182 + Interview Questions in CIRCULAR MOTION IN GENERAL AWARENESS Page 4 InterviewSolution

151.	Consider an object of mass m that moves in a circular orbit with constant velocity v0 along the inside of a cone. Assume the wall of the cone to be frictionless. Find radius of the orbit.(A) \(\frac{v_0^2}{g}\) tan2ϕ(B) \(\frac{v_0^2}{g}\) cos2ϕ(C) \(\frac{v_0^2}{g}\) tanϕ(D) \(\frac{v_0^2}{g}\)
Answer» Answer is (C) \(\frac{v_0^2}{g}\) tanϕ

Discussion

152.	Give the formula of Centripetal acceleration or radial acceleration.
Answer» a = \(\cfrac{\text v^2}r\) = ω²r

Discussion

153.	When a particle moves in a uniform circular motion. It has(A) radial velocity and radial acceleration.(B) tangential velocity and radial acceleration.(C) tangential velocity and tangential acceleration.(D) radial velocity and tangential acceleration.
Answer» Correct option is (B) tangential velocity and radial acceleration.

Discussion

154.	In uniform circular motion,(A) both the angular velocity and the angular momentum vary. (B) the angular velocity varies but the angular momentum remains constant. (C) both the angular velocity and the angular momentum remains constant. (D) the angular momentum varies but the angular velocity remains constant.
Answer» Answer is (C) both the angular velocity and the angular momentum remains constant. L = Iω. In U.C.M., ω = constant ∴ L = constant

154.

In uniform circular motion,(A) both the angular velocity and the angular momentum vary. (B) the angular velocity varies but the angular momentum remains constant. (C) both the angular velocity and the angular momentum remains constant. (D) the angular momentum varies but the angular velocity remains constant.

Answer»

Answer is (C) both the angular velocity and the angular momentum remains constant.

L = Iω. In U.C.M., ω = constant

∴ L = constant

Discussion

155.	The motion of a sphere moving on a rough horizontal surface changes from pure sliding (without rolling) to pure rolling (without slipping). In this process, the force of friction (a) initially acts opposite to the direction of motion and later in the direction of motion (b) causes linear retardation (c) causes angular acceleration (d) stops acting when pure rolling begins
Answer» Correct Answer is: (b, c, & d)

Discussion

156.	When slightly different weights are placed on the two pans of a beam balance, the beam comes to rest at an angle with the horizontal. The beam is supported at a single point P by a pivot. (a) The net torque about P due to the two weights is nonzero at the equilibrium position. (b) The whole system does not continue to rotate about P because it has a large moment of inertia.(c) The centre of mass of the system lies below P. (d) The centre of mass of the system lies above P.
Answer» Correct Answer is: (a, c)

Discussion

157.	A rectangular block has a square base measuring a x a, and its height is h. It moves with a speed v on a smooth horizontal surface.(a) It will topple if v > √(2gh).(b) It will topple if v > √(2ga).(c) It will topple if v > √(2ga2 / h).(d) It will not topple for any value of v.
Answer» Correct Answer is: (d) It will not topple for any value of v. A body topples due to an unbalanced torque about its centre of mass. Here, the two forces acting on the block, viz., mg and N, both pass through the centre of mass and hence produce no torque.

157.

A rectangular block has a square base measuring a x a, and its height is h. It moves with a speed v on a smooth horizontal surface.(a) It will topple if v > √(2gh).(b) It will topple if v > √(2ga).(c) It will topple if v > √(2ga2 / h).(d) It will not topple for any value of v.

Answer»

Correct Answer is: (d) It will not topple for any value of v.

A body topples due to an unbalanced torque about its centre of mass. Here, the two forces acting on the block, viz., mg and N, both pass through the centre of mass and hence produce no torque.

