 5 × 10^-2 m ⇒ \(\frac{F}{A}=\eta \frac{x}{L}\)

Strain θ = \(\frac{F}{A \eta}\)

⇒ \(\frac{1800}{25 \times 10^{-4} \times 2.4 \times 10^6}\)

⇒ \(\frac{180}{25 \times 24}\)

⇒ \(\frac{3}{10}\)

=0.3 radian

⇒ \(\frac{x}{L}=0.3\)

x = 0.3 × 5 × 10^-2

= 1.5 × 10^-2 m

= 1.5 mm

NEET Physics Chapter 4 Elasticity and Viscosity Notes and Solutions

Variation Of Strain With Stress

When a wire is stretched by a load, it is seen that for a small value of the load, the extension produced in the wire is proportional to the load. On removing the load, the wire returns to its original length.

• The wire regains its original dimensions only when the load applied is less or equal to a certain limit. This limit is called the elastic limit.
• Thus, the elastic limit is the maximum stress on whose removal, the bodies regain their original dimensions.
• In the shown figure, this type of behavior is represented by the OB portion of the graph. Till A the stress is proportional to strain and from A to B if deforming forces are removed then the wire comes to its original length but here stress is not proportional to strain.

OA → Limit of Proportionality

OB → Elastic limit

C → Yield Point

CD → Plastic behavior

D → Ultimate point

DE → Fracture

As we go beyond point B, then even for a very small increase in stress, the strain produced is very large. This type of behaviour is observed around point C and at this stage the wire begins to flow like a viscous fluid.

• The point C is called yield point. If the stress is further increased, then the wire breaks off at point D called the breaking point.
• The stress corresponding to this point is called breaking stress or tensile strength of the material of the wire. A material for which the plastic range CD is relatively high is called ductile material.
• These materials get permanently deformed before breaking. The materials for which the plastic range is relatively small are called brittle materials. These materials break as soon as the elastic limit is crossed.

Important Points

• Breaking stress = Breaking force/area of cross-section.
• Breaking stress is constant for a material
• Breaking force depends upon the area of the section of the wire of a given material.
• The working stress is always kept lower than that of breaking stress so the safety factor = breaking stress/working stress may have a large value.
• Breaking strain = elongation or compression/original dimension.
• Breaking strain is constant for the material.

Elastic After Effect

We know that some material bodies take some time to regain their original configuration when the deforming force is removed. The delay in regaining the original configuration by the bodies on the removal of deforming force is called elastic after effect.

• The elastic after-effect is negligibly small for quartz fiber and phosphor bronze. For this reason, the suspensions made from quartz and phosphor bronze are used in galvanometers and electrometers.
• For glass fibre elastic after effect is very large. It takes hours for glass fiber to return to its original state on removal of a deforming force.

Elastic Fatigue

The loss of strength of the material due to repeated strains on the material is called elastic fatigue. That is why bridges are declared unsafe after a long time of their use.

Analogy Of Rod As A Spring

⇒ \(\mathrm{Y}=\frac{\text { stress }}{\text { strain }}\)

⇒ \(\mathrm{Y}=\frac{\mathrm{F} \ell}{\mathrm{A} \Delta \ell}\)

or \(\mathrm{F}=\frac{\mathrm{AY}}{\ell} \Delta \ell\)

⇒ \(\frac{\mathrm{AY}}{\ell}\) = constant, depending on the type of material and geometry of the rod. F = kΔl

where k = \(\frac{\mathrm{AY}}{\ell}\) = equivalent spring constant.

for the system of rods shown in the figure

1. The replaced spring system is shown in the figure
2. Two spring in series]. Figure
3. Represents an equivalent spring system.

Figure (4) represents another combination of rods and their replaced spring system.

Question 1. A mass ‘m’ is attached with rods as shown in the figure. This mass is slightly stretched and released whether the motion of the mass is S.H.M., if yes then find out the time period.

Answer:

⇒ \(k_{e q}=\frac{k_1 k_2}{k_1+k_2}\)

where \(\mathrm{k}_1=\frac{\mathrm{A}_1 \mathrm{Y}_1}{\ell_1}\)

⇒ \(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}_{\mathrm{eq}}}}\)

and \(\mathrm{k}_2=\frac{\mathrm{A}_2 \mathrm{Y}_2}{\ell_2}\)

Elasticity and Viscosity NEET Physics Class 11: Key Concepts

Elastic Potential Energy Stored In A Stretched Wire Or In A Rod

Strain energy stored in equivalent spring

U = \(\frac{1}{2} k x^2\)

where x = \(\frac{\mathrm{F} \ell}{\mathrm{AY}}\)

k = \(\frac{\mathrm{AY}}{\ell}\)

U = \(\frac{1}{2} \frac{A Y}{\ell} \frac{F^2 \ell^2}{A^2 Y^2}\)

⇒ \(\frac{1}{2} \frac{F^2 \ell}{A Y}\)

equation can be re-arranged

⇒ \(\mathrm{U}=\frac{1}{2} \frac{\mathrm{F}^2}{\mathrm{~A}^2} \times \frac{\ell \mathrm{A}}{\mathrm{Y}}\) [lA = volume of rod, F/A = stress]

⇒ \(\mathrm{U}=\frac{1}{2} \frac{(\text { stress) })^2}{\mathrm{Y}} \times \text { volume }\) again, U = \(\frac{1}{2} \frac{\mathrm{F}}{\mathrm{A}} \times \frac{\mathrm{F}}{\mathrm{AY}} \times \mathrm{A} \ell\)

[Stain = \(\frac{\mathrm{F}}{\mathrm{AY}}\)]

⇒ \(\mathrm{U}=\frac{1}{2} \text { stress } \times \text { strain } \times \text { volume }\) again, U = \(\frac{1}{2} \frac{F^2}{A^2 Y^2} A \ell Y\)

⇒ \(U=\frac{1}{2} Y(\text { strain })^2 \times \text { volume }\)

strain energy density ⇒ \(\frac{\text { strain energy }}{\text { volume }}\)

⇒ \(\frac{1}{2} \frac{\text { (stress) }^2}{Y}\)

⇒ \(\frac{1}{2} Y(\text { strain })^2\)

⇒ \(\frac{1}{2} \text { stress } \times \text { strain }\)

Posson’s Ratio (σ)

Within the elastic limit, the ratio between the lateral strain and the linear strain is a constant. This constant is called Poisson’s ratio.

σ = \(\frac{\text { lateral strain }}{\text { longitudinal strain }}\)

Lateral Strain: Change in length of the body from its initial length perpendicular to deforming force lateral strain = \(-\frac{\Delta D}{D}=\frac{D-(D-\Delta D)}{D}=\beta\)

Longitudinal Strain: Change in length of the body from its initial length in the direction of deforming force

Longitudinal strain \(\varepsilon_{\ell}=\frac{\text { change in length }}{\text { original length }}=\frac{\Delta \mathrm{L}}{\mathrm{L}}=\alpha\)

⇒ \(\sigma=\frac{\beta}{\alpha}\)

theoretical limit of σ –1 < σ < 0.5

The experimental value of σ lies between 0.2 and 0.4

Four important Relations between Y, B, η and σ

1. \(\eta=\frac{Y}{2(1+\sigma)}\)
2. \(\frac{9}{Y}=\frac{3}{\eta}+\frac{1}{B}\)
3. \(\sigma=\frac{3 B-2 \eta}{6 B+2 \eta}\)
4. \(B=\frac{Y}{3(1-2 \sigma)}\)

Thermal Stress :

If the temperature of the rod is increased by ΔT, then the change in length

Δl = l α ΔT strain = \(\frac{\text { stress }}{\text { strain }}\)

But due to rigid support, there is no strain. Supports provide force on stresses to keep the length of the rod the same

Y = \(\frac{\text { stress }}{\text { strain }}\)

thermal stress = Y strain = Y α ΔT

⇒ \(\frac{F}{A}=Y \alpha \Delta T\)

F = AY α ΔT

Applications Of Elasticity

Some of the important applications of the elasticity of the materials are discussed below:

1. The material used in bridges loses its elastic strength with time, and bridges are declared unsafe after long use.
2. To estimate the maximum height of a mountain :

The pressure at the base of the mountain = hρg = stress. The elastic limit of a typical rock is 3 × 10⁸ N m^-2

The stress must be less than the elastic limits, otherwise the rock begins to flow.

⇒ \(h<\frac{3 \times 10^8}{\rho g}<\frac{3 \times 10^8}{3 \times 10^3 \times 10}<10^4 \mathrm{~m}\)

∵ \(\left(\rho=3 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3} ; g=10 \mathrm{~ms}^{-2}\right)\)

h = 10km

It may be noted that the height of Mount Everest is nearly 9 km.

Torsion Constant Of A Wire

C = \(\frac{\pi \eta r^4}{2 \ell}\) Where η is the modulus of rigidity r and l are the radius and length of the wire respectively.

1. Toque required for twisting by angle θ, τ = Cθ.
2. Work done in twisting by angle θ, W = \(\frac{1}{2} C \theta^2\)

NEET Physics Chapter 4 Elasticity and Viscosity Notes and Formulas

Viscosity

When a solid body slides over another solid body, a frictional force begins to act between them. This force opposes the relative motion of the bodies.

• Similarly, when a layer of a liquid slides over another layer of the same liquid, a frictional force acts between them which opposes the relative motion between the layers.
• This force is called ‘internal frictional force’. This is due to intermolecular forces. Suppose a liquid is flowing in streamlined motion on a fixed horizontal surface AB.
• The layer of the liquid that is in contact with the surface is at rest, while the velocity of other layers increases with distance from the fixed surface. In the Figure, the lengths of the arrows represent the increasing velocity of the layers.
• Thus there is a relative motion between adjacent layers of the liquid. Let us consider three parallel layers a, b, and c. Their velocities are in increasing order. The layer a tends to retard the layer b, while b tends to retard c.
• Thus each layer tends to decrease the velocity of the layer above it. Similarly, each layer tends to increase the velocity of the layer below it.
• This means that in between any two layers of the liquid, internal tangential forces act which try to destroy the relative motion between the layers. These forces are called ‘viscous forces’.
• If the flow of the liquid is to be maintained, an external force must be applied to overcome the dragging viscous forces. In the absence of the external force, the viscous forces would soon bring the liquid to rest.
• The property of the liquid by virtue of which it opposes the relative motion between its adjacent layers is known as ‘viscosity’.

The property of viscosity is seen in the following examples :

A stirred liquid, when left, comes to rest on account of viscosity. Thicker liquids like honey, coaltar, glycerine, etc. have a larger viscosity than thinner ones like water.

• If we pour coaltar and water on a table, the coaltar will stop flowing soon while the water will flow upto quite a large distance.
• If we pour water and honey in separate funnels, water comes out readily from the hole in the funnel while honey takes enough time to do so.
• This is because honey is much more viscous than water. As honey tends to flow down under gravity, the relative motion between its layers is opposed strongly.
• We can walk fast in the air, but not in water. The reason is again viscosity which is very small for air but comparatively much larger for water.
• The cloud particles fall down very slowly because of the viscosity of air and hence appear floating in the sky.
• Viscosity comes into play only when there is a relative motion between the layers of the same material. This is why it does not act in solids.

Flow Of Liquid In A Tube Critical Velocity

When a liquid flows ‘in a tube, the viscous forces oppose the flow of the liquid, Hence a pressure difference is applied between the ends of the tube which maintains the flow of the liquid.

• If all particles of the liquid passing through a particular point in the tube move along the same path, the flow” of the liquid is called ‘stream-lined flow’.
• This occurs only when the velocity of flow of the liquid is below a certain limiting value called ‘critical velocity’. When the velocity of flow exceeds the critical velocity, the flow is no longer stream-lined but becomes turbulent.
• In this type of flow, the motion of the liquid becomes zig-zag, and eddy-currents are developed in it as shown in Figure.

• Reynold observed that the critical velocity for a liquid flowing in a tube is vc= Rη/ρa. where ρ is density η is the viscosity of the liquid, a is the radius of the tube and R is ‘Reynold’s number’ (whose value for a narrow tube and for water is about 1000).
• When the velocity of the flow of the liquid is less than the critical velocity, then the flow of the liquid is controlled by the viscosity, the density having no effect on it.
• But when the velocity of flow is larger than the critical velocity, then the flow is mainly governed by the density, and the effect of viscosity becomes less important.
• It is because of this reason that when a volcano erupts, then the lava coming out of it flows speedily in spite of being very thick (of large viscosity).

NEET Physics Chapter 4 Elasticity and Viscosity Notes and Solutions

Velocity Gradient And Coefficient Of Viscosity

The property of a liquid by virtue of which an opposing force (internal friction) comes into play whenever there is a relative motion between the different layers of the liquid is called viscosity.

Consider a flow of a liquid over the horizontal solid surface as shown in Figure. Let us consider two layers AB and CD moving with velocities \(\vec{v}\)and \([\vec{v}+d \vec{v}/latex] a distance z and (z + dz) respectively from the fixed solid surface.

According to Newton, the viscous drag or backward force (F) between these layers happens to be.

1. Directly proportional to the area (A) of the layer and
2. Directly proportional to the velocity gradient [latex]\left(\frac{d v}{d x}\right)\) between the layers. Defined as a change in velocity per unit perpendicular separation between the layers. Thus

⇒ \(F \propto A \frac{d v}{d z} \text { or } F=-\eta A \frac{d v}{d z}\)……..(1)

η is called the Coefficient of viscosity. A negative sign shows that the direction of viscous drag (F) is just opposite to the direction of the motion of the liquid.

Similarities And Differences Between Viscosity And Solid Friction

Similarities: Viscosity and solid friction are similar as

1. Both oppose the relative motion. Whereas viscosity opposes the relative motion between two adjacent liquid layers, solid friction opposes the relative motion between two solid layers.
2. Both come into play, whenever there is relative motion between layers of liquid or solid surfaces as the case may be.
3. Both are due to molecular attractions.

Differences between them are given below

Some Applications Of Viscosity

Knowledge of the viscosity of various liquids and gases has been put to use in daily life. Some applications of its knowledge are discussed as under:

1. As the viscosity of liquids varies with temperature, the proper choice of lubricant is made depending on season.
2. Liquids of high viscosity are used in shock absorbers and buffers at railway stations.
3. The phenomenon of viscosity of air and liquid is used to dampen the motion of some instruments.
4. The knowledge of the coefficient of viscosity of organic liquids is used in determining the molecular weight and shape of the organic molecules.
5. It finds an important use in the circulation of blood through arteries and veins of the human body.

Units Of Coefficient Of Viscosity

From the above formula, we have \(\eta=\frac{F}{A\left(\Delta v_x / \Delta z\right)}\)

∴ dimensions of \(\eta=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{L}^2\right]\left[\mathrm{LT}^{-1} / \mathrm{L}\right]}=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{L}^2 \mathrm{~T}^{-1}\right]}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right]\)

Its unit is kg/(meter-second)

In the C.G.S. system, the unit of coefficient of viscosity is dyne s cm–2 and is called poise. In SI the unit of coefficient of viscosity is N sm–2 and is called decompose.

1 decapoise = 1 N sm^-2 = (10⁵ dyne) × s × (10² cm)^-2 = 10 dyne s cm^-2 = 10 poise

Question 1. A man is rowing a boat with a constant velocity ‘v₀’ in a river the contact area of the boat is ‘A’ and the coefficient of viscosity is η. The depth of the river is ‘D’. Find the force required to row the boat.
Answer:

F – F_T= m a_res

As the boat moves with constant velocity a_res = 0

F = F_T

But \(F_T=\eta A \frac{d v}{d z}\)

but \(\frac{d v}{d z}=\frac{v_0-0}{D}=\frac{v_0}{D}\)

then F = F_T = \(\frac{\eta A v_0}{D}\)

Viscosity and Elasticity in NEET Physics: Notes and Applications

Question 2. A cubical block (of side 2m) of mass 20 kg slides on an inclined plane lubricated with the oil of viscosity η = 10^-1 poise with a constant velocity of 10 m/sec. (g = 10 m/sec²) find out the thickness of a layer of liquid.

Answer:

F = F’\(\eta A \frac{d v}{d z}=m g \sin \theta ; \frac{d v}{d z}=\frac{v}{h}\)

∴ 20 × 10 × sin 30° = η × 4 × \(\frac{10}{h}\)

⇒ \(\frac{40 \times 10^{-2}}{100}\) – [η = 10^-1 poise = 10^-2 N-sec-m^-2] = 4 × 10^-3 m = 4 mm

Effect Of Temperature On The Viscosity

The viscosity of liquids decreases with an increase in temperature and increases with a decrease in temperature. That is, \(\eta \propto \frac{1}{\sqrt{T}}\) T. On the other hand, the value of viscosity of gases increases with the increase in temperature and vice-versa. That is, η ∝ T.

Stoke’S Law

Stokes proved that the viscous drag (F) on a spherical body of radius r moving with velocity v in a fluid of viscosity η is given by F = 6 π η r v. This is called Stokes’ law.

Terminal Velocity

When a body is dropped in a viscous fluid, it is first accelerated. As its velocity increases, viscous force also increases so ultimately its acceleration becomes zero and it attains a constant velocity called terminal velocity.

Calculation Of Terminal Velocity

Let us consider a small ball, whose radius is r and density is ρ, is falling freely in a liquid (or gas), whose density is σ and coefficient of viscosity η. When it attains a terminal velocity v. It is subjected to two forces:

1. Effective weight force acting downward

⇒ \(V(\rho-\sigma) g=\frac{4}{3} \pi r^3(\rho-\sigma) g\)

2. A viscous force acting upward

= 6 π η rv.

Since the ball is moving with a constant velocity v i.e., there is no acceleration in it, the net force acting on it must be zero. That is

6πηrv = \(\frac{4}{3} p r^3(\rho-\sigma) g\)

or \(v=\frac{2}{9} \frac{r^2(\rho-\sigma) g}{\eta}\)

Thus, the terminal velocity of the ball is directly proportional to the square of its radius Important point

The air bubble in water always goes up. This is because the density of air (ρ) is less than the density of water (σ).

So the terminal velocity for the air bubble is Negative, which implies that the air bubble will go up. Positive terminal velocity means the body will fall down.

Question 1. A spherical ball is moving with terminal velocity inside a liquid. Determine the relationship between the rate of heat loss with the radius of the ball.
Answer:

Rate of heat loss = power = F × v = 6 π η r v × v = 6 π η r v2 = 6p η r \(\left[\frac{2}{9} \frac{g r^2\left(\rho_0-\rho_{\ell}\right)}{\eta}\right]^2\)

Rate of heat loss α r⁵

Question 2. A drop of water of radius 0.0015 mm is falling in the air. If the coefficient of viscosity of air is 1.8 × 10^-5 kg /(m-s), what will be the terminal velocity of the drop? (density of water = 1.0 × 10³ kg/m² and g = 9.8 N/kg.) The density of air can be neglected.
Answer:

By Stoke’s law, the terminal velocity of a water drop of radius r is given by 2

⇒ \(v=\frac{2}{9} \frac{r^2(\rho-\sigma) g}{\eta}\)

where ρ is the density of water, σ is the density of air, and η the coefficient of viscosity of air. Here σ is negligible and r = 0.0015 mm = 1.5 × 10^-3 mm = 1.5 × 10^-6 m. Substituting the values:

⇒ \(v=\frac{2}{9} \times \frac{\left(1.5 \times 10^{-6}\right)^2 \times\left(1.0 \times 10^3\right) \times 9.8}{1.8 \times 10^{-5}}\)

= 2.72×10^-4

Question 3. A metallic sphere of radius 1.0 × 10^-3 m and density 1.0 × 10⁴ kg/m³ enters a tank of water, after a free fall through a distance of h in the earth’s gravitational field. If its velocity remains unchanged after entering water, determine the value of h. Given : coefficient of viscosity of water = 1.0 × 10^-3 N-s/m², g = 10 m/s² and density of water = 1.0 × 10³ kg/m³.
Answer:

The velocity attained by the sphere in falling freely from a height h is

ν = \(\sqrt{2 \mathrm{gh}}\)….(1)

This is the terminal velocity of the sphere in water. Hence by Stoke’s law, we have 2

ν = \(\frac{2}{9} \frac{r^2(\rho-\sigma) g}{\eta}\)…….(2)

where r is the radius of the sphere, ρ is the density of the material of the sphere σ (= 1.0 × 10³ kg/m³) is the density of water and η is the coefficient of viscosity of water.

∴ ν = \(\frac{2 \times\left(1.0 \times 10^{-3}\right)^2\left(1.0 \times 10^4-1.0 \times 10^3\right) \times 10}{9 \times 1.0 \times 10^{-3}}\)

= 20 m/s

from equation (1), we have h = \(\frac{v^2}{2 g}=\frac{20 \times 20}{2 \times 10}\)

= 20 m

Elasticity and Viscosity Notes for NEET Physics Chapter 4

Applications of Stokes’ Formula

In determining the Electronic Charge by Millikan’s Oil Drop Experiment: Stokes’ formula is used in Millikan’s method for determining the electronic charge. In this method, the formula is applied to find out the radii of small oil drops by measuring their terminal velocity in the air.

The velocity of Raindrops: Raindrops are formed by the condensation of water vapor on dust particles. When they fall under gravity, their motion is opposed by the viscous drag in the air.

• As the velocity of their fall increases, the viscous drag also increases and finally becomes equal to the effective force of gravity.
• The drops then attain a (constant) terminal velocity which is directly proportional to the square of the radius of the drops.
• In the beginning, the raindrops are very small in size and so they fall with such a small velocity that they appear floating in the sky as clouds. As they grow in size by further condensation, then they reach the earth with appreciable velocity,

Parachute: When a soldier with a parachute jumps from a flying airplane, he descends very slowly in the air.

• In the beginning, the soldier falls with gravity acceleration g, but soon the acceleration goes on decreasing rapidly when a parachute is fully opened.
• Therefore, in the beginning, the speed of the falling soldier increases somewhat rapidly but then very slowly.
• Due to the viscosity of air, the acceleration of the soldier becomes ultimately zero and the soldier then falls with a constant terminal speed. The figure shows the speed of the falling soldier with time.