Discussion

158.	A crate of egg is located in the middle of the flat bed of a pick up truck as the truck negotiates an unbanked curve in the road. The curve may be regarded as an arc of circle of radius 35 m. If the coefficient of friction between the crate and the flat bed of the truck is 0.6, the speed with which the truck should turn so that the crate does not slide over the bed is (A) 14.3 m/s (B) 10.3 m/s (C) 12.3 m/s (D) 15.3 m/s
Answer» Answer is (A) 14.3 m/s For the crate not to slide, the centripetal force should be \(\frac{mv^2}{r}\) = μmg ∴ v² = μrg = 0.6 x 35 x 9.8 = 205.8 ∴ v = 14.3 m/s

158.

A crate of egg is located in the middle of the flat bed of a pick up truck as the truck negotiates an unbanked curve in the road. The curve may be regarded as an arc of circle of radius 35 m. If the coefficient of friction between the crate and the flat bed of the truck is 0.6, the speed with which the truck should turn so that the crate does not slide over the bed is (A) 14.3 m/s (B) 10.3 m/s (C) 12.3 m/s (D) 15.3 m/s

Answer»

Answer is (A) 14.3 m/s

For the crate not to slide, the centripetal

force should be \(\frac{mv^2}{r}\) = μmg

∴ v² = μrg = 0.6 x 35 x 9.8 = 205.8

∴ v = 14.3 m/s

Discussion

159.	The minute hand of a clock is 5 cm long. Calculate the linear speed of an ant sitting at the tip
Answer» Linear speed of an ant sitting at the tip = 8.72 x 10^–5 m/s

Discussion

160.	The centripetal acceleration is given by(A) v2/r (B) vr (C) vr2 (D) v/r
Answer» Answer is (A) v²/r

Discussion

161.	Angle between radius vector and centripetal acceleration is(A) 0c (B) πc (C) 2πc (D) none of these
Answer» Answer is (B) π^c The radius vector points outwards while the centripetal acceleration points inwards along the radius.

161.

Angle between radius vector and centripetal acceleration is(A) 0c (B) πc (C) 2πc (D) none of these

Answer»

Answer is (B) π^c

The radius vector points outwards while the centripetal acceleration points inwards along the radius.

Discussion

162.	A body moves along a circular path with certain velocity. What will be the path of body in following figure?(A) Move radially out. (B) Move horizontally out. (C) Fall vertically down. (D) Move tangentially out.
Answer» Answer is (D) Move tangentially out. The instantaneous velocity of a body in U.C.M. is always perpendicular to the radius or along the tangent to the circle at the point.

162.

A body moves along a circular path with certain velocity. What will be the path of body in following figure?(A) Move radially out. (B) Move horizontally out. (C) Fall vertically down. (D) Move tangentially out.

Answer»

Answer is (D) Move tangentially out.

The instantaneous velocity of a body in U.C.M. is always perpendicular to the radius or along the tangent to the circle at the point.

Discussion

163.	The angular speed and linear speed of tip of a second hand of clock acting on it, if its angular speed of rotation is 0.6 rad/s.
Answer» Centripetal force = 0.036 N

Discussion

164.	Calculate the angular speed and linear speed of tip of a second hand of clock if second hand is 5 cm long.
Answer» The angular speed and linear speed of tip of a second hand of clock are 1.07 x 10^–1 rad/s, 5.235 x 10^–3 m/s respectively.

Discussion

165.	A clown is exhibiting a magic trick on the streets wherein he rotates a bucket in a vertical plane without allowing the water in it to spill out. Here, clearly the clown uses centrifugal force to balance the weight of water. This will be possible when,(A) the bucket has r.p.m. = \(\sqrt{\frac{400g}{\pi^2R}}\)(B) the bucket has maximum speed = \(\sqrt{2gR }\)(C) the bucket has r.p.m. = \(\sqrt{\frac{900\,g}{\pi^2\,R}}\)(D) the bucket has r.p.m = \(\sqrt{\frac{3600\,g}{\pi^2R}}\)
Answer» Answer is (C) the bucket has r.p.m. = \(\sqrt{\frac{900\,g}{\pi^2\,R}}\)

Discussion

166.	For a particle moving in a vertical circle, (A) kinetic energy is constant. (B) potential energy is constant. (C) neither K.E. nor P.E. is constant. (D) both kinetic energy and potential energy are constant.
Answer» Answer is (C) neither K.E. nor P.E. is constant.