Elasticity And Viscosity Solids Multiple Choice Question And Answers

Question 1. The diameter of a brass rod is 4 mm and Young’s modulus of brass is 9 × 10¹⁰ N/m². The force required to stretch by 0.1% of its length is:

1. 360 πN
2. 36 N
3. 144 π × 10³ N
4. 36 π × 10⁵ N

Answer: 1. 360 πN

Question 2. A steel wire is suspended vertically from a rigid support. When loaded with a weight in the air, it expands by L_a and when the weight is immersed completely in water, the extension is reduced to L_w. The relative density of the material of the weight is

1. \(\frac{L_a}{L_a-L_w}\)
2. \(\frac{\mathrm{L}_{\mathrm{w}}}{\mathrm{L}_{\mathrm{a}}}\)
3. \(\frac{\mathrm{L}_{\mathrm{a}}}{\mathrm{L}_{\mathrm{w}}}\)
4. \(\frac{L_w}{L_a-L_w}\)

Answer: 2. 1. \(\frac{L_a}{L_a-L_w}\)

Elasticity and Viscosity MCQs for NEET Physics Class 11

Question 3. Two wires of equal length and cross-section area are suspended as shown in the figure. Their Young’s modulus is Y₁ and Y₂ respectively. The equivalent of Young’s modulus will be

1. Y1 + Y₂
2. \(\frac{Y_1+Y_2}{2}\)
3. \(\frac{Y_1 Y_2}{Y_1+Y_2}\)
4. \(\sqrt{Y_1 Y_2}\)

Answer: 2. \(\frac{Y_1+Y_2}{2}\)

Question 4. The load versus elongation graph for four wires of the same materials is shown in the figure. The thinnest wire is represented by the line :

1. OC
2. OD
3. OA
4. OB

Answer: 3. OA

Question 5. A force F is needed to break a copper wire having radius R. Then the force needed to break a copper wire of radius 2 R will be :

1. F/2
2. 2 F
3. 4 F
4. F/4

Answer: 3. 4 F

Question 6. A brass rod of length 2 m and a cross-sectional area of 2.0 cm² is attached end to end to a steel rod of length L and a cross-sectional area of 1.0 cm². The compound rod is subjected to equal and opposite pulls of magnitude 5 × 10⁴ N at its ends. If the elongations of the two rods are equal, then the length of the steel rod (L) is

(Y_Brass = 1.0 × 10¹¹ N/m² and Y_Steel = 2.0 × 1011 N/m² )

1. 1.5 m
2. 1.8 m
3. 1 m
4. 2 m

Answer: 4. 2 m

Question 7. If the ratio of lengths, radii, and Young’s moduli of steel and brass wires in the figure are a, b, and c respectively. Then the corresponding ratio of increase in their lengths would be :

1. \(\frac{2 \mathrm{ac}}{\mathrm{b}^2}\)
2. \(\frac{3 a}{2 b^2 c}\)
3. \(\frac{3 c}{2 a b^2}\)
4. \(\frac{2 a^2 c}{b}\)

Answer: 2. \(\frac{3 a}{2 b^2 c}\)

Question 8. The breaking stress of a wire depends upon

1. Length of the wire
2. The radius of the wire
3. Material of the wire
4. The shape of the cross-section

Answer: 3. Material of the wire

Question 9. The mean distance between the atoms of iron is 3 ×10^-10 m and the interatomic force constant for iron is 7 N /m. Yong’s modulus of elasticity for iron is

1. 2.33 × 10⁵ N/ m²
2. 23.3 × 10¹⁰ N/ m²
3. 233 × 10¹⁰N/ m²
4. 2.33 × 10¹⁰ N/ m²

Answer: 4. 2.33 × 1010 N/ m²

Question 10. An iron rod of length 2m and cross-section area of 50mm² stretched by 0.5mm, when a mass of 250kg is hung from its lower end. Young’s modulus of the iron rod is

1. 19.6 ×10¹⁰ N/ m²
2. 19.6 × 10¹⁵ N/ m²
3. 19.6 × 10¹⁸ N/ m²
4. 19.6 × 10²⁰ N/ m²

Answer: 1. 19.6 ×1010 N/ m²

Question 11. A steel wire of 1 m long and 1 mm² cross-section area is hung from a rigid support. When a weight of 1 kg is hung from it then the change in length will be (given Y = 2 × 10¹¹ N / m²)

1. 0.5 mm
2. 0.25 mm
3. 0.05 mm
4. 5 mm

Answer: 3. 0.05 mm

NEET Physics Chapter 4 Elasticity and Viscosity MCQs with Answers

Question 12. A 1m long metal wire of cross-sectional area 10^-6 m² is fixed at one end from a rigid support and a weight W is hanging at its other end. The graph shows the observed extension of length Δl of the wire as a function of W. Young’s modulus of the material of the wire in SΙ units is

1. 5 × 10⁴
2. 2 × 10⁵
3. 2 × 10¹¹
4. 5 × 10¹¹

Answer: 3. 2 × 10¹¹

Question 13. One end of a horizontal thick copper wire of length 2L and radius 2R is welded to an end of another horizontal thin copper wire of length L and radius R. When the arrangement is stretched by applying forces at two ends, the ratio of the elongation in the thin wire to that in the thick wire is :

1. 0.25
2. 0.50
3. 2.00
4. 4.00

Answer: 3. 2.00

Question 14. The Young’s modulus of a wire of length (L) and radius (r) is Y. If the length is reduced to \(\frac{\mathrm{L}}{2}\) and radius to \(\frac{\mathrm{R}}{2}\) then its Young’s modulus will be

1. \(\frac{\mathrm{Y}}{2}\)s
2. Y
3. 2Y
4. 2Y

Answer: 2. Y

Question 15. A 5m aluminium wire (Y = 7 × 10¹⁰ N/m²) of diameter 3 mm supports a 40 kg mass. In order to have the same elongation in a copper wire (Y= 12 × 10¹⁰ N/m²) of the same length under the same weight, the diameter should be in mm

1. 1.75
2. 2.0
3. 2.3
4. 5.0

Answer: 3. 2.3

Question 16. The following four wires are made of the same material and the same tension is applied to them. Which one will have the maximum increase in length?

1. Length = 100 cm, Diameter = 1mm
2. Length = 50 cm, Diameter = 0.5 mm
3. Length = 200 cm, Diameter = 2mm
4. Length = 300 cm, Diameter = 3 mm

Answer: 2. Length = 50 cm, Diameter = 0.5 mm

Question 17. A catapult’s string made of rubber has a cross-section area of 25 mm2 and a length of 10 cm. To throw a 5 gm pabble it is stretched up to 5 cm and released. The velocity of the projected pabble is (Young coefficient of elasticity of rubber is 5 × 10⁸ N/m²) :

1. 20 m/s
2. 100 m/s
3. 250 m/s
4. 200 m/s

Answer: 3. 250 m/s

Question 18. The diameter of a brass rod is 4 mm and the Young modulus of elasticity is 9 × 10¹⁰ N/m². The force required to increase the length of the rod by 0.10% will be :

1. 360 πN
2. 36 N
3. 144π × 10³ N
4. 36π × 10⁵ N

Answer: 1. 360 πN

Question 19. Two blocks each of mass 2kg are connected as shown in the figure. The breaking stress of the material of the wire is \(\frac{2}{\pi} \times 10^9\) N/m². Find the minimum radius of the wire used if it is not to break.

1. 10^-3 m
2. 10^-4 m
3. 10^-5 m
4. 10^-6 m

Answer: 2. 10^-4 m

Question 20. Four wires of the same material are stretched by the same load. The dimensions of the wires are given below. The one that has the maximum elongation is of

1. Diameter 1 mm and length 1 m
2. Diameter 2 mm and length 2 m
3. Diameter 0.5 mm and length 0.5 m
4. Diameter 3 mm and length 3 m

Answer: 3. Diameter 0.5 mm and length 0.5 m

Question 21. An elevator cable can have a maximum stress of 7 × 10⁷ N/m² for appropriate safety factors. Its maximum upward acceleration is 1.5 m/s². If the cable has to support the total weight of 2000 kg of a loaded elevator, the minimum area of the cross-section of the cable should be (g = 10 m/s²)

1. 3.28 cm²
2. 2.38 cm²
3. 0.328 cm²
4. 8.23 cm²

Answer: 1. 3.28 cm²

Elasticity and Viscosity Multiple Choice Questions for NEET Class 11

Question 22. A 2m long light metal rod AB is suspended from the ceiling horizontally by means of two vertical wires of equal length tied to its ends. One wire is of brass and has a cross-sectional area of 0.2 × 10^-4 m² and the other is of steel with 0.1 × 10^-4 m² cross-sectional area. in order to have equal stresses in the two wires, a weight W is hung from the rod. The position of the weight along the rod from end A should be

1. 66.6 cm
2. 133 cm
3. 44.4 cm
4. 155.6 cm

Answer: 1. 66.6 cm

Question 23. A brass rod of length 2 m and a cross-sectional area of 2.0 cm² is attached end to end to a steel rod of length L and a cross-sectional area of 1.0 cm². The compound rod is subjected to equal and opposite pulls of magnitude 5 × 10⁴ N at its ends. If the elongations of the two rods are equal, then the length of the steel rod (L) is (Y_Brass = 1.0 × 10¹¹ N/m² and Y_Steel = 2.0 × 10¹¹ N/m² )

1. 1.5 m
2. 1.8 m
3. 1 m
4. 2 m

Answer: 4. 2m

Question 24. A square brass plate of side 1.0 m and thickness 0.005 m is subjected to a force F on its smaller opposite edges, causing a displacement of 0.02 cm. If the shear modulus of brass is 0.4 × 10¹¹ N/m², the value of the force F is

1. 4 × 10³ N
2. 400 N
3. 4 × 10⁴ N
4. 1000 N

Answer: 3. 4 × 10⁴ N

Question 25. A solid cylinder of radius r and length l is clamped at its upper end and the lower end is twisted through θ. The shear strain is given by.

1. θ
2. \(\frac{\theta \ell}{\mathrm{r}}\)
3. \(\frac{\theta \mathrm{r}}{\ell}\)
4. 0

Answer: 3. \(\frac{\theta \mathrm{r}}{\ell}\)

Question 26. A 50 kg motor rests on four cylindrical rubber blocks. Each block has a height of 4 cm and a cross-sectional area of 16 cm². The shear modulus of rubber is 2 × 10⁶ N/m². A sideways force of 500 N is applied to the motor. The distance that the motor moves sideways is

1. 0.156 cm
2. 1.56 cm
3. 0.312 cm
4. 0.204 cm

Answer: 1. 0.156 cm

Question 27. If the poison ratio is 0.4 and after increasing the length of the wire by 0.05% then the decrease in its diameter will be :

1. 0.02%
2. 0.1%
3. 0.01%
4. 0.4%

Answer: 1. 0.02%

Question 28. A 50 kg motor rests on four cylindrical rubber blocks. Each block has a height of 4 cm and a cross-sectional area of 16 cm². The shear modulus of rubber is 2 × 10⁶ N/m². A sideways force of 500 N is applied to the motor. The distance that the motor moves sideways is

1. 0.156 cm
2. 1.56 cm
3. 0.312 cm
4. 0.204 cm

Answer: 1. 0.156 cm

NEET Class 11 Physics Elasticity and Viscosity MCQs

Question 29. A metal block is experiencing an atmospheric pressure of 1 × 10⁵ N/m², when the same block is placed in a vacuum chamber, the fractional change in its volume is (the bulk modulus of metal is 1.25 × 10¹¹ N/m²)

1. 4 × 10^-7
2. 2 × 10^-7
3. 8 × 10^-7
4. 1 × 10^-7

Answer: 3. 8 × 10^-7

Question 30. The compressibility of water is 46.4 × 10^-6/atm. This means that

1. The bulk modulus of water is 46.4 × 10⁶ atm
2. The volume of water decreases by 46.4 one-millionths of the original volume for each atmosphere increase in pressure
3. When water is subjected to an additional pressure of one atmosphere, its volume decreases by 46.4%
4. When water is subjected to an additional pressure of one atmosphere, its volume is reduced to 10^-6 of its original volume.

Answer: 2. The volume of water decreases by 46.4 one-millionths of the original volume for each atmosphere increase in pressure

Question 31. If a rubber ball is taken at a depth of 200 m in a pool its volume decreases by 0.1%. If the density of the water is 1 × 10³ kg/m³ and g = 10 m/s², then the volume elasticity in N/m² will be :

1. 108
2. 2 × 108
3. 109
4. 2 × 109

Answer: 4. 2 × 109

Question 32. Two wires of the same material and length but diameter in the ratio 1: 2 are stretched by the same force. The ratio of potential energy per unit volume for the two wires when stretched will be:

1. 1: 1
2. 2: 1
3. 4: 1
4. 16: 1

Answer: 4. 16:1

Question 33. Two wires A and B of the same length and of the same material have the respective radii r₁ and r₂. One end is fixed with a rigid support, and at the other end equal twisting couple is applied. Then the ratio of the angle of twist at the end of A and the angle of twist at the end of B will be

1. \(\frac{r_1^2}{r_2^2}\)
2. \(\frac{r_2^2}{r_1^2}\)
3. \(\frac{r_2^4}{r_1^4}\)
4. \(\frac{r_1^4}{r_2^4}\)

Answer: 3. \(\frac{r_2^4}{r_1^4}\)

Question 34. The upper end of a wire of radius 4 mm and length 100 cm is clamped and its other end is twisted through an angle of 30°. Then angle of the shear is

1. 12°
2. 0.12°
3. 1.2 °
4. 0.012°

Answer: 2. 0.12°

Question 35. A 2m long rod of radius 1 cm which is fixed from one end is given a twist of 0.8 radians. The shear strain developed will be

1. 0.002
2. 0.004
3. 0.008
4. 0.016

Answer: 2. 0.004

Question 36. The relation between γ, η, and K for an elastic material is

1. \(\frac{1}{\eta}=\frac{1}{3 \gamma}+\frac{1}{9 K}\)
2. \(\frac{1}{\mathrm{~K}}=\frac{1}{3 \gamma}+\frac{1}{9 \eta}\)
3. \(\frac{1}{\gamma}=\frac{1}{3 K}+\frac{1}{9 \eta}\)
4. \(\frac{1}{\gamma}=\frac{1}{3 \eta}+\frac{1}{9 K}\)

Answer: 4. \(\frac{1}{\gamma}=\frac{1}{3 \eta}+\frac{1}{9 K}\)

Question 37. The upper end of a wire of radius 4 mm and length 100 cm is clamped and its other end is twisted through an angle of 30º. The angle of the shear is

1. 12º
2. 0.12º
3. 1.2º
4. 0.012º

Answer: 2. 0.12º

Question 38. A rubber ball is brought into 200 m deep water, its volume is decreased by 0.1% then the volume elasticity modulus of the material of the ball will be :

1. 19.6 × 10⁸ N/m²
2. 19.6 × 10^-10 N/m²
3. 19.6 × 10¹⁰ N/m²
4. 19.6 × 10^-8 N/m²

Answer: 1. 19.6 × 10⁸ N/m²

Question 39. The mean density of seawater is ρ, and the bulk modulus is B. The change in density of seawater in going from the surface of the water to a depth h is :

1. \(\frac{\rho g h}{B}\)
2. Bpgh
3. \(\frac{\rho^2 g h}{B}\)
4. \(\frac{B \rho^2}{\mathrm{gh}}\)

Answer: 3. \(\frac{\rho^2 g h}{B}\)

NEET Physics Class 11 Chapter 4: Elasticity and Viscosity MCQs with Solutions

Question 40. A sample of a liquid has an initial volume of 1.5 L. The volume is reduced by 0.2 mL when the pressure increases by 140 kPa. What is the bulk modulus of the liquid?

1. 3.05 × 10⁹ Pa.
2. 1.05 × 10⁹ Pa.
3. 1.05 × 10⁷ Pa.
4. 1.05 × 10¹¹ Pa.

Answer: 2. 1.05 × 10⁹ Pa.

Question 41. If the potential energy of a spring is V on stretching it by 2 cm, its potential energy when it is stretched by 10 cm will be :

1. V/25
2. 5 V
3. V/5
4. 25 V

Answer: 4. 25V

Question 42. If work done in stretching a wire by 1mm is 2J, the work necessary for stretching another wire of the same material, but of double the radius and half the length by 1mm in joule will be –

1. 1/4
2. 4
3. 8
4. 16

Answer: 4. 16

Question 43. According to Hooke’s law of elasticity, if stress is increased, the ratio of stress to strain

1. Increases
2. Decreases
3. Becomes zero
4. Remains constant

Answer: 4. Remains constant

Question 44. Which of the following affects the elasticity of a substance

1. Hammering and annealing
2. Change in temperature
3. Impurity in substance
4. All of these

Answer: 4. All of these

Question 45. Calculate the work done, if a wire is loaded by ‘Mg’ weight and the increase in length is ‘l’

1. Mgl
2. Zero
3. Mgl/2
4. 2Mgl

Answer: 3. Mgl/2

Question 46. An elastic material of Young’s modulus Y is subjected to a stress S. The elastic energy stored per unit volume of the material is

1. \(\frac{2 Y}{S^2}\)
2. \(\frac{S^2}{2 \mathrm{Y}}\)
3. \(\frac{S}{2 \mathrm{Y}}\)
4. \(\frac{S^2}{Y}\)

Answer: 2. \(\frac{S^2}{2 \mathrm{Y}}\)

Question 47. A stretched rubber has

1. Increased kinetic energy
2. Increased potential energy
3. Decreased kinetic energy
4. Decreased potential energy

Answer: 2. Increased potential energy

Question 48. If a spring extends by x on loading, then the energy stored by the spring is (if T is tension in the spring and k is spring constant)

1. \(\frac{T^2}{2 x}\)
2. \(\frac{\mathrm{T}^2}{2 \mathrm{k}}\)
3. \(\frac{2 x}{T}\)
4. \(\frac{2 T^2}{k}\)

Answer: 2. \(\frac{\mathrm{T}^2}{2 \mathrm{k}}\)

Question 49. On stretching a wire, of length L by l using force F the elastic energy stored per unit volume is

1. Fl/2AL
2. FA/2L
3. FL/2A
4. FL/2

Answer: 1. Fl/2AL

Question 50. A wire of length 50 cm and a cross-sectional area of 1 sq. mm is extended by 1 mm. The required work will be (Y = 2 × 10¹⁰ Nm^-2)

1. 6 × 10^-2 J
2. 4 × 10^-2 J
3. 2 × 10^-2 J
4. 1 × 10^-2 J

Answer: 3. 2 × 10^-2 J

Question 51. When a force is applied on a wire of uniform cross-sectional area 3 × 10^-6 m² and length 4m, the increase in length is 1 mm. Energy stored in it will be (Y = 2 × 1011 N / m²)

1. 6250 J
2. 0.177 J
3. 0.075 J
4. 0.150 J

Answer: 3. 0.075 J

Question 52. The elastic energy stored in a wire of Young’s modulus Y is

1. \(\mathrm{Y} \times \frac{\text { Strain }^2}{\text { Volume }}\)
2. Stress × Strain × Volume
3. \(\frac{\text { Stress }^2 \times \text { Volume }}{2 \mathrm{Y}}\)
4. \(\frac{1}{2} Y \times \text { stress } \times \text { Strain } \times \text { Volume }\)

Answer: 3. \(\frac{\text { Stress }^2 \times \text { Volume }}{2 \mathrm{Y}}\)

Question 53. A wire of length 50 cm and a cross-sectional area of 1 sq. mm is extended by 1 mm. The required work will be (Y = 2 × 10¹⁰ Nm^-2)

1. 6 × 10^-2 J
2. 4 × 10^-2 J
3. 2 × 10^-2 J
4. 1 × 10^-2 J

Answer: 3. 2 × 10^-2 J

Question 54. When a load of 5 kg is hung on a wire then an extension of 3 meters takes place, the work done will be :

1. 75 J
2. 60 J
3. 50 J
4. 100 J

Answer: 1. 75 J

Elasticity and Viscosity MCQ Practice Test for NEET Class 11

Question 55. When a load of 5 kg is hung on a wire then an extension of 3 metres takes place, the work done will be :

1. 75 J
2. 60 J
3. 50 J
4. 100 J

Answer: 1. 75 J

Question 56. A wire suspended vertically from one of its ends is stretched by attaching a weight of 200 N to the lower end. The weight stretches the wire by 1mm. Then the elastic energy stored in the wire is?

1. 0.2 J
2. 10J
3. 20J
4. 0.1 J

Answer: 4. 0.1 J

Question 57. If ‘S’ is stress and ‘Y’ is Young’s modulus of the material of a wire, the energy stored in the wire per unit volume is :

1. 2S²Y
2. \(\frac{S^2}{2 \mathrm{Y}}\)
3. \(\frac{2 Y}{S^2}\)
4. \(\frac{S}{2 Y}\)

Answer: 2. \(\frac{S^2}{2 \mathrm{Y}}\)

Question 58. A wire elongates by lmm when a load W is hanged from it. If the wire goes over a pulley and two weights W each are hung at the two ends, the elongation of the wire will be (in mm)

1. l/2
2. l
3. 2l
4. Zero

Answer: 2. 1

Question 59. An oil drop falls through the air with a terminal velocity of 5 × 10^-4 m/s.

1. The radius of the drop will be :

1. 2.5 × 10^-6 m
2. 2 × 10^-6 m
3. 3 × 10^-6 m
4. 4 × 10^-6 m

Answer: 3. 3 × 10^-6 m

2. The terminal velocity of a drop of half of this radius will be (Viscosity of air = \(\frac{18 \times 10^{-5}}{5}\) N-s/m² density of oil = 900 Kg/m³. Neglect density of air as compared to that of oil).

1. 3.25 × 10^-4 m/s
2. 2.10 × 10^-4 m/s
3. 1.5 × 10^-4 m/s
4. 1.25 × 10^-4 m/s

Answer: 4. 1.25 × 10^-4 m/s

Question 60. The terminal velocity of a sphere moving through a viscous medium is :

1. Directly proportional to the radius of the sphere
2. Inversely proportional to the radius of the sphere
3. Directly proportional to the square of the radius of the sphere
4. Inversely proportional to the square of the radius of a sphere

Answer: 3. Directly proportional to the square of the radius of a sphere

Question 61. A sphere is dropped gently into a medium of infinite extent. As the sphere falls, the net force acting downwards on it

1. Remains constant throughout
2. Increases for some time and then becomes constant
3. Decreases for some time and then becomes zero
4. Increases for some time and then decreases.

Answer: 3. Decreases for some time and then becomes zero

Question 62. A solid sphere falls with a terminal velocity of 10 m/s in air. If it is allowed to fall in a vacuum,

1. Terminal velocity will be more than 10 m/s
2. Terminal velocity will be less than 10 m/s
3. Terminal velocity will be 10 m/s
4. There will be no terminal velocity

Answer: 4. There will be no terminal velocity

Question 63. A spherical ball is dropped in a long column of viscous liquid. Which of the following graphs represents the variation of

1. Gravitational force with time
2. Viscous force with time
3. The net force acting on the ball with time.

1. Q, R, P
2. R, Q, P
3. P, Q, R
4. P, R, Q

Answer: 3. P, Q, R

Question 64. A ball of mass m and radius r is released in a viscous liquid. The value of its terminal velocity is proportional to :

1. \(\frac{1}{r}\)
2. \(\frac{\mathrm{m}}{\mathrm{r}}\)
3. \(\sqrt{\frac{m}{r}}\)
4. m only m

Answer: 2. \(\frac{\mathrm{m}}{\mathrm{r}}\)

Question 65. In Poiseuilli’s method of determination of coefficient of viscosity. the physical quantity that requires greater accuracy in measurement is

1. Pressure difference
2. The volume of the liquid collected
3. Length of the capillary tube
4. Inner radius of the capillary tube

Answer: 4. Inner radius of the capillary tube

Question 66. A viscous fluid is flowing through a cylindrical tube. The velocity distribution of the fluid is best represented by the diagram

Answer: 3.

Question 67. What is the velocity υ of a metallic ball of radius r falling in a tank of liquid at the instant when its acceleration is one-half that of a freely falling body? (The densities of metal and of liquid are ρ and σ respectively, and the viscosity of the liquid is η).

1. \(\frac{r^2 g}{9 \eta}(\rho-2 \sigma)\)
2. \(\frac{r^2 g}{9 \eta}(2 \rho-\sigma)\)
3. \(\frac{r^2 g}{9 \eta}(\rho-\sigma)\)
4. \(\frac{2 r^2 g}{9 \eta}(\rho-\sigma)\)

Answer: 1. \(\frac{r^2 g}{9 \eta}(\rho-2 \sigma)\)

Question 68. A tiny spherical oil drop carrying a net charge q is balanced in still air with a vertical uniform electric field of strength \(\). When the field is switched off, the drop is observed to fall with terminal velocity 2 × 10^-3 m s^-1. Given g = 9.8 m s^-2, a viscosity of the air = 1.8 × 10^-5 Ns m^-2, and the density of oil ρ = 900 kg m^-3, the magnitude of q is:

1. 1.6 × 10^-19 C
2. 3.2 × 10^-19 C
3. 4.8 × 10^-19 C
4. 8.0 × 10^-19 C

Answer: 4. 8.0 × 10^-19 C

Question 69. A liquid has only

1. Shear modulus
2. Young’s modulus
3. Bulk modulus
4. All of the above

Answer: 3. Bulk modulus

Question 70. According to Newton, viscous force is given by

F = – ηA \(\frac{d v}{d x}\)

where η = coefficient of viscosity, so dimensions of η will be :

1. [ML^-1T^-2]
2. [MLT^-2]
3. [ML^-1T^-1]
4. [M^-1L2T^-2]

Answer: 3. [ML^-1T^-1]

Question 71. Spherical balls of radius R are falling in a viscous fluid of viscosity η with a velocity ν. The retarding viscous force acting on the spherical ball is:

1. Directly proportional to R but inversely proportional to ν
2. Directly proportional to both radius R and velocity ν
3. Inversely proportional to both radius R and velocity ν
4. Inversely proportional to R but directly proportional to ν

Answer: 2. Directly proportional to both radius R and velocity ν

NEET Class 11 Physics Chapter 4 Elasticity and Viscosity MCQs for Exam Preparation

Question 72. A 20 cm long capillary tube is dipped in water. The water rises upto 8 cm. If the entire arrangement is put in a freely falling elevator, the length of the water column in the capillary tube will be :

1. 8 cm
2. 10 cm
3. 4 cm
4. 20 cm

Answer: 4. 20 cm

Question 73. If the terminal speed of a sphere of gold (density = 19.5 kg/m³) is 0.2 m/s in a viscous liquid then find the terminal speed of a sphere of silver (density = 10.5 kg/m³) of the same size in the same liquid (density = 1.5 kg/m³).

1. 0.2 m/s
2. 0.4 m/s
3. 0.133 m/s
4. 0.1 m/s

Answer: 4. 0.1 m/s

Question 74. A spherical solid ball of volume V is made of a material of density ρ₁. It is falling through a liquid of density ρ₂ (ρ₂< ρ₁). Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speed ν, i.e., F_viscous= – kν²(k > 0). The terminal speed of the ball is

1. \(\frac{\mathrm{Vg} \rho_1}{\mathrm{k}}\)
2. \(\sqrt{\frac{V g \rho_1}{k}}\)
3. \(\frac{{Vg}\left(\rho_1-\rho_2\right)}{k}\)
4. \(\sqrt{\frac{{Vg}\left(\rho_1-\rho_2\right)}{k}}\)

Answer: 4. \(\sqrt{\frac{{Vg}\left(\rho_1-\rho_2\right)}{k}}\)

Question 75. Two hail stones with radii in the ratio of 1: 2 fall from a great height through the atmosphere. Then the ratio of their momenta after they have attained terminal velocity will be

1. 1: 1
2. 1: 4
3. 1: 16
4. 1: 32

Answer: 4. 1:32

Question 76. A space is 2.5 cm wide between two large plane surfaces is filled with oil. The force required to drag a very thin plate of area 0.5 m² just midway through the surfaces at a speed of 0.5 m/sec is 1N. The coefficient of viscosity in kg–sec/m² is :

1. 5 × 10^-2
2. 2.5 × 10^-2
3. 1 × 10^-2
4. 7.5 × 10^-2

Answer: 2. 2.5 × 10^-2

Question 77. A raindrop of radius 1.5 mm, experiences a drag force F = (2 × 10^-5 v) N while falling through the air from a height of 2 km, with a velocity v. The terminal velocity of the raindrop will be nearly (use g = 10 m/s²):

1. 200 m/s
2. 60 m/s
3. 7 m/s
4. 3 m/s

Answer: 3. 7 m/s

Question 78. Two identical rods in geometry but of different materials having coefficients of thermal expansion α₁ and α₂ and Young’s moduli Y₁ and Y₂ respectively are fixed between two rigid massive walls. The rods are heated such that they undergo the same increase in temperature. There is no bending of the rods. If α₁: α₂= 2: 6 the thermal stresses developed in the two rods are equal provided Y₁: Y₂ is equal to :

1. 2 : 3
2. 1: 1
3. 3: 1
4. 4: 9

Answer: 3. 3:1

Question 79. A small steel ball falls through a syrup at a constant speed of 10 cm/s. If the steel ball is pulled upwards with a force equal to twice its effective weight, how fast will it move upwards?

1. 10 cm/s
2. 20 cm/s
3. 5 cm/s
4. – 5 cm/s

Answer: 1. 10 cm/s

Question 80. Two spheres P and Q of equal radii have densities ρ₁ and ρ₂, respectively. The spheres are connected by a massless string and placed in liquids L₁ and L₂ of densities σ₁ and σ₂ and viscosities η₁ and η₂, respectively. They float in equilibrium with the sphere P in L₁ and sphere Q in L₂ and the string is taut (see figure). If sphere P alone in L₂ has terminal velocity VPand Q alone in L₁ has terminal velocity VQ, then

1. \(\frac{\left|\vec{V}_{\mathrm{P}}\right|}{\left|\vec{V}_{\mathrm{Q}}\right|}=\frac{\eta_1}{\eta_2}\)
2. \(\frac{\left|\vec{V}_P\right|}{\left|\vec{V}_Q\right|}=\frac{\eta_2}{\eta_1}\)
3. \(\vec{V}_P \cdot \vec{V}_Q>0\)
4. None of these

Answer: 1. \(\frac{\left|\vec{V}_{\mathrm{P}}\right|}{\left|\vec{V}_{\mathrm{Q}}\right|}=\frac{\eta_1}{\eta_2}\)

Question 81. The force required to stretch a steel wire of 1 cm² cross-section to 1.1 times its length, will be (Y = 2 × 10¹¹ Nm^-2)

1. 2 × 10⁶ N
2. 2 × 10³ N
3. 2 × 10^-6 N
4. 2 × 10^-7 N

Answer: 1. 2 × 10⁶ N

Question 82. A cube is subjected to a uniform volume compression. If the side of the cube decreases by 2%, the bulk strain is –

1. 0.02
2. 0.03
3. 0.04
4. 0.06

Answer: 4. 0.06

Question 83. The following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied?

1. length = 100 cm, diameter = 1 mm
2. length = 200 cm, diameter = 2 mm
3. length = 300 cm, diameter = 3 mm
4. length = 50 cm, diameter = 0.5 mm

Answer: 4. length = 50 cm, diameter = 0.5 mm

Question 84. A copper of fixed volume ‘V’ is drawn into a wire of length ‘l’. When this wire is subjected to a constant force ‘F’, the extension produced in the wire is ‘Δl’. Which of the following graph is a straight line?

1. Δl versus 1/l
2. Δl versus l²
3. Δl versus 1/l²
4. Δl versus l

Answer: 2. Δl versus l²

Question 85. The approximate depth of an ocean is 2700 m. The compressibility of water is 45.4 x 10^-11 Pa^-1 and the density of water is 10³ kg/m³. What fractional compression of water will be obtained at the bottom of the ocean?

1. 1.0 × 10^-2
2. 1.2 × 10^-2
3. 1.4 × 10^-2
4. 0.8 × 10^-2

Answer: 2. 1.2 × 10^-2

Question 86. Two Young’s modulus of steel is twice that of brass. Two wires of the same length and of the same area of cross-section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weights added to the steel and brass wires must be in the ratio of :

1. 2: 1
2. 4: 1
3. 1: 1
4. 1: 2

Answer: 1. 2:1

Question 87. The bulk modulus of a spherical object is ’B’. If it is subjected to uniform pressure ‘p’, the fractional decrease in radius is:

1. \(\frac{p}{B}\)
2. \(\frac{B}{3 p}\)
3. \(\frac{3 p}{B}\)
4. \(\frac{p}{3 B}\)

Answer: 4. \(\frac{p}{3 B}\)

NEET Physics Elasticity and Viscosity: MCQs for Revision

Question 88. Two wires are made of the same material and have the same volume. The first wire has cross-sectional area A and the second wire has cross-sectional area 3A. If the length of the first wire is increased by Δl on applying a force F, how much force is needed to stretch the second wire by the same amount?

1. 9 F
2. F
3. 4 F
4. 6 F

Answer: 1. 9F

Question 89. A small sphere of radius ‘r’ falls from rest in a viscous liquid. As a result, heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity is proportional to :

1. r³
2. r⁴
3. r⁵
4. r²

Answer: 3. r⁵

Question 90. When a block of mass M is suspended by a long wire of length L, the length of the wire becomes (L+ l). The elastic potential energy stored in the extended wire is:

1. \(\frac{1}{2} \mathrm{MgL}\)
2. Mgl
3. MgL
4. \(\frac{1}{2} \mathrm{Mg} \ell\)

Answer: 4. \(\frac{1}{2} \mathrm{Mg} \ell\)

Question 91. Two small spherical metal balls, having equal masses, are made from materials of densitiesρ₁ and ρ₂ (_ρ1 = 8_ρ2) and have radii of 1 mm and 2 mm, respectively. They are made to fall vertically (from rest) in a viscous medium whose coefficient of viscosity equals η and whose density is 0.1ρ₂. The ratio of their terminal velocities would be

1. \(\frac{79}{72}\)
2. \(\frac{19}{36}\)
3. \(\frac{39}{72}\)
4. \(\frac{79}{36}\)

Answer: 4. \(\frac{79}{36}\)

Question 92. A wire of length L, area of cross-section A is hanging from a fixed support. The length of the wire changes to L1 when mass M is suspended from its free end. The expression for Young’s modulus is:

1. \(\frac{M g L}{A\left(L_1-L\right)}\)
2. \(\frac{M g L}{A L}\)
3. \(\frac{M g\left(L_1-L\right)}{A L}\)
4. \(\frac{M g L}{A L_1}\)

Question 93. Two wires are made of the same material and have the same volume. However, wire 1 has a cross-sectional area of A, and wire 2 has a cross-sectional area of 3A. If the length of wire 1 increases by Δx on applying force F, how much force is needed to stretch wire 2 by the same amount?

1. 4F
2. 6F
3. 9F
4. F

Answer: 3. 9F

Question 94. If a ball of steel (density p = 7.8 g cm –3) attains a terminal velocity of 10 cm s-1 when falling in water (Coefficient of Viscosity η_water = 8.5 × 10^-4 Pa.s) then its terminal velocity in glycerine (p = 1.2 g cm^-3, η= 13.2 Pa.s.) would be, nearly :

1. 6.25 × 10^-4 cm s^-1
2. 6.45 × 10^-4 cm s^-1
3. 1.5 ×10^-5 cm s^-1
4. 1.6 ×10^-5 cm s^-1

Answer: 1. 6.25 × 10^-4 cm s^-1

Question 95. A sonometer wire of length 1.5 m is made of steel. The tension in it produces an elastic strain of 1%. What is the fundamental frequency of steel if the density and elasticity of steel are 7.7 × 10³ kg/m³ and 2.2 × 10¹¹ N/m² respectively?

1. 188.5 Hz
2. 178.2 Hz
3. 200.5 Hz
4. 770 Hz

Answer: 2. 178.2 Hz

Question 96. The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by 100°C is : (For steel Young’s modulus is 2 × 10¹¹ N m^-2 and the coefficient of thermal expansion is 1.1 × 10^-5 K^-1)

1. 2.2 × 10⁸ Pa
2. 2.2 × 10⁹ Pa
3. 2.2 × 10⁷ Pa
4. 2.2 × 10⁶ Pa

Answer: 1. 2.2 × 108 Pa

Question 97. A pendulum suspended from a uniform wire of cross-sectional area A has period T. When an additional mass M is added to its bob, the period changes to TM. If Young’s modulus of the 1 material of the wire is Y then \(\frac{1}{\mathrm{Y}}\) is equal to : (g=gravitational acceleration)

1. \(\left[\left(\frac{T_M}{T}\right)^2-1\right] \frac{\mathrm{A}}{\mathrm{Mg}}\)
2. \(\left[\left(\frac{T_{\mathrm{M}}}{\mathrm{T}}\right)^2-1\right] \frac{\mathrm{Mg}}{\mathrm{A}}\)
3. \(\left[1-\left(\frac{T_M}{T}\right)^2\right] \frac{\mathrm{A}}{\mathrm{Mg}}\)
4. \(\left[1-\left(\frac{T}{T_M}\right)^2\right] \frac{\mathrm{A}}{\mathrm{Mg}}\)

Answer: 1. \(\left[\left(\frac{T_M}{T}\right)^2-1\right] \frac{\mathrm{A}}{\mathrm{Mg}}\)

Question 98. A solid sphere of radius r made of a soft material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless piston of area floats on the surface of the liquid, covering the entire cross-section of the cylindrical container. When a mass m is placed on the surface of the piston to dr compress the liquid, the fractional decrement in the radius of the sphere, \(\left(\frac{d r}{r}\right)\) is

1. \(\frac{\mathrm{mg}}{3 \mathrm{Ka}}\)
2. \(\frac{\mathrm{mg}}{\mathrm{Ka}}\)
3. \(\frac{\mathrm{Ka}}{\mathrm{mg}}\)
4. \(\frac{\mathrm{Ka}}{3 \mathrm{mg}}\)

Answer: 1. \(\frac{\mathrm{mg}}{3 \mathrm{Ka}}\)

NEET Physics Class 11 Chapter 4 Elasticity and Viscosity MCQs and Answers

Question 99. A rod, of length L at room temperature and uniform area of cross-section A, is made of a metal having a coefficient of linear expansion α/°C. It is observed that an external compressive force F, is applied on each of its ends, and prevents any change in the length of the rod, when its temperature rises by ΔT K. Young’s modulus, Y, for this metal is :

1. \(\frac{2 \mathrm{~F}}{\mathrm{~A} \alpha \Delta \mathrm{T}}\)
2. \(\frac{F}{A \alpha(\Delta T-273)}\)
3. \(\frac{F}{A \alpha \Delta T}\)
4. \(\frac{F}{2 \mathrm{~A} \alpha \Delta \mathrm{T}}\)

Answer: 3. \(\frac{F}{A \alpha \Delta T}\)

Question 100. A load of mass M kg is suspended from a steel wire of length 2 m and radius 1.0 mm in Searle’s apparatus experiment. The increase in length produced in the wire is 4.0 mm. Now the load is fully immersed in a liquid of relative density 2. The relative density of the material of load is 8. The new value of an increase in the length of the steel wire is :

1. 3.0 mm
2. Zero
3. 5.0 mm
4. 4.0

Answer: 1. 3.0mm

NEET Physics Class 11 Chapter 8 Heat Transfer Introduction

Heat is energy in transit that flows due to temperature difference; from a body at a higher temperature to a body at a lower temperature. This transfer of heat from one body to the other takes place through three processes.

1. Conduction
2. Convection
3. Radiation

Heat Transfer Notes for NEET Physics Class 11

NEET Physics Class 11 Chapter 8 Conduction

The process of transmission of heat energy in which heat is transferred from one particle of the medium to the other, but each particle of the medium stays at its position is called conduction, for example, if you hold an iron rod with one of its end on a fire for some time, the handle will get heated.

• The heat is transferred from the fire to the handle by conduction along the length of the iron rod. The vibrational amplitude of atoms and electrons of the iron rod at the hot end takes on relatively higher values due to the higher temperature of their environment.
• These increased vibrational amplitudes are transferred along the rod, from atom to atom during collision between adjacent atoms. In this way, a region of rising temperature extends itself along the rod to your hand.

Consider a slab of face area A, Lateral thickness L, whose faces have temperatures THand TC(TH> TC).

Now consider two cross sections in the slab at positions A and B separated by a lateral distance of dx. Let the temperature of face A be T and that of face B be T + ΔT. Then experiments show that Q, the amount of heat crossing the area A of the slab at position x in time t is given by

⇒ \(\frac{Q}{t}=-K A \frac{d T}{d x}\)

Here K is a constant depending on the material of the slab and is named the thermal conductivity of the material, and the quantity \(\left(\frac{d T}{d x}\right)\) is called temperature gradient. The (–) sign in the equation heat flows from high to low temperature (ΔT is a –ve quantity)

NEET Physics Class 11 Chapter 8 Steady State

If the temperature of a cross-section at any position x in the above slab remains constant with time (remember, it does vary with position x), the slab is said to be in a steady state.

• Remember steady-state is distinct from thermal equilibrium for which temperature at any position (x) in the slab must be the same.
• For a conductor in a steady state, there is no absorption or emission of heat at any cross-section. (as the temperature at each point remains constant with time).
• The left and right faces are maintained at constant temperatures TH and TC respectively, and all other faces must be covered with adiabatic walls so that no heat escapes through them and the same amount of heat flows through each cross-section in a given Interval of time.

Hence Q₁= Q = Q₂. Consequently, the temperature gradient is constant throughout the slab.

Hence, \(\frac{d T}{d x}=\frac{\Delta T}{L}=\frac{T_f-T_i}{L}=\frac{T_C-T_H}{L}\)

and \(\frac{Q}{t}=-\mathrm{KA} \frac{\Delta T}{L} \Rightarrow \frac{Q}{t}\)

⇒ \(\mathrm{KAQ}\left(\frac{T_H-T_C}{L}\right)\)

Here Q is the amount of heat flowing through a cross-section of slab at any position in a time interval of t.

NEET Physics Chapter 8 Heat Transfer Study Notes

Question 1. One face of an aluminum cube of edge 2 meter is maintained at 100ºC and the other end is maintained at 0ºC. All other surfaces are covered by adiabatic walls. Find the amount of heat flowing through the cube in 5 seconds. (The thermal conductivity of aluminum is 209 W/m–ºC)
Answer:

Heat will flow from the end at 100ºC to the end at 0ºC.

Area of a cross-section perpendicular to the direction of heat flow,

A = 4m²

then \(\frac{Q}{t}=\mathrm{KA} \frac{\left(T_H-T_C\right)}{L}\)

Q = \(\frac{\left(209 \mathrm{~W} / \mathrm{m}^{\circ} \mathrm{C}\right)\left(4 \mathrm{~m}^2\right)\left(100^{\circ} \mathrm{C}-0^{\circ} \mathrm{C}\right)(5 \mathrm{sec})}{2 \mathrm{~m}}\)

= 209 KJ

NEET Physics Class 11 Chapter 8 Thermal Resistance To Conduction

If you are interested in insulating your house from cold weather or for that matter keeping the meal hot in your tiffin box, you are more interested in poor heat conductors, rather than good conductors. For this reason, the concept of thermal resistance R has been introduced.

For a slab of cross-section A, Lateral thickness L, and thermal conductivity K,

Resistance \(R=\frac{L}{K A}\)

In terms of R, the amount of heat flowing through a slab in steady-state (in time t)

⇒ \(\frac{Q}{t}=\frac{\left(T_H-T_L\right)}{R}\)

If we name as thermal current ir

then, \(i_T=\frac{T_H-T_L}{R}\)

This is mathematically equivalent to OHM’s law, with temperature donning the role of electric potential. Hence results derived from OHM’s law are also valid for thermal conduction.

Moreover, for a slab in the steady state, we have seen earlier that the thermal current it remains the same at each cross-section. This is analogous to Kirchoff’s current law in electricity, which can now be very conveniently applied to thermal conduction.

Question 2. Three identical rods of length 1m each, having cross-section area of 1cm2 each and made of Aluminium, copper, and steel respectively are maintained at temperatures of 12ºC, 4ºC, and 50ºC respectively at their separate ends. Find the temperature of their common junction.

[ KCu = 400 W/m-K , K_Al = 200 W/m-K , K_steel = 50 W/m-K ]

Answer:

⇒ \(\mathrm{R}_{\mathrm{Al}}=\frac{L}{K A}=\frac{1}{10^{-4} \times 209}=\frac{10^4}{209}\)

Similarly = \(R_{\text {steel }}=\frac{10^4}{46} \text { and } R_{\text {copper }}=\frac{10^4}{385}\)

Let the temperature of common junction = T then from Kirchoff’s Junction law.

i_Al + i_steel + iCu = 0

⇒ \(\frac{T-12}{R_{A I}}+\frac{T-51}{R_{\text {steel }}}+\frac{T-u}{R_{C u}}=0\)

⇒ (T – 12) 200 + (T – 50) 50 + (T – 4) 400 = 0

⇒ 4(T – 12) + (T – 50) + 8 (T – 4) = 0

⇒ 13T = 48 + 50 + 32 = 130

⇒ T = 10ºC

NEET Physics Class 11 Chapter 8 Growth Of Ice On Ponds

When atmospheric temperature falls below 0°C the water in the lake will start freezing. Let at any time t, the thickness of ice in the lake be y and atmospheric temperature is –θ°C. The temperature of water in contact with the lower surface of ice will be 0ºC.

the area of the lake = A

heat escaping through ice in time dt is

Now due to the escaping of this heat if the thickness of water in contact with the lower surface of ice freezes,

⇒ \(d Q_1=K A \frac{[0-(-\theta)]}{y} d t\)

dQ₂= mL = ρ(dy A)L [as m = ρV = ρA dy]

But as dQ₁= dQ₂, the rate of growth of ice will be

⇒ \(\frac{d y}{d t}=\frac{K \theta}{\rho L} \times \frac{1}{y}\)

and so the time taken by ice to grow a thickness y, \(t=\frac{\rho L}{K \theta} \int_0^y y \quad d y=\frac{1}{2} \frac{\rho L}{K \theta} \quad y^2\)

Time taken to double and triple the thickness will be in the ratio t₁: t₂: t₃:: 1² : 2²: 3², i.e., t₁: t₂: t₃:: 1 : 4: 9 and the time intervals to change thickness from 0 to y, from y to 2y and so on will be in the ratio Δt₁: Δt2: Δt₃: : (1² – 0² ) : (2² – 1² ) : (3² – 2² ), i.e., Δt₁: Δt₂: Δt₃:: 1 : 3: 5.

Class 11 NEET Heat Transfer Notes and Key Concepts

Can you now see how the following facts can be explained by thermal conduction?

1. In winter, iron chairs appear to be colder than the wooden chairs.
2. Ice is covered in gunny bags to prevent melting.
3. Woolen clothes are warmer.
4. We feel warmer in a fur coat.
5. Two thin blankets are warmer than a single blanket of double the thickness.
6. Birds often swell their feathers in winter.
7. A new quilt is warmer than an old one.
8. Kettles are provided with wooden handles.
9. Eskimos make double-walled ice houses.
10. Thermos flask is made double-walled.

NEET Physics Class 11 Chapter 8 Convection

When heat is transferred from one point to the other through the actual movement of heated particles, the process of heat transfer is called convection.

• In liquids and gases, some heat may be transported through conduction. But most of the transfer of heat in them occurs through the process of convection.
• Convection occurs through the aid of the earth’s gravity. Normally the portion of fluid at greater temperature is less dense, while that at lower temperature is denser. Hence hot fluids rise while colder fluids sink, accounting for convection. In the absence of gravity, convection would not be possible.
• Also, the anomalous behavior of water (its density increases with temperature in the range of 0-4ºC) gives rise to interesting consequences vis-a-vis the process of convection. One of these interesting consequences is the presence of aquatic life in temperate and polar waters. The other is the rain cycle.

Can you now see how the following facts can be explained by thermal convection?

1. Oceans freeze top-down and not bottom-up. (this fact is singularly responsible for the presence of aquatic life in temperate and polar waters.)
2. The temperature at the bottom of deep oceans is invariably 4ºC, whether it is winter or summer.
3. You cannot illuminate the interior of a lift in free fall or an artificial satellite of earth with a candle.
4. You can Illuminate your room with a candle.

NEET Physics Class 11 Chapter 8 Radiation

The process of the transfer of heat from one place to another place without heating the intervening medium is called radiation. The term radiation used here is another word for electromagnetic waves. These waves are formed due to the superposition of electric and magnetic fields perpendicular to each other and carry energy.

Properties of Radiation:

1. All objects emit radiation simply because their temperature is above absolute zero, and all objects absorb some of the radiation that falls on them from other objects.
2. Maxwell based on his electromagnetic theory proved that all radiations are electromagnetic waves and their sources are vibrations of charged particles in atoms and molecules.
3. More radiations are emitted at higher temperatures of a body and less at lower temperatures.
4. The wavelength corresponding to the maximum emission of radiation shifts from a longer wavelength to a shorter wavelength as the temperature increases. Due to this, the color of a body appears to be changing. Radiations from a body at NTP have predominantly infrared waves.
5. Thermal radiation travels with the speed of light and moves in a straight line.
6. Radiations are electromagnetic waves and can also travel through a vacuum.
7. Similar to light, thermal radiations can be reflected, refracted, diffracted, and polarized.
8. Radiation from a point source obeys the inverse square law (intensity α ).

Heat Transfer in NEET Physics: Important Formulas and Concepts

NEET Physics Class 11 Chapter 8 Prevost Theory Of Heat Exchange

According to this theory, all bodies radiate thermal radiation at all temperatures. The amount of thermal radiation radiated per unit of time depends on the nature of the emitting surface, its area, and its temperature. The rate is faster at higher temperatures.

• Besides, a body also absorbs part of the thermal radiation emitted by the surrounding bodies when this radiation falls on it. If a body radiates more than what it absorbs, its temperature falls.
• If a body radiates less than what it absorbs, its temperature rises. And if the temperature of a body is equal to the temperature of its surroundings it radiates at the same rate as it absorbs.

NEET Physics Class 11 Chapter 8 Perfectly Black Body And Black Body Radiation (Fery’s Black Body)

A perfectly black body absorbs all the heat radiations of whatever wavelength, is incident on it. It neither reflects nor transmits any of the incident radiation and therefore appears black whatever the color of the incident radiation.

In actual practice, no natural object possesses strictly the properties of a perfectly black body.

• But the lamp-black and platinum black are a good approximation of the black body. They absorb about 99 % of the incident radiation. The most simple and commonly used black body was designed by Fery.
• It consists of an enclosure with a small opening which is painted black from inside. The opening acts as a perfect black body.
• Any radiation that falls on the opening goes inside and has very little chance of escaping the enclosure before getting absorbed through multiple reflections. The cone opposite to the opening ensures that no radiation is reflected directly.

NEET Physics Class 11 Chapter 8 Absorption, Reflection, And Emission Of Radiations

Q = Q_r+ Q_t+ Q_a

1 = \(\frac{Q_r}{Q}+\frac{Q_t}{Q}+\frac{Q_a}{Q}\)

where r = reflecting power , a = absorptive power

and t = transmission power.

1. r = 0, t = 0, a = 1, perfect black body
2. r = 1, t = 0, a = 0, perfect reflector
3. r = 0, t = 1, a = 0, perfect transmitter

NEET Class 11 Heat Transfer: Summary and Key Points

Absorptive Power :

In particular, the absorptive power of a body can be defined as the fraction of incident radiation that is absorbed by the body.

a = \(\frac{\text { Energy absorbed }}{\text { Energy incident }}\)

As all the radiation incident on a black body is absorbed, a = 1 for a black body.