Discussion

167.	When a particle is moved in a vertical circle, (A) it has constant radial and tangential acceleration. (B) it has variable tangential and radial acceleration.(C) it has only constant radial acceleration.(D) it has only constant tangential acceleration.
Answer» Answer is (B) it has variable tangential and radial acceleration.

Discussion

168.	Certain neutron stars are believed to be rotating at about 1 rev/s. If such a star has a radius of 20 km, the acceleration of an object on the equator of the star will be(A) 20 x 108 m/s2 (B) 8 x 105 m/s2 (C) 120 x 105 m/s2 (D) 4 x 108 m/s2
Answer» Answer is (B) 8 x 10⁵ m/s² Using, a = ω² r = 4π² n² r = 4(3.14)² x 1² x 20 x 10³ ∴ a = 8 x 10⁵ m/s²

168.

Certain neutron stars are believed to be rotating at about 1 rev/s. If such a star has a radius of 20 km, the acceleration of an object on the equator of the star will be(A) 20 x 108 m/s2 (B) 8 x 105 m/s2 (C) 120 x 105 m/s2 (D) 4 x 108 m/s2

Answer»

Answer is (B) 8 x 10⁵ m/s²

Using, a = ω² r = 4π² n² r = 4(3.14)² x 1² x 20 x 10³

∴ a = 8 x 10⁵ m/s²

Discussion

169.	At what angular speed should the earth rotate about its axis so that apparent weight of a body on the equator will be zero? What would be the length of the day at that periodic time?[Radius of earth = 6400 km, g = 9.8 m/s2 ]
Answer» Angular speed = 1.237 x 10^–3 rad/s, and length of the day is 5080 s

Discussion

170.	An object follows a curved path. The following quantities may remain constant during the motion(a) speed (b) velocity (c) acceleration (d) magnitude of acceleration.
Answer» (a) speed (d) magnitude of acceleration. Explanation: On a curved path, instantaneous direction of movement changes continuously. So, a vector can not remain constant on a curved path because it has both magnitude and direction. In the above options, velocity and acceleration are vectors, therefore they will change. Speed and magnitude of acceleration do not have directions, so they may remain constant on a curved path, Hence options (a) and (d) are correct.

170.

An object follows a curved path. The following quantities may remain constant during the motion(a) speed (b) velocity (c) acceleration (d) magnitude of acceleration.

Answer»

(a) speed

(d) magnitude of acceleration.

Explanation:

On a curved path, instantaneous direction of movement changes continuously. So, a vector can not remain constant on a curved path because it has both magnitude and direction. In the above options, velocity and acceleration are vectors, therefore they will change. Speed and magnitude of acceleration do not have directions, so they may remain constant on a curved path, Hence options (a) and (d) are correct.

Discussion

171.	An object follows a curved path. The following quantities may remain constant during the motion (a) speed (b) velocity (c) acceleration (d) magnitude of acceleration.
Answer» (a) speed (d) magnitude of acceleration. Explanation: On a curved path, instantaneous direction of movement changes continuously. So, a vector can not remain constant on a curved path because it has both magnitude and direction. In the above options, velocity and acceleration are vectors, therefore they will change. Speed and magnitude of acceleration do not have directions, so they may remain constant on a curved path, Hence options (a) and (d) are correct.

171.

An object follows a curved path. The following quantities may remain constant during the motion (a) speed (b) velocity (c) acceleration (d) magnitude of acceleration.

Answer»

(a) speed

(d) magnitude of acceleration.

Explanation:

On a curved path, instantaneous direction of movement changes continuously. So, a vector can not remain constant on a curved path because it has both magnitude and direction. In the above options, velocity and acceleration are vectors, therefore they will change. Speed and magnitude of acceleration do not have directions, so they may remain constant on a curved path, Hence options (a) and (d) are correct.