Emissive Power:

Energy radiated per unit time per unit area along the normal to the area is known as emissive power.

⇒ \(\frac{Q}{\Delta A \Delta t}\)

(Notice that, unlike absorptive power, emissive power is not a dimensionless quantity).

Spectral Emissive Power (Eλ) :

Emissive power per unit wavelength range at wavelength λ is known as spectral emissive power, Eλ. If E is the total emissive power and Eλ is spectral emissive power, they are related as follows,

⇒ \(\mathrm{E}=\int_0^{\infty} E_\lambda \mathrm{d} \lambda\)

and \(\frac{\mathrm{dE}}{\mathrm{d} \lambda}=\mathrm{E}_\lambda\)

Emissivity:

⇒ \(\mathrm{e}=\frac{\text { Emissive power of } \mathrm{a} \text { body at temperature } \mathrm{T}}{\text { Emissive power of } \mathrm{a} \text { black body at same temperature } \mathrm{T}}\)

⇒ \(\frac{E}{E_0}\)

NEET Physics Class 11 Chapter 8 Kirchoff’s Law

The ratio of the emissive power to the absorptive power for the radiation of a given wavelength is the same for all substances at the same temperature and is equal to the emissive power of a perfectly black body for the same wavelength and temperature.

⇒ \(\frac{E(\text { body })}{a(\text { body })}\) = E(black body)

Hence we can conclude that good emitters are also good absorbers.

Applications Of Kirchhoff’s Law

If a body emits strongly the radiation of a particular wavelength, it must also absorb the same radiation strongly.

1. Let a piece of china with some dark painting on it be first heated to nearly 1300 K and then examined in a dark room. It will be observed that the dark paintings appear much brighter than the white portion. This is because the paintings being better absorbers also emit much more light.
2. The silvered surface of a thermos flask does not absorb much heat from outside. This stops ice from melting quickly. Also, the silvered surface does not radiate much heat from the inside. This prevents hot liquids from becoming cold quickly.
3. A red glass appears red at room temperature. This is because it absorbs green light strongly. However, if it is heated in a furnace, it glows with green light. This is because it emits green light strongly at a higher temperature.
4. When white light is passed through sodium vapors and the spectrum of transmitted light is seen, we find two dark lines in the yellow region. These dark lines are due to absorption of radiation by sodium vapors which it emits when heated.

Fraunhofer lines are dark lines in the spectrum of the sun. When white light emitted from the central core of the sun (photosphere) passes through its atmosphere (chromosphere) radiations of those wavelengths will be absorbed by the gases present there which they usually emit (as a good emitter is a good absorber) resulting in dark lines in the spectrum of sun.

At the time of a solar eclipse, direct light rays emitted from the photosphere cannot reach the earth and only rays from the chromosphere can reach the earth’s surface. At that time we observe bright Fraunhofer lines.

Heat Transfer NEET Physics Notes: Detailed Explanation

NEET Physics Class 11 Chapter 8 Nature Of Thermal Radiations : (Wien’s Displacement Law)

From the energy distribution curve of black body radiation, the following conclusions can be drawn:

1. The higher the temperature of a body, the higher the area under the curve i.e. more amount of energy is emitted by the body at a higher temperature.
2. The energy emitted by the body at different temperatures is not uniform. For both long and short wavelengths, the energy emitted is very small.
3. For a given temperature, there is a particular wavelength (λ_m) for which the energy emitted (Eλ) is maximum.
4. With an increase in the temperature of the black body, the maxima of the curves shift towards shorter wavelengths.

From the study of the energy distribution of black body radiation discussed above, it was established experimentally that the wavelength (λ_m) corresponding to the maximum intensity of emission decreases inversely with an increase in the temperature of the black body. i.e.

λ_m ∝ or λ_m T = b

This is called Wien’s displacement law.

Here b = 0.282 cm-K, is the Wien’s constant.

NEET Physics Chapter 8 Heat Transfer: Study Material

Question 1. Solar radiation is found to have an intensity maximum near the wavelength range of 470 nm. Assuming the surface of the sun to be perfectly absorbing (a = 1), calculate the temperature of the solar surface.
Solution :

Since a =1, the sun can be assumed to be emitting as a black body from Wien’s law for a black body

λ_m. T = b

⇒ T = \(\frac{0.282(\mathrm{~cm}-\mathrm{K})}{\left(470 \times 10^{-7} \mathrm{~cm}\right)}\)

= ~ 6125 K.

NEET Physics Class 11 Chapter 8 Slabs In Parallel And Series

Slabs In Series (in steady state)

Consider a composite slab consisting of two materials having different thicknesses L₁ and L₂, different cross-sectional areas A₁ and A_2, and different thermal conductivities K₁ and K₂. The temperature at the outer surface at the ends is maintained at T_H and T_C, and all lateral surfaces are covered by an adiabatic coating.

Let the temperature at the junction be T, since a steady state has been achieved thermal current through each slab will be equal. Then thermal current through the first slab.

⇒ \(\mathrm{i}=\frac{Q}{t}=\frac{T_H-T}{R_1}\)

or \(\mathrm{T}_{\mathrm{H}}-\mathrm{T}=\mathrm{iR} \mathrm{R}_1\)……(1)

and that through the second slab,

⇒ \(\mathrm{i}=\frac{Q}{t}\)

⇒ \(\frac{T-T_C}{R_2}\)

or \(\mathrm{T}-\mathrm{T}_{\mathrm{c}}=\mathrm{i} \mathrm{R}_2\)…..(2)

adding eqn. (1) and eqn (2)

T_H– T_L= (R₁+ R₂) i

or \(\mathrm{i}=\frac{T_H-T_C}{R_1+R_2}\)

Thus these two slabs are equivalent to a single slab of thermal resistance R1+ R2. If more than two slabs are joined in series and are allowed to attain a steady state, then equivalent thermal resistance is given by

R = R₁+ R₂+ R₃+ ……. (3)

NEET Physics Class 11 Chapter 8 Slabs in Parallel and Series

Question 1. The figure shows the cross-section of the outer wall of a house built in a hill resort to keep the house insulated from the freezing temperature outside. The wall consists of teak wood of thickness L₁ and brick of thickness (L₂= 5L₁), sandwiching two layers of an unknown material with identical thermal conductivities and thickness. The thermal conductivity of teak wood is K1and that of brick is (K₂= 5K). Heat conduction through the wall has reached a steady state with the temperature of three surfaces being known. (T₁= 25ºC, T₂= 20ºC and T₅= –20ºC). Find the interface temperature T₄ and T₃.

Answer: Let the interface area be A. Then thermal resistance of wood,

⇒ \(\mathrm{R}_1=\frac{L_1}{K_1 A}\)

and that of brick wall \(\mathrm{R}_2=\frac{L_2}{K_2 A}=\frac{5 L_1}{5 K_1 A}=\mathrm{R}_1\)

Let the thermal resistance of each sandwiched layer = R. Then the above wall can be visualized as a circuit

thermal current through each wall is the same.

Hence \(\frac{25-20}{R_1}=\frac{20-T_3}{R}\)

⇒ \(\frac{T_3-T_4}{R}=\frac{T_4+20}{R_1}\)

⇒ 25 – 20 = T4+ 20

⇒ T₄= –15ºC

also, 20 – T₃= T₃– T₄

⇒ \(\mathrm{T}_3=\frac{20+T_4}{2}\)

= 2.5ºC

Slabs in Parallel and Series NEET Class 11 Physics Study Notes

Question 2. In example 3, K₁= 0.125 W/m–ºC, K₂= 5K₁= 0.625 W/m–ºC, and the thermal conductivity of the unknown material is K = 0.25 W/mºC. L₁= 4cm, L₂= 5L₁= 20cm and L = 10cm. If the house consists of a single room with a total wall area of 100 m₂, then find the power of the electric heater being used in the room.
Answer:

R₁ = R₂ = \(\frac{\left(4 \times 10^{-2} \mathrm{~m}\right)}{\left(0.125 \mathrm{w} / \mathrm{m}-{ }^{\circ} \mathrm{C}\right)\left(100 \mathrm{~m}^2\right)}\)

= 32 × 10^-4ºC/w

R = \(\frac{\left(10 \times 10^{-2} \mathrm{~m}\right)}{\left(0.25 \mathrm{~W} / \mathrm{m}-{ }^{\circ} \mathrm{C}\left(100 \mathrm{~m}^2\right)\right.}\)

= 40 × 10^-4ºC/w

The equivalent thermal resistance of the entire wall = R₁+ R₂+ 2R = 144×10^-4ºC/W

∴ Net heat current, i.e. amount of heat flowing out of the house per second

⇒ \(\frac{T_H-T_C}{R}=\frac{25^{\circ} \mathrm{C}-\left(-20^{\circ} \mathrm{C}\right)}{144 \times 10^{-40} \mathrm{C} / \mathrm{w}}=\frac{45 \times 10^4}{144}\)watt = 3.12 Kwatt

Hence the heater must supply 3.12 kW to compensate for the outflow of heat.

Slabs In Parallel :

Consider two slabs held between the same heat reservoirs, their thermal conductivities K₁ and K_2, and cross-sectional areas A₁ and A₂ Heat reservoir

then \(\mathrm{R}_1=\frac{L}{K_1 A_1}\), \(\mathrm{R}_2=\frac{L}{K_2 A_2}\)

thermal current through slab 1 and that through slab 2 Net heat current from the hot to cold reservoir

⇒ \(i_1=\frac{T_H-T_C}{R_1}\)

Comparing with i = \(i_1+i_2=\left(T_H-T_C\right)\left(\frac{1}{R_1}+\frac{1}{R_2}\right)\),

⇒ \(\mathrm{i}=\frac{T_H-T_C}{R_{e q}}\) we get, \(\)

⇒ \(\frac{1}{R_{c q}}=\frac{1}{R_1}+\frac{1}{R_2}\)

If more than two rods are joined in parallel, the equivalent thermal resistance is given by

⇒ \(\frac{1}{R_{e q}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\)

Slabs in Parallel and Series: Key Formulas for NEET Physics

Question 3. Figure shows a copper rod joined to a steel rod. The rods have equal length and equal cross sectional area. The free end of the copper rod is kept at 0ºC and that of the steel rod is kept at 100ºC. Find the temperature of the junction of the rod. Conductivity of copper = 390 W/mºC Conductivity of steel = 46 W/m ºC 0ºC Copper Steel 100ºC
Answer:

Heat current in first rod (copper) = \(\frac{390 \times A(A-\theta)}{\ell}\)

Here θ is the temperature of the junction and A and l are the area and length of the copper rod.

Heat current in second rod (steel) = \(\frac{46 \times A(\theta-100)}{\ell}\)

In series combination. heat current remains the same. So,

⇒ \(\frac{390 \times A(0-\theta)}{\ell}\)

⇒ \(\frac{46 \times A(\theta-100)}{\ell}\)

⇒ -390 θ = 46θ-4600

⇒ 436 θ = 4600 θ = 10.6ºC

Question 4. An aluminum rod and a copper rod of equal length 1m and cross-sectional area 1cm₂ are welded together as shown in the figure. One end is kept at a temperature of 20ºC and the other at 60ºC. Calculate the amount of heat taken out per second from the hot end. The thermal conductivity of aluminum is 200 W/mºC and of copper is 390 W/mºC.

⇒ \(20^{\circ} \mathrm{C} \begin{array}{|l|}
\hline \text { Aluminium } \\
\hline \text { Copper } \\
\hline\end{array} 60^{\circ} \mathrm{C}\)

Answer: Heat current through the \(\frac{200 \times\left(1 \times 10^{-4}\right)}{1}\) = (60-20)

Heat current through the copper rod = \(\frac{390 \times\left(1 \times 10^{-4}\right)}{1}\) . (60-20)

Total heat = 200 × 10–4 × 40 + 390 × 10–4× 40 = 590 × 40 × 10–4= 2.36 Joule

Question 5. The three rods shown in the figure have identical geometrical dimensions. Heat flows from the hot end at the rate of 40W in an arrangement

1. Find the rate of heat flow when the rods are joined in arrangement
2. The thermal conductivity of aluminum and copper are 200 W/mºC and 400 W/mºC respectively.

Answer:

In the arrangement

The three rods are joined in series. The rate of flow of heat,

⇒ \(\frac{d \theta}{d t}=\frac{K A\left(\theta_1-\theta_2\right)}{\ell}=\frac{\theta_1-\theta_2}{R}\)

But, R = R₁+ R₂+ R₃[In series]

∴ 40 = \(\frac{100-0}{R_1+R_2+R_1}\)

40 = \(\frac{100}{\frac{\ell}{K_1 A}+\frac{\ell}{K_2 A}+\frac{\ell}{K A}}\)

40 = \(\frac{100}{\frac{\ell}{A}\left[\frac{2}{K_1}+\frac{1}{K_2}\right]}\)

⇒ \(\frac{\ell}{A}\left[\frac{2}{200}+\frac{1}{400}\right]=\frac{100}{40}\)

⇒ \(\frac{\ell}{A}\)

= 200 per m

In Figure two rods are all in parallel and the resultant of both is in series with the first rod

∴ \(\frac{d Q}{d t}=\frac{\theta_1-\theta_2}{R}\)

But R = \(\mathrm{R}_1+\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}}\)

⇒ \(\frac{d Q}{d t}=\frac{100-0}{\frac{\ell}{K_1 A}+\frac{1}{\frac{K_1 A}{\ell}+\frac{K_2 A}{\ell}}}\)

⇒ \(\frac{100-0}{\frac{\ell}{A}\left[\frac{1}{K_1}+\frac{1}{K_1+K_2}\right]}\)

⇒ \(\frac{600 \times 100}{200 \times 4}\)

= 75W

Question 6. A metal rod of length 20cm and diameter 2 cm is covered with a nonconducting substance. One of its ends is maintained at 100ºC while the other end is put at 0ºC. It is found that 25 g of ice melts in 5 min. Calculate the coefficient of thermal conductivity of the metal. Latent heat of ice = 80 cal gram^-1
Answer:

Here, the length of the rod,

Δx = 20 cm = 20 × 10^-2m

Diameter = 2cm,

Radius = r = 1 cm = 10^-2m

Area of cross-section

a = πr²= π(10^-2)²= π × 10^-4 sq. m

ΔT = 100 – 0 = 100ºC

Mass of ice melted, m = 25g

As L = 80 cal/ g

∴ Heat conducted, ΔQ = mL = 25 × 80 = 2000 cal = 2000 × 4.2 J

Δt = 5 min = 300 s

From = \(\frac{\Delta Q}{\Delta t}=\mathrm{KA} \frac{\Delta T}{\Delta x}\)

K = \(\frac{2000 \times 4.2 \times 20 \times 10^{-2}}{300 \times 10^{-4} \pi \times 100}\)

= 1.78Js^-1m^-1ºC^-1

Slabs in Parallel and Series NEET Physics: Important Questions

Question 7. Two thin concentric shells made from copper with radius r₁ and r₂(r₂> r₁) have a material of thermal conductivity K filled between them. The inner and outer spheres are maintained at temperatures T_H and T_C respectively by keeping a heater of power P at the center of the two spheres. Find the value of P.
Answer:

Heat flowing per second through each cross-section of the sphere = P = i. Thermal resistance of the spherical shell of radius x and thickness dx,

dR = \(\frac{d x}{K .4 \pi x^2}\)

⇒ \(\mathrm{R}=\int_n^2 \frac{d x}{4 \pi x^2 \cdot K}=\frac{1}{4 \pi K}\left(\frac{1}{r_1}-\frac{1}{r_2}\right)\)

thermal current i = P = \(\frac{T_H-T_C}{R}=\frac{4 \pi K\left(T_H-T_C\right) r_1 r_2}{\left(r_2-r_1\right)}\)

NEET Physics Class 11 Chapter 8 Stefan-Boltzmann’S Law

According to this law, the amount of radiation emitted per unit of time from area A of a black body at absolute temperature T is directly proportional to the fourth power of the temperature.

u = σ A T⁴

where σ is Stefan’s constant = 5.67 x 10^-8 W/m² k⁴

A body that is not black absorbs and hence emits less radiation than that given by the equation.

For such a body, u e A T⁴

where e = emissivity (which is equal to absorptive power) which lies between 0 to 1

With the surroundings of temperature T0, net energy radiated by an area A per unit of time.

⇒ \(\Delta u=u-u_0=e \quad \sigma A\left(T^4-T_0^4\right)\)

NEET Physics Class 11 Chapter 8 Stefan-Boltzmann’s Law Notes

Question 2. A black body at 2000K emits radiation with λ_m= 1250 nm. If the radiation coming from the star SIRUS λ_mis 71 nm, then the temperature of this star is …….
Answer:

Using Wien’s displacement law

⇒ \(\frac{T_2}{T_1}=\frac{\left(\lambda_m\right)_1}{\left(\lambda_m\right)_2}\)

∴ \(\mathrm{T}_2=\frac{2000 \times 1250 \times 10^{-3}}{71 \times 10^{-9}}\)

T₂ = 35.211 K

Question 3. At 1600 K maximum radiation is emitted at a wavelength of 2µm. Then the corresponding wavelength at 2000 K will be –
Answer:

Using T₁= T₂

∴ \(\frac{\lambda_{m_2} T_1}{T_2}\)

∴ \(\frac{2 \times 10^{-6} \times 1600}{2000}\)

= 1.6 µm

Question 4. If the temperature of a body is increased by 50%, then the increase in the amount of radiation emitted by it will be
Answer: Percentage increase in the amount of radiation emitted

∴ \(\frac{E_2-E_1}{E_1} \times 100\)

⇒ \(\frac{\left(1.5 T_1\right)^4-T_1^4}{T_1^4} \times 100\)

⇒ \(\frac{E_2-E_1}{E_1} \times 100\)

⇒ \(\left[(1.5)^4-1\right] \times 10\)

⇒ \(\frac{E_2-E_1}{E_1} \times 100\)

= 400%

Class 11 NEET Physics Stefan-Boltzmann’s Law Study Notes

Question 5. A blackened platinum wire of length 5cm and perimeter 0.02 cm is maintained at a temperature of 300K. Then at what rate the wire is losing its energy? (Take σ = 57 × 10^-8units)
Solution :

The rate of radiation heat loss is

⇒ \(\frac{d Q}{d t}=\mathrm{eA} \sigma \mathrm{T}^4 \text { (watts) }\)

for blackened surface e = 1

and A = (2πr)l = P_erimeter × length

∴ A = 0.02 × 5 × 10^-4

Thus

⇒ \(\frac{d Q}{d t}\)= 0.02 × 5 × 10^-4× 5.7 × 10^-8× (300)⁴

⇒ \(\frac{d Q}{d t}\)= 46.2W

Stefan-Boltzmann’s Law for NEET Physics: Practice Problems

Question 6. A hot black body emits energy at the rate of 16 J m^-2 s^-1 and its most intense radiation corresponds to 20,000 Å. When the temperature of this body is further increased and its most intense radiation corresponds to 10,000 Å, then find the value of energy radiated in Jm^-2 s^-1.
Answer:

Wein’s displacement law is :

λ_m.T = b i.e. \(\mathrm{T} \propto \frac{1}{\lambda_m}\)

Here, λ_m becomes half.

∴ Temperature doubles.

Also E = σT⁴

⇒ \(\frac{e_1}{e_2}=\left(\frac{T_1}{T_2}\right)^4\)

⇒ \(\mathrm{E}_2=. \mathrm{E}_1 \frac{e_1}{e_2}=\left(\frac{T_1}{T_2}\right)^4 \mathrm{e}_1=(2)^4 .16\)

= 16.16 = 256 J m^-2 s^-1

NEET Physics Class 11 Chapter 8 Newton’s Law Of Cooling

For a small temperature difference between a body and its surroundings, the rate of cooling of the body is directly proportional to the temperature difference and the surface area exposed.

⇒ \(\frac{d Q}{d t} \propto\left(\theta-\theta_0\right)\), where θ and θ0are temperature corresponding to object and surroundings.

From above expression \(\frac{d \theta}{d t}=-k\left(\theta-\theta_0\right)\)

This expression represents Newton’s law of cooling. It can be derived directly from Stefan’s law, which gives,

⇒ \(\mathrm{k}=\frac{4 \mathrm{e} \sigma \theta_0^3}{\mathrm{mc}} \mathrm{A}\)

Now \(\frac{d \theta}{d t}=-k\left[\begin{array}{ll}
\theta & \left.-\theta_0\right]
\end{array}\right.\)

⇒ \(\int_{\theta_i}^{\theta_f} \frac{d \theta}{\left(\theta-\theta_0\right)}=\int_0^t-k d t\)

where \(\theta_{\mathrm{i}}=\ln \frac{\left(\theta_f-\theta_0\right)}{\left(\theta_i-\theta_0\right)}\) initial temperature of object and

θf = final temperature of object.

⇒ –kt ⇒ (θf− θ0) = (θi– θ0) e–kt

⇒ θf= θ0+ (θi– θ0) e –kt

Limitations of Newton’s Law of Cooling:

1. The difference in temperature between the body and surroundings must be small
2. The loss of heat from the body should be by radiation only.
3. The temperature of the surroundings must remain constant during the cooling of the body.

NEET Physics Class 11 Chapter 8 Newton’s Law of Cooling

Approximate Method For Applying Newton’s Law Of Cooling

Sometimes when we need only approximate values from Newton’s law, we can assume a constant rate of cooling, which is equal to the rate of cooling corresponding to the average temperature of the body during the interval.

⇒ \(\left\langle\frac{d \theta}{d t}\right\rangle-\mathrm{k}\left(<\theta>-\theta_0\right)\)

If θ_i and θ_f are the initial and final temperatures of the body then,

⇒ \(<\theta>=\frac{\theta_i+\theta_f}{2}\)

Remember equation is only an approximation and an equation must be used for exact values.

Comparison Of Specific Heat Of Two Liquids Using Newton’s Law Of Cooling:

If an equal volume of two liquids of densities and specific heats ρ₁, s_1, and ρ₂, s₂ respectively are filled in calorimeters having the same surface area and finish, cool from the same initial temperature θ₁ to the same final temperature θ₂ with the same temperature of surroundings, i.e., θ₀, in time intervals t₁and t₂ respectively and water equivalent of calorimeter is w. According to Newton’s law of cooling

⇒ \(\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\text {liq }}=\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\text {water }}\)

⇒ \(\frac{\left(w+m_1 s_1\right)\left(\theta_1-\theta_2\right)}{t_1}\)

⇒ \(\frac{\left(w+m_2 s_2\right)\left(\theta_1-\theta_2\right)}{t_2}\)

or \(\frac{w+m_1 s_1}{t_1}=\frac{w+m_2 s_2}{t_2}\)

If the water equivalent of calorimeter w is negligible then

⇒ \(\frac{m_1 s_1}{t_1}=\frac{m_2 s_2}{t_2}\)

So, \(\frac{m_1 s_1}{m_2 s_2}=\frac{t_1}{t_2} \quad \text { or } \frac{\rho_1 s_1}{\rho_2 s_2}=\frac{t_1}{t_2}\)(v₁= v₂, volume are equal) with the help of this eqn. we can find the specific heat of the liquid.

Question 1. A body at a temperature of 40ºC is kept in a surrounding constant temperature of 20ºC. It is observed that its temperature falls to 35ºC in 10 minutes. Find how much more time will it take for the body to attain a temperature of 30ºC.
Answer:

Δθ_f = Δθ_ie^-kt

for the interval in which temperature falls from 40 to 35º C.