Discussion

172.	A square plate lies in the xy plane with its centre at the origin and its edges parallel to the x and y axes. Its moments of inertia about the x, y and z axes are Ix, Iy and Iz respectively, and about a diagonal it is ID.(a) Ix = Iy = 1/2 Iz(b) Ix = Iy = 2Iz(c) ID = Ix(d) ID = Iz
Answer» Correct Answer is: (a) I_x = I_y= 1/2 I_z, (c) I_D = I_x

Discussion

173.	In the figure, the blocks have unequal masses m1 and m2 (m1 > m2). m1 has a downward acceleration a. The pulley P has a radius r, and some mass. The string does not slip on the pulley.(a) The two sections of the string have unequal tensions. (b) The two blocks have accelerations of equal magnitude. (c) The angular acceleration of P is a/r.(d) a < (m1 - m2 / m1 + m2) g
Answer» Correct Answer is: (a, b, c, & d)

Discussion

174.	A particle P of mass m is attached to a vertical axis by two strings AP and BP of length l each. The separation AB = l. P rotates around the axis with an angular velocity ω. The tensions in the two strings are T1 and T2.(a) T1 = T2(b) T1 + T2 = mω2 l(c) T1 - T2 = 2mg(d) BP will remain taut only if ω ≥ √(2g/l).
Answer» Correct Answer is: (b, c, & d) △ABP is equilateral. Let T₁ and T₂ be the tensions in PA and PB respectively. T₁cos 60° = T₂cos 60° + mg or T₁ - T₂ = 2mg. T₁cos 30° + T₂cos 30° = mω² lcos 30° or T₁ + T₂ = mω² l. ∴ T₂ = 1/2 m [ω²l - 2g]. For T₂ ≥ 0, ω ≥ √(2g/l).

174.

A particle P of mass m is attached to a vertical axis by two strings AP and BP of length l each. The separation AB = l. P rotates around the axis with an angular velocity ω. The tensions in the two strings are T1 and T2.(a) T1 = T2(b) T1 + T2 = mω2 l(c) T1 - T2 = 2mg(d) BP will remain taut only if ω ≥ √(2g/l).

Answer»

Correct Answer is: (b, c, & d)

△ABP is equilateral. Let T₁ and T₂ be the tensions in PA and PB respectively.

T₁cos 60° = T₂cos 60° + mg or T₁ - T₂ = 2mg.

T₁cos 30° + T₂cos 30° = mω² lcos 30° or T₁ + T₂ = mω² l.

∴ T₂ = 1/2 m [ω²l - 2g].

For T₂ ≥ 0, ω ≥ √(2g/l).

Discussion

175.	Two particles A and B are located at distances rA and rB respectively from the centre of a rotating disc such that rA > rB. In this case, if angular velocity ω of rotation is constant then (A) both A and B do not have any acceleration. (B) both A and B have same acceleration. (C) A has greater acceleration than B. (D) B has greater acceleration than A.
Answer» Answer is (C) A has greater acceleration than B. As ω is constant, acceleration is due to the change in direction of velocity = ω²r As r_A> r_B ⇒ a_A > a_B

175.

Two particles A and B are located at distances rA and rB respectively from the centre of a rotating disc such that rA > rB. In this case, if angular velocity ω of rotation is constant then (A) both A and B do not have any acceleration. (B) both A and B have same acceleration. (C) A has greater acceleration than B. (D) B has greater acceleration than A.

Answer»

Answer is (C) A has greater acceleration than B.

As ω is constant, acceleration is due to the change in direction of velocity = ω²r

As r_A> r_B ⇒ a_A > a_B

Discussion

176.	For a particle moving in a circle, (A) the resultant force on the particle must be towards the centre. (B) the cross product of tangential acceleration and angular velocity will be zero. (C) direction of angular acceleration and angular velocity must be same. (D) the resultant force must be away from the centre.
Answer» Answer is (A) the resultant force on the particle must be towards the centre.