(35 – 20) = (40 – 20) e^-k.10

⇒ \(\mathrm{e}^{-10 \mathrm{k}}=\frac{3}{4}\)

⇒ \(K=\frac{\ln \frac{4}{3}}{10}\)

for the next interval (30 – 20) = (35 – 20)e^-kt

⇒ \(\mathrm{e}^{-10 \mathrm{k}}=\frac{2}{3}\)

⇒ \(\mathrm{kt}=\mathrm{ln} \frac{3}{2}\)

⇒ \(\frac{\left(\ln \frac{4}{3}\right) t}{10}=\ell n\)

⇒ \(\mathrm{t}=10 \frac{\left(\ln \frac{3}{2}\right)}{\left(\ln \frac{4}{3}\right)}\) minute = 14.096 min

Alternative : (by approximate method) for the interval in which temperature falls from 40 to 35ºC 40 35

<θ> = \(\frac{40+35}{2}\) = 37.5ºC

⇒ \(\left\langle\frac{d \theta}{d t}\right\rangle=-\mathrm{k}\left(<\theta>-\theta_0\right)\)

⇒ \(\frac{\left(35^{\circ} \mathrm{C}-40^{\circ} \mathrm{C}\right)}{10(\mathrm{~min})}\)

= –K(37.5ºC – 20ºC)

K = \(\frac{1}{35}\)(min^-1)

for the interval in which temperature falls from 35ºC to 30ºC

<θ> = \(\frac{\left(30^{\circ} \mathrm{C}-35^{\circ} \mathrm{C}\right)}{t}\) = (32.5º) C

= – K(32.5ºC – 20ºC)

⇒ required time, t = \(\)

⇒ \(\frac{5}{12.5} \times 35 \mathrm{~min}\)

= 14 min

Newton’s Law of Cooling NEET Class 11 Notes

Question 2. Two liquids of the same volume are cooled under the same conditions from 65ºC to 50ºC. The time taken is 200sec and 480sec. If the ratio of their specific heats is 2 : 3 then find the ratio of their densities. (neglect the water equivalent of a calorimeter)
Answer:

From Newton’s law of cooling

⇒ \(\left(\frac{m_1 s_1+w_1}{t_1}\right)\left(\theta_1-\theta_2\right)\)

⇒ \(\left(\frac{m_2 s_2+w_2}{t_2}\right)\left(\theta_1-\theta_2\right)\)

here w₁= w₂= 0

⇒ \(\frac{m_1 s_1}{t_1}=\frac{m_2 s_2}{t_2}\)

⇒ \(\frac{V d_1 s_1}{t_1}=\frac{V d_2 s_2}{t_2}\)

⇒ \(\frac{d_1}{d_2}=\frac{t_1 s_2}{t_2 s_1}\)

⇒ \(\frac{200}{480} \times \frac{3}{2}\)

⇒ \(\frac{5}{8}\)

Question 3. A calorimeter of water equivalent to 5 × 10^-3 kg contains 25 × 10^-3 kg of water. It takes 3 minutes to cool from 28°C to 21°C. When the same calorimeter is filled with 30 × 10^-3 kg of turpentine oil then it takes 2 minutes to cool from 28°C to 21°C. Find out the specific heat of turpentine oil.
Answer:

⇒ \(\mathrm{R}_{\text {water }}=\mathrm{R}_{\text {turpentine }} \frac{\left(m_1+w\right)}{t_1}=\frac{\left(m_2 s_2+w\right)}{t_2}\)

or \(\frac{\left(25 \times 10^{-3}+5 \times 10^{-3}\right)}{3}=\frac{30 \times 10^{-3} s_2+5 \times 10^{-3}}{2}\)

10 = \(\frac{30 s_2+5}{2}, 20\)

⇒ \(30 s_2+5\)

∴ specific heat of turpentine s₂= 1/2 = 0.5 kcal/kg/°C

Question 4. A man, the surface area of whose skin is 2m², is sitting in a room where the air temperature is 20°C. If his skin temperature is 28°C, find the rate at which his body loses heat. The emissivity of his skin = 0.97.
Answer:

Absolute room temperature (T₀) = 20 + 273 = 293 K

Absolute skin temperature (T) = 28 + 273 = 301 K

Rate of heat loss = σ e A (T⁴ – T₀⁴)

= 5.67 × 10^-8 × 0.97 × 2 × {(301)⁴ – (293)⁴} = 92.2 W

Question 5. Compare the rate of loss of heat from a metal sphere of the temperature 827°C, with the rate of loss of heat from the same sphere at 427 °C, if the temperature of surroundings is 27°C.
Answer:

Given : T₁= 827 °C = 1100 K, T₂= 427 °C = 700 K and T₀= 27 °C = 300 K

According to Steafan’s law of radiation, \(\frac{d Q}{d t}\)

⇒ \(\sigma \mathrm{Ae}\left(\mathrm{T}^4-\mathrm{T}_0{ }^4\right)\)

⇒ \(\frac{\left(\frac{d Q}{d t}\right)_1}{\left(\frac{d Q}{d t}\right)_2}=\frac{\left(T_1^4-T_0^4\right)}{\left(T_2^4-T_0^4\right)}\)

⇒ \(\frac{\left[(1100)^4-(300)^4\right]}{\left[(700)^4-(300)^4\right]}\) = 6.276

⇒ \(\left(\frac{d Q}{d t}\right)_1:\left(\frac{d Q}{d t}\right)_2\) = 6.276:1

Question 6. Two spheres of the same material have radii of 6 cm and 9 cm respectively. They are heated to the same temperature and allowed to cool in the same enclosure. Compare their initial rates of emission of heat and initial rates of fall of temperature.
Answer:

Given : r₁= 6 cm r₂= 9 cm,

∴ \(\frac{r_1}{r_2}=\frac{2}{3}\)

According to Stefan’s law of radiation, the rate of emission of heat by an ordinary body,

⇒ \(\mathrm{R}_{\mathrm{h}}=\left(\frac{d Q}{d t}\right)=\sigma \mathrm{AeT}^4\)

or \(R_h \propto r^2\) (A=4σr²)

⇒ \(\frac{R_{h 1}}{R_{h 2}}=\left(\frac{r_1}{r_2}\right)^2=\left(\frac{2}{3}\right)^2=\frac{4}{9}\)

⇒ \(\frac{R_{F T 1}}{R_{F T 2}}=\frac{r_2}{r_1}=\frac{3}{2}\)

[Rate of fall of temp. \(\mathrm{R}_{\mathrm{FT}}, \frac{d \theta}{d t}, \frac{\sigma A e\left(T^4-T_0^4\right)}{m s J}\) = or \(R_{F T} \propto \frac{A}{m} \text { or } R_{F T} \propto \frac{1}{r}\)]

∴ Initial rates of emission of heat are in the ratio 4: 9 and initial rates of fall of temperature are in the ratio 3: 2.

Question 7. The filament of an evacuated light bulb has a length of 10 cm, a diameter of 0.2 mm, and an emissivity of 0.2, calculate the power it radiates at 2000 K. (σ = 5.67 × 10^-8 W/m² K⁴)
Answer:

l = 10 cm = 0.1 m, d = 0.2 mm, r = 0.1 mm = 1 × 10^-4 m, e = 0.2, T = 2000 K, σ = 5.67 × 10^-8 W/m² K⁴

According to Stefan’s law of radiation, the rate of emission of heat for an ordinary body (filament), E = σAeT⁴ = σ(2 π r l) eT⁴

= 5.67 × 10^-8 × 2 × 3.14 × 1 × 10^-4 × 0.1 × 0.2 × (2000)⁴

= 11.4 W

∴ Power radiated by the filament = 11.4 W [A = 2πrl]

NEET Physics Chapter 8 Newton’s Law of Cooling Study Notes

Question 8. The energy radiated from a black body at a temperature of 727°C is E. By what factor does the radiated energy increase if the temperature is raised to 2227°C?
Solution :

⇒ \(\frac{E_2}{E_1}=\left[\frac{T_2}{T_1}\right]^4\)

⇒ \(\left[\frac{2227+273}{727+273}\right]^4=\left[\frac{2500}{1000}\right]^4\)

= 39

Question 9. An ice box made of 1.5 cm thick styrofoam has dimensions of 60 cm × 30cm. It contains ice at 0ºC and is kept in a room at 40ºC. Find the rate at which the ice is melting. Latent heat of fusion of ice = 3.36 × 10⁵J/kg. and thermal conductivity of stryrofoam = 0.4 W/m-ºC.
Answer:

The total surface area of the walls

= 2(60 cm × 60 cm + 60 cm × 30 cm + 60 cm × 30 cm)

= 1.44 m²

The thickness of the wall = 1.5 cm = 0.015m

The rate of heat flow into the box is

⇒ \(\frac{\Delta Q}{\Delta t}=\frac{K A\left(\theta_1-\theta_2\right)}{x}\)

⇒ \(\frac{\left(0.04 \mathrm{~W} / \mathrm{m}-{ }^{\circ} \mathrm{C}\right)\left(1.44 \mathrm{~m}^2\right)\left(40^{\circ} \mathrm{C}\right)}{0.015 \mathrm{~m}}\)

= 154 W.

The rate at which the ice melts is = 0.46 g/s

Question 10. A black body emits 10 watts per cm² at 327ºC. The sun radiates 10⁵ watts per cm². Then what is the temperature of the sun?
Answer:

∴ \(\frac{E_{\text {sun }}}{E_{\text {body }}}=\left(\frac{E_{\text {sun }}}{E_{\text {body }}}\right)^4\)

∴ \(\frac{T_{\text {sun }}}{T_{\text {body }}}=\left(\frac{10^5}{10}\right)^{1 / 4}\)

= 6000 K

∴ T_sun = 6000 K

Question 11. A bulb made of a tungsten filament of a surface is 0.5 cm² is heated to a temperature of 3000k when operated at 220V. The emissivity of the filament is e = 0.35 and take σ = 5.7 × 10^-8 mks units. Then the wattage of the bulb is …..(calculate)
Answer:

The emissive power watt/m² is

E = eσT⁴

Therefore the power of the bulb is

P = E x area (Watts)

∴ P = eσT⁴A

∴ P = 0.35 × 0.5 × 10^-4× 5, 7 × 10^-8× (3000)⁴

⇒ P = 80.8 W

Question 12. In the above example, if the temperature of the filament falls to 2000k due to a drop in mains voltage, then what will be the wattage of the bulb?
Answer:

Now the power of the bulb will be such that

⇒ \(\frac{P_2}{P_1}=\left(\frac{T_2}{T_1}\right)^4\)

Thus \(P_2=P_1 \times\left(\frac{2}{3}\right)^4\)

∴ \(P_2=80.8 \times \frac{16}{81}\)

Thus P₂= P1x

∴ P₂= 80.8 x

⇒ P₂= 15.96

Question 13. A liquid takes 30 seconds to cool from 95ºC to 90ºC and 70 seconds to cool from 55 to 50ºC. Find the room temperature and the time it will take to cool from 50ºC to 45ºC
Answer:

From the first date

⇒ \(\frac{95-90}{30}=\mathrm{k}\left(\frac{95 \times 90}{2}-\theta_0\right)=\mathrm{k}\)….(1)

From the second data \(\frac{55-50}{70}=\mathrm{k}\left(\frac{55-50}{0}-\theta_0\right)\) = k….(2)

Dividing (1) and (2) we get

⇒ \(\frac{7}{3}=\frac{92.5-\theta_0}{52.5-\theta_0}\)

⇒ \(\theta_0=22.5^{\circ}\) ……….(3)

Let the time taken in cooling from 50ºC to 45ºC is t, then

⇒ \(\frac{50-45}{t}\)

⇒ \(\mathrm{k}\left(\frac{50-45}{2}-\theta_0\right)\)….(4)

Using θ₀= 22.5ºC, and dividing (1) by (2) we get

⇒ \(\frac{t}{30}=\left(\frac{92.5-22.5}{47.5-22.5}\right)\)

t = 84 sec

Class 11 NEET Physics Newton’s Law of Cooling Key Concepts

Question 14. A blackened metal disc is held normal to the sun’s rays, Both of its surfaces are exposed to the atmosphere if the distance of the earth from the sun is 216 times the radius of the sun and the temperature of the sun is 6000K, the temperature of the disc in the steady state will be
Answer:

In the steady state, the heat received from the sun will be equal to the heat radiated out. Heat received from the sun will be on one side only and it will radiate from both sides.

∴ \(\mathrm{A} \sigma\left(\frac{R s}{d}\right)^2 \mathrm{~T}^4\)=2AσT⁴

⇒ \(\frac{R s}{d}=\frac{1}{216}\)

∴ T′ = \(\frac{T}{(216)^{1 / 2} 2^{1 / 4}}\)

⇒ \(\frac{6000}{14.7 \times 1.189}\)

= 345K

∴ T′ = 70ºC

Question 15. Behaving like a black body sun emits maximum radiation at wavelength 0.48µm. The mean radius of the sun is 6.96 × 10⁸m. Stefan’s constant is 5.67 × 10^-8wm^-2k^-4and wien’s constant is 0.293 cm-k. The loss of mass per second by the emission of radiation from the sun is
Answer: Using Wien’s law

T = T = \(\frac{b}{\lambda_m}=\frac{0.293 \times 10^{-2}}{0.48 \times 10^{-6}}\)

= 6104 K

Energy given out by sun per second

= AσT⁴=4π (6.96 × 10⁸)²× 5.67 × 10^-8(6104)⁴

⇒ 49.285 × 10²⁵J

Loss of mass per second

m = \(\frac{E}{c^2}=\frac{49.285 \times 10^{25}}{9 \times 10^{16}}\)

m = 5.4×10⁹kg/s

Newton’s Law of Cooling Practice Problems for NEET Physics

Question 16. 50g of water and an equal volume of alcohol (relative density 0.8) are placed one after the other in the same calorimeter. They are found to cool from 60ºC to 50ºC in 2 minutes and 1 minute respectively if the water equivalent of the calorimeter is 2g then what is the specific heat of the alcohol?
Answer:

Given t_w= 2min, telco = 1 min

m_w= 50g, malco = 50 × 0.8 = 40g

S_w= 1 in cgs units, w = 2g

Therefore,

⇒ \(\mathrm{S}_{\mathrm{alco}}=\frac{1}{m_{\text {alco }}}\left\{\frac{t_{\text {alco }}}{t_w}\left[m_w+W\right]-W\right\}\)

⇒ \(\mathrm{S}_{\mathrm{alco}}=\frac{1}{40}\left\{\frac{1}{2}[50+2]-2\right\}\)

⇒ \(\mathrm{S}_{\mathrm{alco}}=\frac{24}{50}\)

S_alco = 0.6 cgs units = 0.6 cal/gºC

NEET Physics Class 11 Chapter 8 Heat Transfer Multiple Choice Question And Answers

Question 1. A wall has two layers A and B, each made of different material. Both the layers have the same thickness. The thermal conductivity for A is twice that of B. Under steady state, the temperature difference across the whole wall is 36°C. Then the temperature difference across layer A is

1. 6°C
2. 12°C
3. 18°C
4. 24°C

Answer: 2. 12°C

Question 2. A heat flux of 4000 J/s is to be passed through a copper rod of length 10 cm and an area of cross-section 100 sq. cm. The thermal conductivity of copper is 400 W/mC. The two ends of this rod must be kept at a temperature difference of–

1. 1ºC
2. 10ºC
3. 100ºC
4. 1000ºC

Answer: 3. 100ºC

NEET Physics Class 11 Chapter 8 Heat Transfer Multiple Choice Questions and Answers

Question 3. If two conducting slabs of thickness d₁ and d₂, and thermal conductivity K₁ and K₂ are placed together face to face as shown in the figure the steady state temperatures of outer surfaces are θ₁ and θ₂. The temperature of the common surface is–

1. \(\frac{K_1 \theta_1 d_1+K_2 \theta_2 d_2}{K_1 d_1+K_2 d_2}\)
2. \(\frac{K_1 \theta_1+K_2 \theta_2}{K_1+K_2}\)
3. \(\frac{K_1 \theta_1+K_2 \theta_2}{\theta_1+\theta_2}\)
4. \(\frac{K_1 \theta_1 d_2+K_2 \theta_2 d_1}{K_1 d_2+K_2 d_1}\)

Answer: 4. \(\frac{K_1 \theta_1 d_2+K_2 \theta_2 d_1}{K_1 d_2+K_2 d_1}\)

Question 4. Which of the following qualities suit a cooking utensil?

1. High specific heat and low thermal conductivity
2. High specific heat and high thermal conductivity
3. Low specific heat and low thermal conductivity
4. Low specific heat and high thermal conductivity

Answer: 4. Low specific heat and high thermal conductivity

Question 5. The lengths and radii of two rods made of the same material are in the ratios 1: 2 and 2 : 3 respectively; If the temperature difference between the ends of the two rods is the same, then in the steady state, the amount of heat flowing per second through them will be in the ratio:

1. 1 : 3
2. 4 : 3
3. 8: 9
4. 3: 2

Answer: 3. 8: 9

Question 6. Two metal cubes with 3 cm-edges of copper and aluminum are arranged as shown in the figure (K_CU =385 W/m-K, K_AL = 209 W/m-K).

1. The total thermal current from one reservoir to the other is :

1. 1.42 × 10³ W
2. 2.53 × 10³ W
3. 1.53 × 10⁴ W
4. 2.53 × 10⁴ W

Answer: 1. 1.42 × 10³ W

2. The ratio of the thermal current carried by the copper cube to that carried by the aluminum cube is: –

1. 1.79
2. 1.69
3. 1.54
4. 1.84

Answer: 4. 1.84

Question 7. Two rods having thermal conductivities in the ratio of 5 : 3 and having equal length and equal cross section are joined face-to-face (series combination). If the temperature of the free end of the first rod is 100ºC and the free end of the second rod is 20ºC, the temperature of the junction is–

1. 50ºC
2. 70ºC
3. 85ºC
4. 90ºC

Answer: 2. 70ºC

Question 8. One end of a metal rod of length 1.0m and area of cross-section 100 cm² is maintained at 100ºC. If the other end of the rod is maintained at 0ºC, the quantity of heat transmitted through the rod per minute will be (coefficient of thermal conductivity of the material of rod = 100W/Kg/K)

1. 3 × 103 J
2. 6 × 103 J
3. 9 × 103 J
4. 12 × 103 J

Answer: 2. 6 × 103 J

Heat Transfer MCQs for NEET Physics Class 11 with Answers

Question 9. The coefficients of thermal conductivity of a metal depends on

1. Temperature difference between the two sides
2. Thickness of the metal plate
3. Area of the plate
4. None of the above

Answer: 4. None of the above

Question 10. Two identical square rods of metal are welded end to end as shown in the figure

1. Assume that 10 cal of heat flows through the rods in 2 min. Now the rods are welded as shown in the figure.
2. The time it would take for 10 cal to flow through the rods now, is

1. 0.75 min
2. 0.5 min
3. 1.5 min
4. 1 min

Answer: 2. 0.5 min

Question 11. The area of cross-section of two rods of equal lengths are A₁ and A₂ and thermal conductivities are K₁ and K₂. Specific heats are S₁ and S₂. Condition for equal heat flow is–

1. \(\mathrm{K}_1=\mathrm{K}_2\)
2. \(\mathrm{K}_1 \mathrm{~S}_1=\mathrm{K}_2 \mathrm{~S}_2\)
3. \(\frac{K_1}{A_1 S_1}=\frac{K_2}{A_2 S_2}\)
4. \(\mathrm{K}_1 \mathrm{~A}_1=\mathrm{K}_2 \mathrm{~A}_2\)

Answer: 4. \(\mathrm{K}_1 \mathrm{~A}_1=\mathrm{K}_2 \mathrm{~A}_2\)

Question 12. If two metallic plates of equal thickness, equal cross-section area, and thermal conductivities K₁ and K₂ are put together face to face (series combination) and a common plate is constructed, then the equivalent thermal conductivity of this plate will be

1. \(\frac{K_1 \quad K_2}{K_1+K_2}\)
2. \(\frac{2 K_1 K_2}{K_1+K_2}\)
3. \(\frac{\left(K_1^2+K_2^2\right)^{3 / 2}}{K_1 K_2}\)
4. \(\frac{\left(K_1^2+K_2^2\right)^{3 / 2}}{2 K_1 K_2}\)

Answer: 2. \(\frac{2 K_1 K_2}{K_1+K_2}\)

Question 13. Consider a compound slab consisting of two different materials having equal thicknesses, equal cross-section area, and thermal conductivities k and 2k respectively. If they are connected in parallel combination, the equivalent thermal conductivity of the slab is–

1. \(\sqrt{2}\)
2. 3k
3. \(\frac{4}{3} \mathrm{k}\)
4. \(\frac{2}{3} \mathrm{k}\)

Answer: 3. \(\frac{4}{3} \mathrm{k}\)

Question 14. The two ends of a rod of length L and a uniform cross-sectional area A kept at two temperatures T₁ and T₂(T₁>T₂ ). The rate of heat transfer,\(\frac{d Q}{d t}\) through the rod in a steady state is given by

1. \(\frac{d Q}{d t}=\frac{K L\left(T_1-T_2\right)}{A}\)
2. \(\frac{d Q}{d t}=\frac{K\left(T_1-T_2\right)}{L A}\)
3. \(\frac{d Q}{d t}=K L A\left(T_1-T_2\right)\)
4. \(\frac{d Q}{d t}=\frac{K A\left(T_1-T_2\right)}{L}\)

Answer: 4. \(\frac{d Q}{d t}=\frac{K A\left(T_1-T_2\right)}{L}\)

Question 15. A square is made of four rods of the same material one of the diagonals of a square is at a temperature difference of 100°C, then the temperature difference of the second diagonal :

1. 0°C
2. \(\frac{100}{\ell}\)
3. \(\frac{100}{2 \ell}\)
4. 100°C

Answer: 1. 0°C

Question 16. Three rods made of the same material and having the same cross-section are joined as shown in Fig. Each rod is of the same length. The left and right ends are kept at 0°C and 90°C respectively. The temperature of the junction of the three rods will be:

1. 45°C
2. 60°C
3. 30°C
4. 20°C

Answer: 2. 60°C

NEET Physics Chapter 8 Heat Transfer MCQs and Answer Key

Question 17. Two containers, one containing ice at 0°C and the other containing boiling water at 100°C are connected by two identical rods. When rods are in parallel the rate of heat transfer is Q₁ and when rods are in series, the rate of heat transfer is Q₂. Then Q₂ /Q₁ will be:

1. 2: 1
2. 1: 2
3. 4: 1
4. 1: 4

Answer: 4. 1: 4

Question 18. 2 litre water at 27°C is heated by a 1 kW heater in an open container. On average heat is lost to surroundings at the rate of 160 J/s. The time required for the temperature to reach 77°C is

1. 8 min 20 sec
2. 10 min
3. 7 min
4. 14 min

Answer: 1. 8 min 20 sec

Question 19. If the temperature difference on the two sides of a wall increases from 100°C to 200°C, its thermal conductivity

1. Remains unchanged
2. Is doubled
3. Is halved
4. Becomes four times

Answer: 1. Remains unchanged

Question 20. A cylindrical rod having temperature T₁ and T₂ at its ends. The rate of flow of heat is Q₁cal/sec. If all the linear dimensions are doubled keeping the temperature constant then the rate of flow of heat Q₂ will be–

1. 4Q₁
2. 2Q₁
3. \(\frac{Q_1}{4}\)
4. \(\frac{Q_1}{2}\)

Answer: 2. 2Q₁

Question 21. One end of a thermally insulated rod is kept at a temperature of T₁ and the other at T₂. The rod is composed of two sections of lengths L₁ and L₂ and thermal conductivities k₁ and k₂ respectively. The temperature at the interface of the sections is

1. \(\frac{\left(\begin{array}{ll}
   K_2 & L_2 T_1+K_1 L_1 T_2
   \end{array}\right)}{\left(K_1 L_1+K_2 L_2\right)}\)
2. \(\frac{\left(\begin{array}{ll}
   K_2 L_1 T_1+K_1 L_2 T_2
   \end{array}\right)}{\left(K_2 L_1+K_1 L_2\right)}\)
3. \(\frac{\left(\begin{array}{ll}
   K_1 & L_2 T_1+K_2 L_1 T_2
   \end{array}\right)}{\left(K_1 L_2+K_2 L_1\right)}\)
4. \(\frac{\left(\begin{array}{ll}
   K_1 L_1 T_1+K_2 L_2 T_2
   \end{array}\right)}{\left(K_1 L_1+K_2 L_2\right)}\)

Answer: 3. \(\frac{\left(\begin{array}{ll}
K_1 & L_2 T_1+K_2 L_1 T_2
\end{array}\right)}{\left(K_1 L_2+K_2 L_1\right)}\)

Question 22. Three rods A, B, and C of the same length and same cross-section area are joined as shown in the figure. Their thermal conductivities are in the ratio 1: 2: 1.5. If the open ends of A and C are at 200°C and 18°C respectively, the temperature at the junction of A and B in equilibrium is-

1. 156°C
2. 116°C
3. 74°C
4. 148°C

Answer: 2. 116°C

Question 23. In the above question, the temperature at the junction of B and C will be

1. 124°C
2. 124°K
3. 74°C
4. 74°K

Answer: 3. 74°C

Question 24. The ends of the two rods of different materials with their thermal conductivities, radii of cross-section, and lengths in the ratio 1: 2 are maintained at the same temperature difference. If the rate of flow of heat in the larger rod is 4 cal/sec., that in the shorter rod will be (in cal/sec)

1. 1
2. 2
3. 8
4. 16

Answer: 2. 2

Question 25. The coefficients of thermal conductivity of copper, mercury, and glass are respectively K_c, K_m, and Kg such that K_c> K_m> K_g. If the same quantity of heat is to flow per second per unit area of each and corresponding temperature gradients are X_c, X_m, and X_g.

1. X_c= X_m= K_g
2. X_c> X_m> X_g
3. X_c< X_m< X_g
4. X_m< X_c< X_g

Answer: 3. X_c< X_m< X_g

Question 26. A compound slab is composed of two parallel layers of different materials, with thicknesses of 3 cm and 2 cm. The temperatures of the outer faces of the compound slab are maintained at 100°C and 0°C. If conductivities are 0.036 cal/cm-sec-°C and 0.016 cal/cm-sec-°C then the temperature of the junction is-

1. 40°C
2. 60°C
3. 100°C
4. 50°C

Answer: 2. 60°C

Question 27. The intensity of heat radiation by a point source measured by a thermopile placed at a distance d is Ι, If the distance of the thermopile is doubled then the intensity of radiation will be

1. Ι
2. 2Ι
3. \(\frac{\mathrm{I}}{4}\)
4. \(\frac{\mathrm{I}}{2}\)

Answer: 3. \(\frac{\mathrm{I}}{4}\)

Question 28. Two rods of copper and brass of the same length and area of cross-section are joined as shown. One end is kept at 100°C and the other at 0°C. The temperature at the mid-point will be

1. More if A is at 100°C and B at 0°C
2. More if A is at 0°C and B at 100°C
3. Will be the same in both the above cases, but not 50°C
4. 50°C in both the above cases

Answer: 1. More if A is at 100°C and B at 0°C

Question 29. Two identical square rods of metal are welded end to end as shown in Fig.

1. 20 cal. of heat flows through it in 4 min. If the rods are welded as shown in Fig.
2. The same amount of heat will flow through the rods in

1. 1 min.
2. 2 min.
3. 3 min.
4. 16 min.

Answer: 1. 1 min.

Question 30. A wall consists of alternating blocks with length ‘d’ and coefficient of thermal conductivity k₁ and k₂. The cross sectional area of the blocks is the same. The equivalent coefficient of thermal conductivity of the wall between left and right is:-

1. \(\mathrm{K}_1+\mathrm{K}_2\)
2. \(\frac{\left(K_1+K_2\right)}{2}\)
3. \(\frac{K_1 K_2}{K_1+K_2}\)
4. \(\frac{2 K_1 K_2}{K_1+K_2}\)

Answer: 2. \(\frac{\left(K_1+K_2\right)}{2}\)

Class 11 NEET Physics Heat Transfer Multiple Choice Questions

Question 31. Five rods of the same dimensions are arranged as shown in the fig. They have thermal conductivities, k₁, k₂, k₅, k_4, and k₃ when points A and B are maintained at different temperatures. No heat flows through the central rod if-

1. \(k_1 k_4=k_2 k_3\)
2. \(\mathrm{k}_1=\mathrm{k}_4 \text { and } \mathrm{k}_2=\mathrm{k}_3\)
3. \(\frac{k_1}{k_4}=\frac{k_2}{k_3}\)
4. \(k_1 k_2=k_3 k_4\)

Answer: 4. \(k_1 k_2=k_3 k_4\)

Question 32. Three metal rods made of copper, aluminum, and brass, each 20 cm long and 4 cm in diameter, are placed end to end with aluminum between the other two. The free ends of copper and brass are maintained at 100 and 0°C respectively. Assume that the thermal conductivity of copper is twice that of aluminum and four times that of brass. The equilibrium temperatures of the copper-aluminium and aluminium-brass junctions are respectively.

1. 68 °C and 75 °C
2. 75 °C and 68 °C
3. 57 °C and 86 °C
4. 86 °C and 57 °C

Answer: 3. 57 °C and 86 °C

Question 33. The coefficient of thermal conductivity of copper is nine times that of steel. In the composite cylindrical bar shown in the figure, what will be the temperature at the junction of copper and steel?

1. 75ºC
2. 67ºC
3. 33ºC
4. 25ºC

Answer: 1. 75ºC

Question 34. The heat conduction coefficient of copper is 9 times the heat conduction coefficient of steel. The junction temperature of the combined cylindrical rod shown in the figure will be.

1. 75°C
2. 67°C
3. 33°C
4. 25°C

Answer: 1. 75°C

Question 35. Water is usually heated by

1. Conduction
2. Convection
3. Radiation
4. All the above processes

Answer: 2. Convection

Question 36. In natural convection a heated portion of a liquid moves because-

1. Its molecular motion becomes aligned
2. Of molecular collisions within it
3. Its density is less than that of the surrounding fluid
4. Of currents of the surrounding fluid

Answer: 3. Its density is less than that of the surrounding fluid

Question 37. It is hotter at the same distance over the top of the fire than it is on the side of it mainly because

1. Heat is radiated upwards
2. Air conducts heat upwards
3. Convection takes more heat upwards
4. Conduction, convection, and radiation all contribute significantly to transferring heat upwards

Answer: 3. Convection takes more heat upwards

Question 38. Ventilators are provided at the top of the room

1. To bring oxygen for breathing
2. So that sunlight may enter the room
3. To maintain convection currents to keep the air fresh in the room
4. To provide an outlet for carbon dioxide

Answer: 3. To maintain convection currents to keep the air fresh in the room

Question 39. The mode of transmission of heat in which heat is carried by moving particles is:

1. Wave motion
2. Convection
3. Conduction
4. Radiation

Answer: 2. Convection

Question 40. The temperature of a piece of metal is increased from 27°C to 327°C. The rate of emission of heat by radiation by a metal will become-

1. Double
2. Four times
3. Eight times
4. Sixteen times

Answer: 4. Sixteen times

Question 41. Radiation emitted by a surface is directly proportional to-

1. The third power of its temperature
2. The fourth power of its temperature
3. Twice the power of its temperature
4. None of above

Answer: 2. Fourth power of its temperature

Question 42. If the temperature of the surface of the sun becomes half then the energy emitted by it to the earth per second will reduce to –

1. 1/2
2. 1/4
3. 1/16
4. 1/64

Answer: 3. 1/16

Question 43. If the distance between point sources and the screen is doubled then the intensity of light becomes-

1. Four times
2. Doubled
3. Half
4. One fourth

Answer: 4. One fourth

Question 44. At T = 200K a black body emits maximum energy at a wavelength of 14 μm. Then at T = 1000K, the body will emit maximum energy at a wavelength of-

1. 70 mm
2. 70 μm
3. 2.8 μm
4. 2.8 mm

Answer: 3. 2.8 μm

Question 45. If the temperature of a black body is raised by 50%, then the energy emitted per second will be increased by an order of-

1. 50%
2. 100%
3. 200%
4. 400%

Answer: 4. 400%

Question 46. What represents the color of the star-

1. Density
2. Distance
3. Energy
4. Temperature

Answer: 4. Temperature

Question 47. The black body spectrum is-

1. Continuous spectrum with black lines
2. Continuous spectrum with black bands
3. Continuous spectrum
4. None of the above

Answer: 3. Continuous spectrum

Question 48. There is a black spot on the body. If the body is heated and carried in a dark room then it glows more. This can be explained based on-

1. Newton’s law of cooling
2. Vien’s law
3. Kirchoff’s law
4. Stefan’s

Answer: 3. Kirchoff’s law

Question 49. A heated body emits radiation which has maximum intensity at frequency νm. If the temperature of the body is doubled:

1. The maximum intensity radiation will be at frequency 2 νm
2. The maximum intensity radiation will be at frequency νm.
3. The total emitted energy will increase by a factor of 2.
4. None of these

Answer: 1. The maximum intensity radiation will be at frequency 2 νm

Question 50. If λ_m denotes the wavelength at which the radiative emission from a black body at a temperature T K is maximum, then-

1. \(\lambda_{\mathrm{m}} \propto \mathrm{T}^4\)
2. \(\lambda_{\mathrm{m}}\) is independent of T
3. \(\lambda_{\mathrm{m}} \propto \mathrm{T}\)
4. \(\lambda_m \propto \mathrm{T}^{-1}\)

Answer: 4. \(\lambda_m \propto \mathrm{T}^{-1}\)

Question 51. A black body at 1227°C emits radiations with maximum intensity at a wavelength of 5000 Å. If the temperature of the body is increased by 1000°C, the maximum intensity will be at

1. 4000 Å
2. 5000Å
3. 6000 Å
4. 3000Å

Answer: 4. 3000Å

Question 52. A black body is at 727°C. It emits energy at a rate which is proportional to

1. (277)²
2. (1000)⁴
3. (1000)²
4. (727)⁴

Answer: 2. (1000)⁴

Heat Transfer in NEET Physics Class 11: MCQs and Answer Key

Question 53. If the temperature of the body increases by 10%, then the increase in radiated energy of the body is :

1. 10%
2. 40%
3. 46%
4. 1000%

Answer: 3. 46%

Question 54. Infrared radiations are detected by

1. Spectrometer
2. Pyrometer
3. Nanometer
4. Photometer

Answer: 2. Pyrometer

Question 55. The plots of intensity vs. wavelength for three black bodies at temperatures T₁, T_2, and T₃ respectively are as shown. Their temperatures are such that-

1. T₁> T₂> T₃
2. T₁> T₃> T₂
3. T2> T₃> T₁
4. T₃> T₂> T₁

Answer: 2. T₁> T₃> T₂

Question 56. In which of the following phenomenon heat convection does not take place

1. Land and sea breeze
2. Boiling of water
3. Heating of glass surface due to filament of the bulb
4. The air around the furnace

Answer: 3. Heating of glass surface due to a filament of the bulb

Question 57. The energy radiated by a black body is directly proportional to :

1. T²
2. T^-2
3. T⁴
4. T

Answer: 3. T⁴

Question 58. When a substance is gradually heated, its initial color is

1. Red
2. Green
3. Yellow
4. White

Answer: 1. Red

Question 59. If the temperature becomes double, the emitted radiation will be :

1. 16 times
2. 8 times
3. times
4. 32 times

Answer: 1. 16 times

Question 60. If at temperature T₁= 1000 K, the wavelength is 1.4 × 10^-6 m, then at what temperature the wavelength will be 2.8 × 10^-6m?

1. 2000 K
2. 500 K
3. 250 K
4. None of these

Answer: 2. 500 K

Question 61. A black body is heated from 27°C to 927°C the ratio of radiations emitted will be :

1. 1: 256
2. 1: 64
3. 1: 16
4. 1: 4

Answer: 1. 1: 256

Question 62. Water is used to cool the radiators of engines in cars because :

1. Of its low boiling point
2. Of its high specific heat
3. Of its low-density
4. Of its easy availability

Answer: 2. Of its high specific heat

Question 63. The color of the star indicates its :

1. Temperature
2. Distance
3. Velocity
4. Size

Answer: 1. Temperature

Question 64. The means of energy transfer in a vacuum are:

1. Irradiation
2. Convection
3. Radiation
4. Conduction

Answer: 3. Radiation

Question 65. The temperature of the black body increases from T to 2T. The factor by which the rate of emission will increase is

1. 4
2. 2
3. 16
4. 8

Answer: 3. 16

Question 66. Let there be four articles having colors blue, red, black, and white. When they are heated together and allowed to cool, which article will cool at the earliest?

1. Blue
2. Red
3. Black
4. White

Answer: 2. Red

Question 67. A piece of red glass when heated in dark to red hot states will appear to be :

1. White
2. Red
3. Green
4. Invisible

Answer: 3. Green

Question 68. What is the mode of heat transfer by which a hot cup of coffee loses most of its heat?

1. Condition
2. Convection
3. Evaporation
4. Radiation

Answer: 1. Condiction

NEET Physics Chapter 8 Heat Transfer: MCQs Practice and Answers

Question 69. Which one of the following processes depends on gravity :

1. Conduction
2. Convection
3. Radiation
4. None of the above

Answer: 2. Convection

Question 70. For a black body at a temperature of 727°C, its radiating power is 60 watts and the temperature of surrounding is 227°C. If the temperature of the black body is changed to 1227°C then its radiating power will be

1. 304 W
2. 320 W
3. 240 W
4. 120 W

Answer: 2. 320 W

Question 71. Wien’s displacement law expresses a relation between-

1. Wavelength corresponds to maximum energy and temperature.
2. Radiation energy and wavelength
3. Temperature and wavelength
4. Color of light and temperature

Answer: 1. Wavelength corresponds to maximum energy and temperature.

Question 72. The unit of Stefan’s constant is-

1. Watt-m²-K⁴
2. Watt-m²/K⁴
3. Watt/m²-K
4. Watt/m²K⁴

Answer: 4. Watt/m²K⁴

Question 73. Which of the following radiations has the least wavelength?

1. γ-rays
2. β-rays
3. α-rays
4. X-rays

Answer: 1. γ-rays

Question 74. If the temperature of the sun were to increase from T to 2T and its radius from R to 2R, then the ratio of the radiant energy received on Earth to what it was previously would be

1. 4
2. 16
3. 32
4. 64

Answer: 4. 64

Question 75. Assuming the sun to be a spherical body of radius R at a temperature of T K, evaluate the total radiant power, incident on Earth, at a distance r from the Sun. (earth radius = r₀)

1. \(\frac{R^2 \sigma T^4}{r^2}\)
2. \(\frac{4 \pi r_0^2 \quad R^2 \sigma T^4}{r^2}\)
3. \(\frac{\pi r_0^2 \quad R^2 \sigma T^4}{r^2}\)
4. \(\frac{r_0^2 \quad R^2 \sigma T^4}{4 \pi r^2}\)

Answer: 3. \(\frac{\pi r_0^2 \quad R^2 \sigma T^4}{r^2}\)

Question 76. The energy emitted per second by a black body at 1227ºC is E. If the temperature of the black body is increased to 2727ºC, the energy emitted per second in terms of E is –

1. 16 E
2. E
3. 4E
4. 2E

Answer: 1. 16 E

Question 77. Temp. of a black body is 3000 k. When the black body cools, then change in wavelength Δ λ = 9 microns corresponding to maximum energy density. Now temp. of a black body is-

1. 300 K
2. 2700 K
3. 270 K
4. 1800 K

Answer: 1. 300 K

Question 78. If the radius of the sun is RS, the radius of the orbit of the earth about the sun is Reand σ is Stefan’s constant, then the amount of radiation falling per second on a unit area of the earth’s surface is-

1. \(\left(\frac{R_s}{R_e}\right)^2 \sigma \mathrm{T}^4\)
2. \(\left(\frac{R_e}{R_s}\right)^2 \sigma \mathrm{T}^4\)
3. \(\frac{\sigma}{T^4}\left(\frac{R_s}{R_e}\right)^2\)
4. \(\left(\frac{R_e}{R_s}\right)^2 \frac{T^4}{\sigma}\)

Answer: 1. \(\left(\frac{R_s}{R_e}\right)^2 \sigma \mathrm{T}^4\)

Question 79. Which of the following surfaces will absorb maximum radiant energy?

1. Black
2. Rough
3. Smooth white
4. Rough black

Answer: 4. Rough black

Question 80. After heating two pieces of iron, they are taken to a dark room. One of them appears red and another appears blue, then-

1. The temperature of the red piece will be higher.
2. The temperature of the blue piece will be higher.
3. The temperature of both pieces will be the same.
4. Nothing can be said about their temp.

Answer: 2. The temperature of the blue piece will be higher.

Question 81. If the temperature of a lamp is about 600K, then the wavelength at which maximum emission takes place will be (Wien’s constant b = 3 × 10^-3 m-K)

1. 500 A°
2. 5000 A°
3. 50000 A°
4. 500000 A°

Answer: 3. 50000 A°

Question 82. The rate of cooling of a sphere of thermal capacity 1000 cal/K is 400 J/s, and its rate of fall of temperature is-

1. 0.095 K/min
2. 0.62 K/min
3. 2.8 K/min
4. 5.7 K/min

Answer: 4. 5.7 K/min

Question 83. If maximum spectral emissivity at temperature T₁ K is at wavelength λ₁, then the wavelength of maximum emissivity at temperature T₂ K will be

1. \(\frac{\lambda_1 T_2}{2}\)
2. \(\lambda_1\left(\frac{T_1}{T_2}\right)^4\)
3. \(\lambda_1\left(\frac{T_1}{T_2}\right)^5\)
4. \(\frac{\lambda_1 \mathrm{~T}_1}{\mathrm{~T}_2}\)

Answer: 4. \(\frac{\lambda_1 \mathrm{~T}_1}{\mathrm{~T}_2}\)

Question 84. The spectral emissive power of a black body at a temperature of 6000K is maximum at λ_m= 5000 A°. If the temperature is increased by 10%, then the decrease in λ_m will be

1. 2.5%
2. 5.0%
3. 7.5%
4. 10%

Answer: 4. 10%

Question 85. The rate of emission of energy by a unit area of a body is 10 watts and that of the sun is 106 watts. The emissive power of the body is 0.1. If the temperature of the sun is 6000K, then the temperature of the body will be

1. 6000K
2. 600K
3. 60010K
4. (600 10)K

Answer: 2. 600K

Heat Transfer NEET Physics Class 11: Multiple Choice Questions

Question 86. The ratio of masses of two copper spheres of identical surfaces is 8: 1. If their temperatures are 2000K and 1000K respectively then the ratio of energies radiated per second by the two is-

1. 128: 1
2. 64: 1
3. 16: 1
4. 4: 1

Answer: 2. 64: 1

Question 87. A solid body is heated upto very high temperatures. As we go on heating, its brightness increases and it appears white at the end. The sequence of the color observed as the temperature of the body increases will be

1. Yellow, green, red, white.
2. Green, yellow, red, white.
3. Red, green, yellow, white.
4. Red, yellow, green, white.

Answer: 4. Red, yellow, green, white.

Question 88. The effective area of a black body is 0.1 m² and its temperature is 100 K. The amount of radiation emitted by it per minute is –

1. 1.34 cal
2. 8.1 cal
3. 5.63 cal
4. 1.34 J

Answer: 2. 8.1 cal

Question 89. What is the energy of emitted radiation from the Sun when the temperature is doubled-

1. 2
2. 4
3. 8
4. 16
5. Answer: 4. 16

Question 90. Newton’s law of cooling is a special case of

1. Wien’s displacement law
2. Kirchoff’s law
3. Stefan’s law
4. Planck’s law

Answer: 3. Stefan’s law

Question 91. A hot liquid is kept in a big room. Its temperature is plotted as a function of time. Which of the following curves may represent the plot?

1. a
2. c
3. d
4. b

Answer: 1. a

Question 92. A body takes 4 minutes to cool from 100°C to 70°C. To cool from 70°C to 40°C it will take-(room temperature is 15°C)

1. 7 minutes
2. 6 minutes
3. 5 minutes
4. 4 minutes

Answer: 1. 7 minutes

Question 93. A cup of tea cools from 80°C to 60°C in one minute. The ambient temperature is 30°C. In cooling from 60°C to 50°C it will take-

1. 30 seconds
2. 60 seconds
3. 96 seconds
4. 48 seconds

Answer: 4. 48 seconds

Question 94. A hot liquid cools from 70°C to 60°C in 5 minutes. The time needed by the same liquid to cool from 60°C to 50°C will be

1. Less than 5 minutes
2. More than 5 minutes
3. Equal to 5 minutes
4. Less or more than 5 minutes depends on the density of the liquid

Answer: 2. More than 5 minutes

Question 95. Which of the following is a true statement?

1. A good absorber is a bad conductor
2. Each body emits and absorbs radiation at each temperature
3. In a black body energy of emitted radiation is equal for all wavelength
4. Planck’s law gives the relation between the maximum wavelength of black body radiation and its temperature.

Answer: 2. Each body emits and absorbs radiation at each temperature

Question 96. A body takes 10 minutes to cool down from 62°C to 50°C. If the temperature of the surroundings is 26°C then in the next 10 minutes temperature of the body will be :

1. 38°C
2. 40°C
3. 42°C
4. 44°C

Answer: 3. 42°C

Question 97. A body cools from 60°C to 50°C in 10 minutes. If the room temperature is 25°C and assuming Newton’s law of cooling to hold good, the temperature of the body at the end of the next 10 minutes will be :

1. 45°C
2. 41.67°C
3. 40°C
4. 38.5°C

Answer: 2. 41.67°C

Question 98. Two spheres of radii in the ratio 1: 2 and densities in the ratio 2: 1 and of the same specific heat, are heated to the same temperature and left in the same surrounding. Their rate of cooling will be in the ratio :

1. 2: 1
2. 1: 1
3. 1: 2
4. 1: 4

Answer: 2. 1: 1

Question 99. The formation of ice is started in a lake with water at 0°C. When the atmospheric temperature is –10°C. If the time taken for 1 cm of ice to be formed is 7 hours, the time taken for the thickness of ice to increase from 1cm to 2 cm is :

1. Less than 7 hours
2. 7 hours
3. More than 14 hours
4. More than 7 hours but less than 14 hours

Answer: 3. More than 14 hours

Question 100. Two circular discs A and B with equal radii are blackened. They are heated to the same temperature and are cooled under identical conditions. What inference do you draw from their cooling curves?

1. A and B have the same specific heats
2. The specific heat of A is less
3. The specific heat of B is less
4. Nothing can be said

Answer: 2. Specific heat of A is less

Question 101. According to Newton’s law of cooling, the rate of cooling of a body is proportional to (Δθ)n, where Δθ is the difference between the temperature of the body and the surroundings, and n is equal to

1. 2
2. 3
3. 4
4. 1

Answer: 4. 1

Chapter 8 Heat Transfer MCQs for NEET Physics Class 11

Question 102. A liquid cools down from 70° C to 60°C in 5 min. The time taken to cool it from 60°C to 50°C will be

1. 5 min
2. Lesser than 5 min
3. Greater than 5 min
4. Lesser or greater than 5 minutes depending upon the density of the liquid

Answer: 3. Greater than 5 min

Question 103. The heat capacities of three liquids A, B, and C of the same volumes are in the ratio 3: 2: 1. They are allowed to cool in the same surroundings and same conditions for the same temperature difference. Which of these will cool first?

1. A
2. B
3. C
4. All will cool at the same time

Answer: 3. C

Question 104. The temperature of a room is 30°C. A body kept in it takes 4 minutes to cool from 61°C to 59°C. The time taken by the body to cool from 51°C to 49°C will be

1. 4 min.
2. 5 min.
3. 6 min.
4. 8 min.

Answer: 3. 6 min.

Question 105. Two cylindrical conductors A and B of the same metallic material have their diameters in the ratio 1: 2 and lengths in the ratio 2: 1. If the temperature difference between their ends is the same, the ratio of heat conducted respectively by A and B per second is,

1. 1: 2
2. 1: 4
3. 1: 16
4. 1: 8

Answer: 4. 1: 8

Question 106. According to Kirchoff’s law-

1. a_λe_λ= E_λ
2. E_λa_λ= e_λ
3. a_λ= e_λE_λ
4. E_λ, a_λ, e_λ= const.

Answer: 2. E_λa_λ= e_λ

Question 107. A spherical solid black body of radius ‘r’ radiates power ‘H’ and its rate of cooling is ‘C’. If the density is constant then which of the following is/are true?

1. H ∝ r and c ∝ r²
2. H ∝ r² and c ∝
3. H ∝ r and c ∝ 2
4. H ∝ r² and c ∝ r²

Answer: 2. H ∝ r² and c ∝

Question 108. Which of the following is nearest to Blackbody-

1. An enclosure with a small hole
2. Carbon black
3. Ebonite
4. None of these

Answer: 1. An enclosure with a small hole

Question 109. Which of the following processes is reversible?

1. Transfer of heat by radiation
2. Electrical heating of nichrome wire
3. Transfer of heat by conduction
4. Isothermal compression

Answer: 4. Isothermal compression

Question 110. Assuming the sun to have a spherical outer surface of radius r, radiating like a black body at temperature t°C, the power received by a unit surface, (normal to the incident rays) at a distance R from the center of the sun is (considering solar constant to be uniform)

1. \(\frac{4 \pi r^2 t^4}{R^2}\)
2. \(\frac{r^2 \sigma(t+273)^4}{4 \pi R^2}\)
3. \(\frac{16 \pi^2 r^2 \sigma t^4}{R^2}\)
4. \(\frac{r^2 \sigma(t+273)^2}{R^2}\)

Answer: 4. \(\frac{r^2 \sigma(t+273)^2}{R^2}\)

Question 111. Which of the following is more close to a black body?