Discussion

177.	A chain of mass M and radius R placed on a smooth table is revolving with v about a vertical axis coinciding with the symmetry axis of the chain. Find tension in the chain.(A) \(\frac{Mv^2}{2R}\)(B) \(\frac{Mv^2}{R}\)(C) \(\frac{Mv^2}{2\pi R}\) (D) \(\frac{3Mv^2}{2R}\)
Answer» Answer is (C) \(\frac{Mv^2}{2\pi R}\)

Discussion

178.	A helicopter was designed at MIT in which there was a different system to keep the helicopter stable at a height. The radius of the masses and the angular velocity of rotation can be varied (see figure). The mass of each bob is 5 kg and the number of bobs is 8.During the test run, empty helicopter of mass 400 kg is to be kept stable. Due to small malfunction, the radius had to be fixed at 1 m and could not be changed. Find the value of ω required to successfully complete the test run. [Take g = 10 m/s2 ] (A) 5 rad/s (B) 2.5 rad/s (C) 10 rad/s (D) 20 rad/s
Answer» Answer is (C) 10 rad/s

Discussion

179.	The difference in tensions in the string at lowest and highest points in the path of the particle of mass ‘m’ performing vertical circular motion is (A) 2 mg(B) 4 mg(C) 6 mg(D) 8 mg
Answer» Correct option is (C) 6 mg

Discussion

180.	When a car is going round a circular track, the resultant of all the forces on the car in an inertial frame is (A) acting away from the centre. (B) acting towards the centre. (C) zero. (D) acting tangential to the track.
Answer» Answer is (B) acting towards the centre.

Discussion

181.	ABCDE is a smooth iron track in the vertical plane. The sections ABC and CDE are quarter circles. Points B and D are very close to C. M is a small magnet of mass m. The force of attraction between M and the track is F, which is constant and always normal to the track. M starts from rest at A.(a) If M is not leave the track at C then F ≥ 2mg. (b) At B, the normal reaction of the track is F - 2mg. (c) At D, the normal reaction of the track is F + 2mg. (d) The normal reaction of the track is equal to F at some point between A and B.
Answer» Correct Answer sis: (a, b, c, & d) At C, 1/2 mv² = mgr or mv²/r = 2mg. If N = the normal reaction of the track, F - N = mv²/r = 2mg or F = N + 2mg. As N ≥ 0, F ≥ 2mg Taking the speeds at B and D to be the same as at C, at B, F - N = 2mg or N = F - 2mg (Centre of curvature is O.) at D, N - F = 2mg or N = F + 2mg (Centre of curvature is O'.) For ∠AOM = θ, mg(r - rcos θ) = 1/2 mv² or mv²/r = 2mg (1 - cos θ) or 2mg (1 - cos θ) = mgcos θ + F - N. For F = N, cos θ = 2/3.

181.

ABCDE is a smooth iron track in the vertical plane. The sections ABC and CDE are quarter circles. Points B and D are very close to C. M is a small magnet of mass m. The force of attraction between M and the track is F, which is constant and always normal to the track. M starts from rest at A.(a) If M is not leave the track at C then F ≥ 2mg. (b) At B, the normal reaction of the track is F - 2mg. (c) At D, the normal reaction of the track is F + 2mg. (d) The normal reaction of the track is equal to F at some point between A and B.

Answer»

Correct Answer sis: (a, b, c, & d)

At C, 1/2 mv² = mgr or mv²/r = 2mg.

If N = the normal reaction of the track,

F - N = mv²/r = 2mg or F = N + 2mg.

As N ≥ 0, F ≥ 2mg

Taking the speeds at B and D to be the same as at C,

at B, F - N = 2mg or N = F - 2mg (Centre of curvature is O.)

at D, N - F = 2mg or N = F + 2mg (Centre of curvature is O'.)

For ∠AOM = θ,

mg(r - rcos θ) = 1/2 mv² or mv²/r = 2mg (1 - cos θ)

or 2mg (1 - cos θ) = mgcos θ + F - N.

For F = N, cos θ = 2/3.

Discussion

182.	The force required to keep a body in uniform circular motion is (A) centripetal force. (B) centrifugal force. (C) frictional force. (D) breaking force.
Answer» Answer is (A) centripetal force.

Discussion

Explore topic-wise InterviewSolutions in .

Consider an object of mass m that moves in a circular orbit with constant velocity v0 along the inside of a cone. Assume the wall of the cone to be frictionless. Find radius of the orbit.(A) \(\frac{v_0^2}{g}\) tan2ϕ(B) \(\frac{v_0^2}{g}\) cos2ϕ(C) \(\frac{v_0^2}{g}\) tanϕ(D) \(\frac{v_0^2}{g}\)

Give the formula of Centripetal acceleration or radial acceleration.