1. Blackboard paint
2. Green leaves
3. Black holes
4. Red roses

Answer: 1. Blackboard paint

Question 112. A black body is at a temperature of 2800 K. The energy of radiation emitted by this object with a wavelength between 499 nm and 500 nm is U₁, between 999 nm and 1000 nm is U_2, and between 1499 nm and 1500 nm is U₃. The Wien constant b = 2.88 × 106 nm K. Then

1. U₁= 0
2. U₃= 0
3. U₁> U₂
4. U₂>U₁

Answer: 4. U₂>U₁

Question 113. The temperature of bodies X and Y vary with time as shown in the figure. If the emissivity of bodies X and Y are e_X and e_Y and absorptive powers are A_X and A_Y, (assume other conditions are identical for both) then:

1. e_Y> e_X, A_Y> A_X
2. e_Y< e_X, A_Y< A_X
3. e_Y> e_X, A_Y< A_X
4. e_Y< e_X, A_Y> A_X

Answer: 1. e_Y> e_X, A_Y> A_X

Question 114. Three discs of the same material A, B, and C of radii 2 cm, 4 cm, and 6 cm respectively are coated with carbon black. Their wavelengths corresponding to maximum spectral radiancy are 300, 400, and 500 nm respectively then maximum power will be emitted by

1. A
2. B
3. C
4. Same for all

Answer: 2. B

Chapter 8 Heat Transfer MCQs for NEET Physics Class 11

Question 115. Three graphs marked 1, 2, and 3 represent the variation of maximum emissive power and wavelength of radiation of the sun, a welding arc, and a tungsten filament. Which of the following combinations is correct

1. 1- tungsten filament, 2 → welding arc, 3 → sun
2. 2- tungsten filament, 3 → welding arc, 1 → sun
3. 3- tungsten filament, 1 → welding arc, 2 → sun
4. 2- tungsten filament, 1 → welding arc, 3 → sun

Answer: 1. 1- tungsten filament, 2 → welding arc, 3 → sun

Question 116. Two rectangular blocks, having identical dimensions, can be arranged either in configuration Ι or in configuration ΙΙ as shown in the figure, One of the blocks has a thermal conductivity of k, and the other 2k. The temperature difference between the ends along the x-axis is the same in both configurations. It takes 9s to transport a certain amount of heat from the hot end to the cold end in configuration 1. The time to transport the same amount of heat in the configuration 2 is:

1. 2.0 s
2. 3.0 s
3. 4.5 s
4. 6.0 s

Answer: 1. 2.0 s

Question 117. Parallel rays of light of intensity Ι = 912 Wm^-2 are incident on a spherical black body kept in surroundings of temperature 300 K. Take Stefan-Biltzmann constant σ = 5.7 × 10^-8 Wm^-2 K^-4 and assume that the energy exchange with the surroundings is only through radiation. The final steady state temperature of the black body is close to:

1. 330 K
2. 660 K
3. 990 K
4. 1550 K

Answer: 1. 330 K

Question 118. Two spherical stars A and B emit blackbody radiation. The radius of A is 400 times that of B and A emits 10⁴ times the power emitted from B. The ratio \(\left(\frac{\lambda_A}{\lambda_B}\right)\) to their wavelengths λ_A and λ_B at which the peaks occur in their respective radiation curves is:

1. 1
2. 2
3. 3
4. 4

Answer: 2. 2

Question 119. The earth radiates in the infrared region of the spectrum. The spectrum is correctly given by :

1. Rayleigh-Jeans law
2. Planck’s law of radiation
3. Stefan’s law of radiation
4. Wien’s law

Answer: 1. Rayleigh-Jeans law

Question 120. Two spheres of different materials one having a radius double of other and a wall thickness 1/4 of the other are filled with ice. If the time required to completely melt the ice is 25 min. for a larger radius sphere and 16 min. for a smaller radius sphere, then the ratio of the thermal conduction coefficient for the material of larger radius to that of the thermal conduction coefficient for the material of smaller radius sphere will be.

1. 4: 5
2. 5: 4
3. 8: 25
4. 1: 25

Answer: 3. 8: 25

Question 121. A black body is at room temperature. It is placed in a furnace, and it is observed that

1. In the beginning, it is seen most black, and later on, it is seen as the brightest.
2. It is always seen as black.
3. It can’t be resolved at any times
4. In the beginning, it is seen mostly black, and later on it can’t be resolved.

Answer: 1. In the beginning it is seen as the most black and later on it is seen brightest.

Question 122. If a liquid takes 30 sec. in cooling of 95°C to 90°C and 70 sec. in cooling of 55°C to 50°C then temp. of the room is-

1. 16.5 °C
2. 22.5 °C
3. 28.5 °C
4. 32.5 °C

Answer: 2. 22.5 °C

Question 123. A body takes 2 minutes in cooling from 365K to 361K. If the room temperature is 293K, then the time taken to cool from 344K to 342K will be

1. 1 min.
2. 1.2 min.
3. 1.4 min.
4. 1.8 min.

Answer: 3. 1.4 min.

Question 124. The reflection and absorption coefficients of a given surface at 0°C for a fixed wavelength are 0.5 (each). At the same temperature and wavelength, the transmission (coefficient) of the surface will be

1. 0.5
2. 1.0
3. Zero
4. In between zero and one

Answer: 3. Zero

Question 125. The earth receives radiation from the sun at the rate of 1400 watts/m². The distance from the center of the sun to the surface of the earth is 1.5 × 1011 m and the radius of the sun is 7.0 × 108 m. Treating the sun as a black body the temperature of the sun will be

1. 6000K
2. 5500K
3. 5800K
4. 6200K

Answer: 3. 5800K

Question 126. The rate of cooling of a heated solid sphere is R cal/min. If it is divided into two hemispheres the rate of cooling at the same temperature will become-

1. 1.25R cal/min.
2. 1.5R cal/min.
3. 1.75R cal/min.
4. 2.5R cal/min.

Answer: 2. 1.5R cal/min.

Question 127. Equal volumes of a liquid of relative density 1.02 and water are allowed to cool from 80°C to 60°C in the same surroundings. The times taken are 8 mts and 15 mts respectively. The specific heat of the liquid in cal/gm-°C is-

1. 0.52
2. 0.81
3. 1.02
4. 1.23

Answer: 1. 0.52

Question 128. Two identical calorimeters of negligible heat capacities are filled with two liquids A & B whose densities are in the ratio 4 : 3. The ratio of times taken in cooling from 80°C to 75°C is 5: 6. The ratio of their specific heats is-

1. 1: 2
2. 5: 6
3. 4 : 3
4. 5: 8

Answer: 4. 5: 8

Question 129. Blackened metal foil receives heat from a heated sphere placed at a distance r from it. It is found that foil receives power P. If the temperature and the distance of the sphere are doubled, then the power received by the foil will be

1. P
2. 2P
3. 8P
4. 4P

Answer: 4. 4P

NEET Physics Class 11 Heat Transfer MCQs and Solutions

Question 130. The temperature of the two outer surfaces of a composite slab, consisting of two materials K and 2K, and thickness x and 4x, respectively, are T₂ and T₁(T₂> T₁). The rate of heat transfer through the slab,\(\left(\frac{A\left(T_2-T_1\right) K}{x}\right) f\) with f equal to–

1. 1
2. 1/2
3. 2/3
4. 1/3

Answer: 4. 1/3

Question 131. If the radius of a star is R and it acts as a black body, what would be the temperature of the star, in which the rate of energy production is Q?

1. Q /4πR²σ
2. (Q /4πR²σ)–1/2
3. (4πR²Q/σ)1/4
4. (Q/ 4πR²σ)1/4

Answer: 4. (Q/ 4πR²σ)1/4

Question 132. A slab of stone of area 0.36 m² and thickness 0.1 m is exposed on the lower surface to steam at 100°C. A block of ice at 0°C rests on the upper surface of the slab. In one hour 4.8 kg of ice is melted. The thermal conductivity of slab is :(Given latent heat of fusion of ice = 3.36 × 105 J kg^-1) :

1. 1.24 J/m/s/°C
2. 1.29 J/m/s/°C
3. 2.05 J/m/s/°C
4. 1.02 J/m/s/°C

Answer: 1. 1.24 J/m/s/°C

Question 133. A piece of iron is heated in a flame. It first becomes dull red then becomes reddish yellow and finally turns to white hot. The correct explanation for the above observation is possible by using :

1. Wien’s Displacement Law
2. Kirchoff’s Law
3. Newton’s Law of Cooling
4. Stefan’s Law

Answer: 1. Wien’s displacement Law

Question 134. A certain quantity of water cools from 700C to 600C in the first 5 minutes and to 540C in the next 5 minutes. The temperature of the surroundings is;

1. 450C
2. 200C
3. 420C
4. 100C

Answer: 1. 450C

Question 135. On observing light from three different stars P, Q and R, it was found that the intensity of the violet colour is maximum in the spectrum of P, the intensity of the green colour is maximum in the spectrum of R and the intensity of the red colour is maximum in the spectrum in the spectrum of Q. If T_P, T_Q, and T_R are the respective absolute temperature of P, Q, and R, then it can be concluded from the above observations that :

1. T_P> T_R> T_Q
2. T_P< T_R< T_Q
3. T_P< T_Q<T_R
4. T_P >T_QT_R

Answer: 1. T_P> T_R> T_Q

Question 136. The two ends of a metal rod are maintained at temperatures 100ºC and 110ºC. The rate of heat flow in the rod is found to be 4.0 J/s. If the ends are maintained at temperatures 200ºC and 210ºC, the rate of heat flow will be :

1. 16.8 J/s
2. 8.0 J/s
3. 4.0 J/s
4. 44.0 J/s

Answer: 3. 4.0 J/s

Question 137. The coefficient of linear expansion of brass and steel rods are α₁ and α₂. Lengths of brass and steel rods are l₁ and l₂ respectively. If (l₂– l₁) is maintained the same at all temperatures, which one of the following relations holds good?

1. α₁l₁= α₂l₂
2. α₁l₂= α₂l₁
3. α₁l_2² = α₂l_1²
4. α_1²l₂= α_2² l₁

Answer: 1. α₁l₁= α₂l₂

Question 138. A refrigerator works between 4°C and 30°C. it is required to remove 600 calories of heat every second to keep the temperature of the refrigerated space constant. The power required is : (Take 1 cal = 4.2 Joules)

1. 2365 W
2. 2.365 W
3. 23.65 W
4. 236.5 W

Answer: 4. 236.5 W

Question 139. A piece of ice falls from a height h so that it melts completely. Only one-quarter of the heat produced is absorbed by the ice and all energy of ice gets converted into heat during its fall. The value of h is : [Latent heat of ice is 3.4 × 10⁵ J/Kg and g = 10 N/kg]

1. 68 km
2. 34 km
3. 544 km
4. 136 km

Answer: 4. 136 km

Question 140. A block body is at a temperature of 5760 K. The energy of radiation emitted by the body at wavelength 250 nm is U₁ at wavelength 500 nm is U₂ and that at 1000 nm is U₃. Wien’s constant, b = 2.88 × 10⁶ nmK. Which of the following is correct?

1. U₂> U₁
2. U₁= 0
3. U₃= 0
4. U₁> U₂

Answer: 1. U₂> U₁

Question 141. Two identical bodies are made of a material for which the heat capacity increases with temperature. One of these is at 100ºC, while the other one is at 0ºC. If the two bodies are brought into contact, then assuming no heat loss, the final common temperature is

1. 0º C
2. 50º C
3. More than 50º C
4. Less than 50º C but greater than 0º C

Answer: 3. More than 50º C

NEET Physics Class 11 Heat Transfer MCQs and Solutions

Question 142. A body cools from a temperature of 3T to 2T in 10 minutes. The room temperature is T. Assume that Newton’s law of cooling is applicable. The temperature of the body at the end of next 10 minutes will be

1. T
2. \(\frac{7}{4} \mathrm{~T}\)
3. \(\frac{3}{2} \mathrm{~T}\)
4. \(\frac{4}{3} \mathrm{~T}\)

Answer: 3. \(\frac{3}{2} \mathrm{~T}\)

Question 143. Two rods A and B of different materials are welded together as shown in the figure. Their thermal conductivities are K₁ and K₂. The thermal conductivity of the composite rod will be

1. \(\frac{\mathrm{K}_1+\mathrm{K}_2}{2}\)
2. \(\frac{3\left(\mathrm{~K}_1+\mathrm{K}_2\right)}{2}\)
3. \(\mathrm{K}_1+\mathrm{K}_2\)
4. \(2\left(\mathrm{~K}_1+\mathrm{K}_2\right)\)

Answer: 1. \(\frac{\mathrm{K}_1+\mathrm{K}_2}{2}\)

Question 144. A spherical black body with a radius of 12 cm radiates 450-watt power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watts would be :

1. 225
2. 450
3. 1000
4. 1800

Answer: 4. 1800

Question 145. The power was radiated by a black body in P and it radiated maximum energy at wavelength, λ₀. If the temperature of the black body is now changed so that it radiates maximum energy at wavelength \(\frac{3}{4} \lambda_0\) the power radiated by it becomes nP. The value of n is :

1. \(\frac{3}{4}\)
2. \(\frac{81}{256}\)
3. \(\frac{256}{81}\)
4. \(\frac{4}{3}\)

Answer: 3. \(\frac{256}{81}\)

Question 146. A copper rod of 88 cm and an aluminum rod of unknown length have their increase in length independent of an increase in temperature. The length of aluminium rod is (α_Cu = 1.7 × 10^-5 K^-1 and α_Al = 2.2 × 10^-5 K^-1)

1. 68 cm
2. 6.8 cm
3. 113.9 cm
4. 88 cm

Answer: 1. 68 cm

Question 147. The unit of thermal conductivity is :

1. W m^-1 K^-1
2. J m K^-1
3. J m^-1 K^-1
4. W m K^-1

Answer: 1. W m^-1 K^-1

Question 148. An object kept in a large room having an air temperature of 25ºC takes 12 minutes to cool from 80ºC to 70ºC. The time taken to cool the same object from 70º to 60ºC would be nearly

1. 10 min
2. 12 min
3. 20 min
4. 15 min

Answer: 4. 15 min

Question 149. A deep rectangular pond of surface area A, containing water (density = ρ, specific heat capacity = s), is located in a region where the outside air temperature is at a steady value of –26ºC. The thickness of the ice layer in this pond, at a certain instant, is x. Taking the thermal conductivity of ice as K, and its specific latent heat of fusion as L, the rate of increase of the thickness of the ice layer, at this instant, would be given by

1. 26K/ρx(L-4s)
2. 26K/(ρx2L)
3. 26K/(ρxL)
4. 26K/ρx(L+4s)

Answer: 3. 26K/(ρxL)

Question 150. Three stars A, B, and C have surface temperatures T_A, T_B, and T_C respectively. Star A appears bluish, star B appears reddish, and star C is yellowish. Hence

1. T_A> T_B> T_C
2. T_B> T_C> T_A
3. T_C> T_B> T_A
4. T_A> T_C> T_B

Answer: 4. T_A> T_C> T_B

Question 151. An ideal gas equation can be written as \(\mathrm{P}=\frac{\rho R T}{M_0}\) where ρ and M₀ are respectively,

1. Mass density is, the mass of the gas
2. Number density, molar mass
3. Mass density, molar mass
4. Number density, the mass of the gas

Answer: 3. Mass density, molar mass

Question 152. A cylinder contains hydrogen gas at a pressure of 245 K Pa and a temperature of 270C density is (R=8.3 J mol^-1 K^-1)

1. 0.02 kg/m³
2. 0.5 kg/m³
3. 0.2 kg/m³
4. 0.1 / kg m

Answer: 3. 0.2 kg/m³

Question 153. A cup of coffee cools from 90°C to 80°C in two minutes, when the room temperature is 20°C. The time taken by a similar cup of coffee to cool from 80°C to 60°C at room temperature same at 20°C is

1. \(\frac{13}{5} t\)
2. \(\frac{10}{13} t\)
3. \(\frac{5}{13} t\)
4. \(\frac{13}{10} t\)

Answer: 1. \(\frac{13}{5} t\)

Question 154. A flask containing air at 27ºC is corked up at atmospheric pressure. The cork can be forced out by a pressure of 2.5 atmospheres. To what temperature the flask should be heated to do that?

1. 150 K
2. 300 K
3. 600 K
4. 750 K

Answer: 4. 750 K

Question 155. 1 kcal of heat flowing through a rod of iron per second. When the rod is cut down to 4 pieces then what will be the heat flowing through each piece per second having the same differential temperature (temperature gradient)?

1. (1/2) kcal
2. (1/4) kcal
3. 1 kcal
4. (1/15) kcal

Answer: 3. 1 kcal

Question 156. Black holes in orbit around a normal star are detected from the earth due to the frictional heating of infalling gas into the black hole, which can reach temperatures greater than 106K. Assuming that the infalling gas can be modelled as a blackbody radiator then the wavelength of maximum power lies

1. In the visible region
2. In the X-ray region
3. In the microwave region
4. In the gamma-ray region of the electromagnetic spectrum.

Answer: 2. In the X-ray region

Question 157. Two conductors having the same width and length, thickness d₁ and d₂ thermal conductivity K₁ and K₂ are placed one above the other. Find the equivalent thermal conductivity.

1. \(\frac{\left(d_1+d_2\right)\left(K_1 d_2+K_2 d_1\right)}{2\left(K_1+K_2\right)}\)
2. \(\frac{\left(d_1-d_2\right)\left(K_1 d_2+K_2 d_1\right)}{2\left(K_1+K_2\right)}\)
3. \(\frac{K_1 d_1+K_2 d_2}{d_1+d_2}\)
4. \(\frac{K_1+K_2}{d_1+d_2}\)

Answer: 3. \(\frac{K_1 d_1+K_2 d_2}{d_1+d_2}\)

Question 158. A long metallic bar carries heat from one of its ends to the other end under a steady-state. The variation of temperature θ along the length x of the bar from its hot end is best described by which of the following figures

Answer: 1

Question 159. If a piece of metal is heated to temperature θ and then allowed to cool in a room which is at temperature θ₀, the graph between the temperature T of the metal and time t will be closest to:

Answer: 3

Question 160. Three rods of Copper, brass, and steel are welded together to form a Y-shaped structure. Area of cross section of each rod = 4 cm². The end of the copper rod is maintained at 100°C whereas ends of brass and steel are kept at 0°C. Lengths of the copper, brass, and steel rods are 46, 13, and 12 cm respectively. The rods are thermally insulated from surroundings except at the ends. Thermal conductivities of copper, brass, and steel are 0.92, 0.26, and 0.12 CGS units respectively. The rate of heat flow through copper rod is:

1. 1.2 cal/s
2. 2.4 cal/s
3. 4.8 cal/s
4. 6.0 cal/s

Answer: 4. 6.0 cal/s

Question 161. An ideal gas undergoes a quasi-static, reversible process in which its molar heat capacity C remains constant. If during this process the relation of pressure P and volume V is given by PVⁿ = constant, then n is given by (Here C_p and C_v are molar specific heat at constant pressure and constant volume, respectively ) :

1. \(n=\frac{C-C_p}{C-C_V}\)
2. \(n=\frac{C_p-C}{C-C_V}\)
3. \(n=\frac{C-C}{C-C p}\)
4. \(n=\frac{C_p}{C_V}\)

Answer: 1. \(n=\frac{C-C_p}{C-C_V}\)

Question 162. Temperature difference of 120°C is maintained between two ends of a uniform rod AB of length 2L. Another bent rod PQ, of the same cross-section as AB and length \(\frac{3 L}{2}\), is connected across AB (see figure). In a steady state, the temperature difference between P and Q will be close to :

1. 75°C
2. 45°C
3. 60°C
4. 35°C

Answer: 2. 45°C

Question 163. A heat source at T = 10³ K is connected to another heat reservoir at T = 10² K by a copper slab that is 1m thick. Given that the thermal conductivity of copper is 0.1 WK^-1 m^-1, the energy flux through it in the steady state is :

1. 200 Wm^-2
2. 90 Wm^-2
3. 65 Wm^-2
4. 120 Wm^-2

Answer: 2. 90 Wm^-2

Question 164. A thermometer graduated according to a linear scale reads a value of x₀ when in contact with ice. What is the temperature of an object in °C, if this thermometer in contact with the object reads x₀/2?

1. 35
2. 60
3. 25
4. 40

Answer: 3. 25

Question 165. A cylinder of radius R is surrounded by a cylindrical of inner radius R and outer radius 2R. The thermal conductivity of the material of the inner cylinder is K₁ and that of the outer cylinder is K₂. Assuming no loss of heat, the effective thermal conductivity of the system for the heart flowing along the length of the cylinder is:

1. \(\mathrm{K}_1+\mathrm{K}_2\)
2. \(\frac{2 K_1+3 K_2}{2}\)
3. \(\frac{\mathrm{K}_1+\mathrm{K}_2}{2}\)
4. \(\frac{\mathrm{K}_1+3 \mathrm{~K}_2}{4}\)

Answer: 4. \(\frac{\mathrm{K}_1+3 \mathrm{~K}_2}{4}\)

Question 166. Two rods A and B of identical dimensions are at a temperature of 30°C. If A is heated upto 180°C and B upto T°C, then the new lengths are the same. If the ratio of the coefficients of linear expansion of A and B is 4 : 3, then the value of T is :

1. 270°C
2. 200°C
3. 230°C
4. 250°C

Answer: 3. 230°C

Circular Motion

Fundamental parameter of circular motion

Radius Vector: The vector joining the center of the circle and the center of the particle performing circular motion is called the radius vector.

It has constant magnitude and variable direction

Angular Displacement (δθ or θ)

The angle described by the radius vector is called angular displacement.

Infinitesimal angular displacement is a vector quantity. However, finite angular displacement is a scalar quantity.

S.I Unit Radian

Dimension: M⁰L⁰T⁰

1 radian = \(\frac{360}{2 \pi}\)

No. Of revolution = \(\frac{\text { angular displacement }}{2 \pi}\)

In 1 revolution Δθ = 360º = 2π radian

In N revolution Δθ = 360º × N = 2πN radian

Clockwise rotation is taken as a negative

Anticlockwise rotation is taken as a positive

NEET Physics Class 11 Chapter 2 Circular Motion Solutions

Question 1. If a particle completes one and a half revolutions along the circumference of a circle then its angular displacement is –

1. 0
2. π
3. 2π
4. 2π

Answer: = 3π

Angular Velocity (ω):

• The rate of change of angular displacement with time is called angular velocity. It is a vector quantity.
• The angle traced per unit time by the radius vector is called angular speed.

Instantaneous angular velocity = \(=\omega=\lim _{\delta t \rightarrow 0} \frac{\delta \theta}{\delta \mathrm{t}} \text { or } \omega=\frac{\mathrm{d} \theta}{\mathrm{dt}}\)

Average angular velocity

= \(\bar{\omega}=\frac{\theta_2-\theta_1}{\mathrm{t}_2-\mathrm{t}_1}=\frac{\Delta \theta}{\Delta \mathrm{t}}\)

S.I. Unit: rad/sec

Angular Velocity Dimension: M⁰L⁰T^-1

Angular Velocity Direction: Infinitesimal angular displacement, angular velocity, and angular acceleration are vector quantities whose direction is given by the right-hand rule.

Right-hand Rule: Imagine the axis of rotation to be held in the right hand with fingers curled around the axis and the thumb stretched along the axis. If the curled fingers denote the sense of rotation, then the thumb denotes the direction of the angular velocity (or angular acceleration of infinitesimal angular displacement.

Angular Acceleration (a):

The rate of change of angular velocity with time is called angular acceleration. Average angular acceleration

⇒ \(\bar{\alpha}=\frac{\omega_2-\omega_1}{t_2-t_1}=\frac{\Delta \omega}{\Delta t}\)

Instantaneous angular acceleration

⇒ \(\alpha=\frac{d \omega}{d t}=\frac{d^2 \theta}{d t^2}\)

It is a vector quantity, whose direction is along the change in direction of angular velocity.

S.I. Unit: radian/sec²

Dimension: M⁰L⁰T^-2

Relation Between Angular Velocity And Linear Velocity:

Suppose the particle moves along a circular path from point A to point B in infinitesimally small time δt. As, δt → 0, δθ → 0

∴ arc AB = chord AB i.e. displacement of the particle is along a straight line.

∴ Linear velocity, v = \(v=\lim _{\delta t\rightarrow 0} \frac{\delta s}{\delta t}\)

But, δs = r.δθ

∴ v = \(v=\lim _{\delta t \rightarrow 0} \frac{r \cdot \delta s}{\delta t}=r \lim _{\delta t \rightarrow 0} \frac{\delta \theta}{\delta t}\)

But, \(\lim _{\delta t \rightarrow 0} \frac{\delta \theta}{\delta \mathrm{t}}=\omega\) = angular velocity

V= r. ω [for circular motion only]

i.e. (linear velocity) = (Radian) × (angular velocity)

In vector notation, \(\vec{v}=\vec{\omega} \times \vec{r}\) [in general]

The linear velocity of a particle performing circular motion is the vector product of its angular velocity and radius vector.