When a particle moves in a uniform circular motion. It has(A) radial velocity and radial acceleration.(B) tangential velocity and radial acceleration.(C) tangential velocity and tangential acceleration.(D) radial velocity and tangential acceleration.

A rectangular block has a square base measuring a x a, and its height is h. It moves with a speed v on a smooth horizontal surface.(a) It will topple if v &gt; √(2gh).(b) It will topple if v &gt; √(2ga).(c) It will topple if v &gt; √(2ga2 / h).(d) It will not topple for any value of v.

The minute hand of a clock is 5 cm long. Calculate the linear speed of an ant sitting at the tip

The centripetal acceleration is given by(A) v2/r (B) vr (C) vr2 (D) v/r

Angle between radius vector and centripetal acceleration is(A) 0c (B) πc (C) 2πc (D) none of these

A body moves along a circular path with certain velocity. What will be the path of body in following figure?(A) Move radially out. (B) Move horizontally out. (C) Fall vertically down. (D) Move tangentially out.

The angular speed and linear speed of tip of a second hand of clock acting on it, if its angular speed of rotation is 0.6 rad/s.

Calculate the angular speed and linear speed of tip of a second hand of clock if second hand is 5 cm long.

For a particle moving in a vertical circle, (A) kinetic energy is constant. (B) potential energy is constant. (C) neither K.E. nor P.E. is constant. (D) both kinetic energy and potential energy are constant.

When a particle is moved in a vertical circle, (A) it has constant radial and tangential acceleration. (B) it has variable tangential and radial acceleration.(C) it has only constant radial acceleration.(D) it has only constant tangential acceleration.

Certain neutron stars are believed to be rotating at about 1 rev/s. If such a star has a radius of 20 km, the acceleration of an object on the equator of the star will be(A) 20 x 108 m/s2 (B) 8 x 105 m/s2 (C) 120 x 105 m/s2 (D) 4 x 108 m/s2

At what angular speed should the earth rotate about its axis so that apparent weight of a body on the equator will be zero? What would be the length of the day at that periodic time?[Radius of earth = 6400 km, g = 9.8 m/s2 ]

An object follows a curved path. The following quantities may remain constant during the motion(a) speed (b) velocity (c) acceleration (d) magnitude of acceleration.

An object follows a curved path. The following quantities may remain constant during the motion (a) speed (b) velocity (c) acceleration (d) magnitude of acceleration.

A square plate lies in the xy plane with its centre at the origin and its edges parallel to the x and y axes. Its moments of inertia about the x, y and z axes are Ix, Iy and Iz respectively, and about a diagonal it is ID.(a) Ix = Iy = 1/2 Iz(b) Ix = Iy = 2Iz(c) ID = Ix(d) ID = Iz

A chain of mass M and radius R placed on a smooth table is revolving with v about a vertical axis coinciding with the symmetry axis of the chain. Find tension in the chain.(A) \(\frac{Mv^2}{2R}\)(B) \(\frac{Mv^2}{R}\)(C) \(\frac{Mv^2}{2\pi R}\) (D) \(\frac{3Mv^2}{2R}\)

The difference in tensions in the string at lowest and highest points in the path of the particle of mass ‘m’ performing vertical circular motion is (A) 2 mg(B) 4 mg(C) 6 mg(D) 8 mg

When a car is going round a circular track, the resultant of all the forces on the car in an inertial frame is (A) acting away from the centre. (B) acting towards the centre. (C) zero. (D) acting tangential to the track.

The force required to keep a body in uniform circular motion is (A) centripetal force. (B) centrifugal force. (C) frictional force. (D) breaking force.

A rectangular block has a square base measuring a x a, and its height is h. It moves with a speed v on a smooth horizontal surface.(a) It will topple if v > √(2gh).(b) It will topple if v > √(2ga).(c) It will topple if v > √(2ga2 / h).(d) It will not topple for any value of v.