Circular Motion NEET Physics Class 11 Solutions

Relation Between Angular Acceleration And Linear Acceleration

For perfect circular motion, we know

v = ω r

on differentiating with respect to time

we get \(\frac{d v}{d t}=r \frac{d \omega}{d t}\)

a = r α

In vector form = \(\overrightarrow{\mathrm{a}}=\vec{\alpha} \times \overrightarrow{\mathrm{r}}\) ×(linear acc.) = (angular acc) × (radius)

Types Of Circular Motion

Uniform Circular Motion: The motion of a particle along the circumference of a circle with a constant speed is called uniform circular motion. Uniform circular motion is an accelerated motion. In the case of uniform circular motion :

Speed remains constant. v = constant

and v = ω r

angular velocity ω = constant

Motion will be periodic with time period = \(T=\frac{2 \pi}{\omega}=\frac{2 \pi r}{v}\)

Frequency Of Uniform Circular Motion: The number of revolutions performed per unit of time by the particle performing uniform circular motion is called the frequency (n)

∴ n = \(n=\frac{1}{T}=\frac{v}{2 \pi r}=\frac{\omega}{2 \pi}\)

S.I. unit of frequency is Hz.

As ω = constant, from ω = ω₀+ αt

angular acceleration α = 0

As a_t = αr, tangential acc. a_t= 0

As a_t = 0, a = \(a=\left(a_r^2+a_t^2\right)^{1 / 2}\) yields a = ar, i.e. acceleration is not zero but along radius towards the center and has magnitude

⇒ \(\mathrm{a}=\mathrm{a}_{\mathrm{r}}=\left(\mathrm{v}^2 / \mathrm{r}\right)=\mathrm{r} \omega^2\)

Speed and magnitude of acceleration are constant. but their directions are always changing so velocity and acceleration are not constant.

The direction of \(\overrightarrow{\mathrm{v}}\) is always along the tangent while that of \(\overrightarrow{\mathrm{a}_{\mathrm{r}}}\) along the radius \(\vec{v} \perp \vec{a}_r\)

If the moving body comes to rest, i.e. \(\vec{v} \rightarrow 0\), and if radial acceleration vanishes, the body will fly off along the tangent. So a tangential velocity and a radial acceleration (hence force) is a must for uniform circular motion.

As \(\vec{F}=\frac{m v^2}{r}\) ≠ 0, the body is not in equilibrium and the linear momentum of the particle moving on the circle is not conserved. However, as the force is control, i.e.,

⇒ \(\vec{\tau}=0\), so angular momentum is conserved, i.e.,

⇒ \(\overrightarrow{\mathrm{p}}\) ≠ constant but

⇒ \(\overrightarrow{\mathrm{L}}\) = constant

The work done by a centripetal force is always zero as it is perpendicular to velocity and hence displacement. By work-energy theorem as work done = change in kinetic energy ΔK = 0

So K (kinetic energy) remains constant

For example., Planets revolving around the sun, the motion of an electron around the nucleus in an atom

In one-dimensional motion, acceleration is always parallel to velocity and changes only the magnitude of the velocity vector.

In uniform circular motion, acceleration is always perpendicular to velocity and changes only the direction of the velocity vector.

In the more general case, like projectile motion, acceleration is neither parallel nor perpendicular to the figure that summarizes these three cases.

If a particle moving with uniform speed v on a circle of radius r suffers angular displacement θ in time Δt then change in its velocity.

⇒ \(\Delta \vec{v}=\Delta \vec{v}_2-\Delta \vec{v}_1\)

⇒ \(\vec{v}_1=\vec{v}_1 \hat{i}\)

⇒ \(\vec{v}_2=\vec{v}_2 \cos \theta \hat{i}+\vec{v}_2 \sin \theta \hat{j}\)

⇒ \(\Delta \vec{v}=\left(\vec{v}_2 \cos \theta-\vec{v}_1\right) \hat{i}+\vec{v}_2 \sin ^2 \hat{j}\)

⇒ \(|\Delta \vec{v}|=\sqrt{\left(\vec{v}_2 \cos \theta-\vec{v}_1\right)^2+\vec{v}_2 \sin ^2}\)

⇒ \(|\Delta \vec{v}|=\sqrt{2 v^2-2 v^2 \cos \theta}=\sqrt{2 v^2(1-\cos \theta)}=\sqrt{2 v^2\left(2 \sin ^2 \frac{\theta}{2}\right)}\)

⇒ \(v_1=v_2=v\)

⇒ \(|\Delta \vec{v}|=2 v \sin \frac{\theta}{2}\)

Question 1. A particle is moving in a circle of radius r centered at O with constant speed v. What is the change in velocity in moving from A to B? Given ∠AOB = 40º.
Answer:

⇒ \(|\Delta \vec{v}|=2 v \sin 40^{\circ} / 2=2 \mathrm{v} \sin 20^{\circ}\)

Non-Uniform Circular Motion:

A circular motion in which both the direction and magnitude of the velocity change is called nonuniform circular motion.

• A merry-go-round is spinning up from rest to full speed, or a ball whirling around in a vertical circle. The acceleration is neither parallel nor perpendicular to the velocity.
• We can resolve the acceleration vector into two components:

Radial Acceleration: ar perpendicular to the velocity ⇒ changes only the directions of velocity Acts just like the acceleration in a uniform circular motion.

⇒ \(a_c=\text { or } \quad a_r=\frac{v^2}{r}\)

Centripetal force: \(F_c=\frac{m v^2}{r}=m \omega^2 r\)

Tangential acceleration: a_r parallel to the velocity (since it is tangent to the path)

⇒ changes in the magnitude of the velocity act just like one-dimensional acceleration

⇒ \(a_t=\frac{d v}{d t}\)

Tangential acceleration : \(a_t=\frac{d v}{d t}\) where \(v=\frac{d s}{d t}\) and s = length of arc

Tangential acceleration: F_t = ma_t

The net acceleration vector is obtained by vector addition of these two components.

⇒ \(a=\sqrt{a_r^2+a_t^2}\)

In non-uniform circular motion :

speed \(|\vec{v}|\) ≠ constant angular velocity ω ≠ constant

i.e. speed ≠ constant i.e. angular velocity ≠ constant

In any instant

⇒ v = magnitude of the velocity of a particle

⇒ r = radius of circular path

⇒ ω = angular velocity of a particle

then, at that instant v = r ω

The net force on the particle

⇒ \(\vec{F}=\vec{F}_c+\vec{F}_t \Rightarrow F=\sqrt{F_c^2+F_t^2}\)

If θ is the angle made by F = Fc,

then tan θ = \(=\frac{F_t}{F_c} \Rightarrow \theta=\tan^{-1}\left[\frac{F_{\mathrm{t}}}{F_c}\right]\)

[Note angle between Fcand Ftis 90º] Angle between F and Ftis (90º – θ)

Net acceleration: \(a=\sqrt{a_c^2+a_1^2}=\frac{F_{\mathrm{net}}}{m}\)

The angle made by ‘a’ with ac, tan θ = \(\frac{a_t}{a_c}=\frac{F_t}{F_c}\)

Special Note:

In both uniform and non-uniform circular motion Fc is perpendicular to velocity.

So work done by centripetal force will be zero in both cases.

In uniform circular motion Ft= 0, as = at= 0, so work done will be zero by tangential force.

But in non-uniform circular motion Ft≠ 0, the work done by tangential force is non-zero.

Rate of work done by net force in non-uniform circular motion = rate of work done by tangential force

⇒ \(P=\frac{d W}{d t}=\vec{F}_t \cdot \vec{v}=\vec{F}_t \cdot \frac{d \vec{x}}{d t}\)

In a circle tangent and radius are always normal to each other, so

⇒ \(\vec{a}_{\mathrm{t}} \perp \vec{a}_{\mathrm{r}}\)

Net acceleration in case of circular motion \(a=a_r^2=a_t^2\)

Here it must be noted that at governs the magnitude of \(\vec{v}\) while ar its direction of motion so that

If ar = 0 and at = 0 a → 0 ⇒ motion is uniform translatory

If ar = 0 and at ≠ 0 a → at ⇒ motion is accelerated translatory

If ar ≠ 0 and at = 0 a → ar ⇒ motion is uniform circular

If ar ≠ 0 and at ≠ 0 a → \(a \rightarrow \sqrt{a_r^2+a_1^2}\) ⇒ motion is non-uniform circular.

NEET Class 11 Physics Chapter 2 Circular Motion: Solutions and Explanations

Question 2. A road makes a 90º bend with a radius of 190 m. A car enters the bend moving at 20 m/s. Finding this too fast, the driver decelerates at 0.92 m/s². Determine the acceleration of the car when its speed rounding the bend has dropped to 15 m/s.
Answer:

Since it is rounding a curve, the car has a radial acceleration associated with its changing direction, in addition to the tangential deceleration that changes its speed. We are given that at = 0.92 m/s²; since the car is slowing down, the tangential acceleration is directed opposite the velocity.

The radial acceleration is \(a_r=\frac{v^2}{r}=\frac{(15 \mathrm{~m} / \mathrm{s})^2}{190 \mathrm{~m}}=1.2 \mathrm{~m} / \mathrm{s}^{21}\)

Magnitude of net acceleration,

⇒ \(a=\sqrt{a_r^2+a_t^2}=\left[(1.2 \mathrm{~m} / \mathrm{s})^2+(0.92 \mathrm{~m} / \mathrm{s})^2\right]^{1 / 2}=1.5 \mathrm{~m} / \mathrm{s}^2\)

and points at an angle \(\theta=\tan ^{-1}\left(\frac{\mathrm{a}_{\mathrm{r}}}{\mathrm{a}_{\mathrm{t}}}\right)=\tan ^{-1}\left(\frac{1.2 \mathrm{~m} / \mathrm{s}^2}{0.92 \mathrm{~m} / \mathrm{s}^2}\right)=53^{\circ}\)

relative to the tangent line to the circle.

Question 3. A particle is constrained to move in a circular path of radius r = 6m. Its velocity varies with time according to the relation v = 2t (m/s). Determine its

1. Centripetal acceleration,
2. Tangential acceleration,
3. Instantaneous acceleration at
   1. t = 0 sec. and
   2. t = 3 sec.

Answer:

At = 0,

v = 0, Thus a_r = 0

but \(\frac{d v}{d t}=2\) thus at = 2 m/s2 and a = \(\sqrt{a_t^2+a_r^2}=2 \mathrm{~m} / \mathrm{s}^2\)

At t = 3 sec. v = 6 m/s so \(a_r=\frac{v^2}{r}=\frac{(6)^2}{6}=6 \mathrm{~m} / \mathrm{s}^2\)

and \(a_t=\frac{d v}{d t}=2 \mathrm{~m} / \mathrm{s}^2\) Therefore, \(a=a=\sqrt{2^2+6^2}=\sqrt{40} \mathrm{~m} / \mathrm{s}^2\)

Question 4. The kinetic energy of a particle moving along a circle of radius r depends on the distance covered as K = As² where A is a constant. Find the force acting on the particle as a function of s.
Answer:

According to the given Question

⇒ \(\frac{1}{2} m v^2=A s^2 \text { or } v=s \sqrt{\frac{2 A}{m}}\) ……….(1)

So \(a_{\mathrm{r}}=\frac{\mathrm{v}^2}{\mathrm{r}}=\frac{2 \mathrm{A} \mathrm{s}^2}{\mathrm{mr}}\) ………….(2)

Furthermore as at = \(a_t=\frac{d v}{d t}=\frac{d v}{d s} \cdot \frac{d s}{d t}=v \frac{d v}{d s}\) …………(3)

from eqn. (1), ⇒ ……….. (4)

Substitute values from eqn(1) and eqn(4) in eqn(3)

⇒ \(a_t=\left[s \sqrt{\frac{2 A}{m}}\right]\left[\sqrt{\frac{2 A}{m}}\right]=\frac{2 A s}{m}\)

so \(a=\sqrt{a_r^2+a_t^2}=\sqrt{\left[\frac{2 A s^2}{m r}\right]^2+\left[\frac{2 A s}{m}\right]^2}\)

i.e. \(\mathrm{a}=\frac{2 \mathrm{As}}{\mathrm{m}} \sqrt{1+[\mathrm{s} / \mathrm{r}]^2}\)

so \(\mathrm{F}=\mathrm{ma}=2 \mathrm{As} \sqrt{1+[\mathrm{s} / \mathrm{r}]^2}\)

NEET Physics Class 11 Circular Motion Key Solutions

Question 5. A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration varies with time t as ac= k²rt², where k is a constant. Determine the power delivered to a particle by the forces acting on it.
Answer:

If v is instantaneous velocity, centripetal acceleration ac = \(a_c=\frac{v^2}{r} \Rightarrow=\frac{v^2}{r} k^2 r^2 \Rightarrow v=k r t\)

In circular motion work done by centripetal force is always zero and work is done only by tangential force.

∵ Tangent acceleration \(a_t=\frac{d v}{d t}=\frac{d}{d t}(k r t)=k r\)

∴ Tangential force Ft= mat= mkr

Power P = \(F_t v=(m k r)(k r t)=m k^2 r^2 t\)

Circular Motion Multiple Choice Question And Answers

Question 1. Two racing cars of masses m₁ and m₂ are moving in circles of radii r₁ and r₂ respectively; their speeds are such that they each make a complete circle in the same time t. The ratio of the angular speed of the first to the second car is :

1. m₁: m₂
2. r₁: r₂
3. 1: 1
4. m₁r₁: m₂r₂

Answer: 1. m₁: m₂

Question 2. A wheel is at rest. Its angular velocity increases uniformly and becomes 80 radians per second after 5 seconds. The total angular displacement is :

1. 800 rad
2. 400 rad
3. 200 rad
4. 100 rad

Answer: 3. 200 rad

Question 3. When a particle moves in a circle with a uniform speed

1. Its velocity and acceleration are both constant
2. Its velocity is constant but the acceleration changes
3. Its acceleration is constant but the velocity changes
4. Its velocity and acceleration both change

Answer: 4. Its velocity and acceleration both change

Circular Motion MCQs for NEET Physics Class 11 with Solutions

Question 4. The relation between an angular velocity, the position vector and the linear velocity of a particle moving in a circular path is.

1. \(\vec{\omega} \times \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{v}}\)
2. \(\vec{\omega} \cdot \vec{r}=\vec{v}\)
3. \(\overrightarrow{\mathbf{r}} \times \vec{\omega}=\overrightarrow{\mathrm{v}}\)
4. \(\vec{\omega} \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{v}}\)

Answer: 1. \(\vec{\omega} \times \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{v}}\)

Question 5. A wheel is of diameter 1m. If it makes 30 revolutions/sec., then the linear speed of a point on its circumference will be.

1. 30 π m/s
2. π m/s
3. 60π m/s
4. π/2 m/s

Answer: 1. 30π m/s

Question 6. In a uniform circular motion

1. Both the angular velocity and the angular momentum vary
2. The angular velocity varies but the angular momentum remains constant.
3. Both the angular velocity and the angular momentum stay constant
4. The angular momentum varies but the angular velocity remains constant.

Answer: 3. Both the angular velocity and the angular momentum stay constant

Question 7. The angular speed of a flywheel making 120 revolutions/minute is.

1. 2π rad/s
2. 4π2 rad/s
3. π rad/s
4. 4π rad/s

Answer: 4. 4π rad/s

Question 8. The angular velocity of the second needle in a watch is-

1. \(\frac{\pi}{30}\)
2. 2π
3. π
4. \(\frac{60}{\pi}\)

Answer: 1. \(\frac{\pi}{30}\)

Question 9. The average acceleration vector for a particle having a uniform circular motion is- 

1. A constant vector of magnitude \(\frac{v^2}{r}\)
2. A vector of magnitude \(\frac{v^2}{r}\) directed normal to the plane of the given uniform circular motion.
3. Equal to the instantaneous acceleration vector at the start of the motion.
4. A null vector.

Answer: 4. A null vector.

Question 10. The angular velocity of the minute hand of a clock is:

1. \(\frac{\pi}{30} \mathrm{rad} / \mathrm{s} \)
2. π rad/s
3. 2π rad/s
4. \(\frac{\pi}{1800} \mathrm{rad} / \mathrm{s}\)

Answer: 4. \(\frac{\pi}{1800} \mathrm{rad} / \mathrm{s}\)

Question 11. The second hand of a watch has a length of 6 cm. The speed of the endpoint and magnitude of the difference of velocities at two perpendicular positions will be :

1. 2π and 0 mm/s
2. \(2 \sqrt{2}\) π and 4.44 mm/s
3. \(2 \sqrt{2}\) π and 2π mm/s
4. 2π and \(2 \sqrt{2}\) π mm/s

Answer: 4. 2π and \(2 \sqrt{2}\) π mm/s

Question 12. An aeroplane revolves in a circle above the surface of the earth at a fixed height with a speed of 100 km/hr. The change in velocity after completing 1/2 revolution will be.

1. 200 km/hr
2. 150 km/hr
3. 300 km/hr
4. 400 km/hr

Answer: 1. 200 km/hr

Question 13. A particle moving on a circular path travels the first one-third part of the circumference in 2 sec and the next one-third part in 1 sec. The average angular velocity of the particle is (in rad/sec) –

1. \(\frac{2 \pi}{3}\)
2. \(\frac{\pi}{3}\)
3. \(\frac{4 \pi}{9}\)
4. \(\frac{5 \pi}{3}\)

Answer: 3. \(\frac{4 \pi}{9}\)

NEET Class 11 Physics Circular Motion MCQs and Answer Key

Question 14. A grind-stone starts revolving from rest, if its angular acceleration is 4.0 rad/sec² (uniform) then after 4 sec. What are its angular displacement and angular velocity respectively –

1. 32 rad, 16 rad/sec
2. 16 rad, 32 rad/sec
3. 64 rad, 32 rad/sec
4. 32 rad, 64 rad/sec

Answer: 1. 32 rad, 16 rad/sec

Question 15. Angular displacement of any particle is given θ = ω₀t +\(\frac{1}{2}\)αt² where ω₀ and α are constant if ω₀ = 1 rad/sec, α = 1.5 rad/sec² then in t = 2 sec. angular velocity will be (in rad/sec)

1. 1
2. 5
3. 3
4. 4

Answer: 4. 4

Question 16. A particle of mass M is revolving along a circle of radius R and another particle of mass m is revolving in a circle of radius r. If the periods of both particles are the same, then the ratio of their angular velocities is:

1. 1
2. \(\frac{R}{r}\)
3. \(\frac{r}{R}\)
4. \(\sqrt{\frac{R}{r}}\)

Answer: 1. 1

Question 17. In a uniform circular motion

1. Velocity and acceleration remain constant
2. Kinetic energy remains constant
3. Speed and acceleration changes
4. Only velocity changes and acceleration remains constant

Answer: 2. Kinetic energy remains constant

Question 18. Which of the following statements is false for a particle moving in a circle with a constant angular speed?

1. The velocity vector is tangent to the circle
2. The acceleration vector is tangent to the circle
3. The acceleration vector points to the center of the circle
4. The velocity and acceleration vectors are perpendicular to each other

Answer: 2. The acceleration vector is tangent to the circle

Question 19. A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane, it follows that

1. Its velocity is constant
2. Its acceleration is constant
3. Its kinetic energy is constant
4. It moves in a straight line

Answer: 3. Its kinetic energy is constant

Question 20. A wheel is subjected to uniform angular acceleration about its axis. Initially, its angular velocity is zero. In the first 2 seconds, it rotates through an angle θ1. In the next 2 sec, it rotates through an additional angle θ \(\frac{\theta_2}{\theta_1}\) is

1. 1
2. 2
3. 3
4. 5

Answer: 3. 3

Question 21. If the equation for the displacement of a particle moving on a circular path is given by (θ)= 2t³ + 0.5, where θ is in radians and t in seconds, then the angular velocity of the particle after 2 sec from its start is

1. 8 rad/sec
2. 12 rad/sec
3. 24 rad/sec
4. 36 rad/sec

Answer: 3. 24 rad/sec

Question 22. For a particle in a non-uniform accelerated circular motion

1. Velocity is radial and acceleration is transverse only
2. Velocity is transverse and acceleration is radial only
3. Velocity is radial and acceleration has both radial and transverse components
4. Velocity is transverse and acceleration has both radial and transverse components

Answer: 4. Velocity is transverse and acceleration has both radial and transverse components

Question 23. Two particles P and Q are located at distances rP and rQ respectively from the axis of a rotating disc such that rP > rQ :

1. Both P and Q have the same acceleration
2. Both P and Q do not have any acceleration
3. P has greater acceleration than Q
4. Q has greater acceleration than P

Answer: 3. P has greater acceleration than Q

Question 24. Let ar and at represent radial and tangential acceleration. The motion of a particle may be circular if :

1. a_r = 0, a_t = 0
2. a_r = 0, a_t ≠ 0
3. a_r ≠ 0, a_t = 0
4. None of these

Answer: 3. a_r ≠ 0, a_t = 0

Question 25. A stone tied to one end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If a stone makes 25 revolutions in 14 seconds, the magnitude of the acceleration of the stone is :

1. 850 cm/s²
2. 996 cm/s²
3. 720 cm/s²
4. 650 cm/s²

Answer: 2. 996 cm/s²

Question 26. A body is moving in a circular path with acceleration a. If its velocity gets doubled, find the ratio of acceleration after and before the change :

1. 1: 4
2. 4: 1
3. 2: 1
4. 2: 1

Answer: 2. 4: 1

NEET Physics Class 11 Circular Motion MCQs with Solutions

Question 27. A spaceman in training is rotated in a seat at the end of a horizontal arm of length 5m. If he can withstand acceleration upto 9 g then what is the maximum number of revolutions per second permissible? (Take g = 10 m/s²)

1. 13.5 rev/s
2. 1.35 rev/s
3. 0.675 rev/s
4. 6.75 rev/s

Answer: 3. 0.675 rev/s

Question 28. A particle of mass m is moving in a uniform circular motion. The momentum of the particle is

1. Constant over the entire path
2. Constantly changes and direction of change is along the tangent
3. Constantly changes and direction of change are along the radial direction
4. Constantly change and direction of change are along a direction which is the instantaneous vector sum of the radial and tangential direction

Answer: 3. Constantly changes and direction of change is along the radial direction

Question 29. A particle is going in a uniform helical and spiral path separately as shown in the figure with constant speed.

1. The velocity of the particle is constant in both cases
2. The acceleration of the particle is constant in both cases
3. The magnitude of acceleration is constant in (1) and decreasing in (2)
4. The magnitude of acceleration is decreasing continuously in both cases

Answer: 3. The magnitude of acceleration is constant in (1) and decreasing in (2)

Question 30. A car is traveling with linear velocity v on a circular road of radius r. If the speed is increasing at the rate of ‘a’ meter/sec², then the resultant acceleration will be –

1. \(\sqrt{\left[\frac{v^2}{r^2}-a^2\right]}\)
2. \(\sqrt{\left.\frac{v^4}{r^2}+a^2\right]}\)
3. \(\sqrt{\left[\frac{v^4}{r^2}-a^2\right]}\)
4. \(\sqrt{\left.\frac{v^2}{r^2}+a^2\right]}\)

Answer: 2. \(\sqrt{\left.\frac{v^4}{r^2}+a^2\right]}\)

Question 31. If the mass, speed & radius of rotation of a body moving on a circular path are increased by 50% then to keep the body moving in a circular path increase in force required will be –

1. 225%
2. 125%
3. 150%
4. 100%

Answer: 2. 125%

Question 32. A motorcycle driver doubles its velocity when he is having a turn. The force exerted outwardly will be.

1. Double
2. Half
3. 4 times
4. 1/4 times

Answer: 3. 4 times

Question 33. For a particle in circular motion, the centripetal acceleration is

1. Less than its tangential acceleration
2. Equal to its tangential acceleration
3. More than its tangential acceleration
4. May be more or less than its tangential acceleration

Answer: 4. May be more or less than its tangential acceleration

Question 34. If the radii of circular paths of two particles of the same masses are in the ratio of 1: 2, then in order to have the same centripetal force, their speeds should be in the ratio of:

1. 1: 4
2. 4: 1
3. 1 : \(\sqrt{2}\)
4. \(\sqrt{2}\): 1

Answer: 3. 1 : \(\sqrt{2}\)

Question 35. On a horizontal smooth surface, a mass of 2 kg is whirled in a horizontal circle by means of a string at an initial angular speed of 5 revolutions per minute. Keeping the radius constant the tension in the string is doubled. The new angular speed is near:

1. 14 rpm
2. 10 rpm
3. 2.25 rpm
4. 7 rpm