Q: A curved highway of radius 100\text{ m} is banked for a design speed of 72\text{ km/h}. Taking g = 10\text{ m/s}^2, the optimum angle of banking is:

\tan^{-1}(0.4). v = 72\text{ km/h} = 20\text{ m/s}. For an ideally banked road, \tan\theta = \dfrac{v^2}{rg} = \dfrac{400}{100 \times 10} = 0.4, so \theta = \tan^{-1}(0.4).

Q: A vehicle negotiates an unbanked circular curve of radius 50\text{ m} on a level road. If the coefficient of friction is \mu = 0.2 and g = 9.8\text{ m/s}^2, the maximum speed without skidding is:

9.9\text{ m/s}. On a level road friction alone supplies the centripetal force: \mu mg = \dfrac{mv^2}{r}, so v_{\max} = \sqrt{\mu r g} = \sqrt{0.2 \times 50 \times 9.8} = \sqrt{98} \approx 9.9\text{ m/s}.

Q: In a conical pendulum, a bob of mass m revolves in a horizontal circle while the string makes an angle \theta with the vertical. The tension in the string is:

\dfrac{mg}{\cos\theta}. The bob has no vertical acceleration, so the vertical component of tension balances weight: T\cos\theta = mg \Rightarrow T = \dfrac{mg}{\cos\theta}. The horizontal component T\sin\theta supplies the centripetal force.

Q: The optimum angle of banking of a curved road is independent of:

The mass of the vehicle. From \tan\theta = \dfrac{v^2}{rg}, the banking angle depends on speed, radius and g only. The mass cancels out, which is why one banking angle serves cars and trucks alike.

Q: A car takes a turn of radius 20\text{ m} on a level road with coefficient of friction 0.5. The maximum speed with which it can turn is (g = 10\text{ m/s}^2):

10\text{ m/s}. v_{\max} = \sqrt{\mu r g} = \sqrt{0.5 \times 20 \times 10} = \sqrt{100} = 10\text{ m/s}.

Question 1

A curved highway of radius 100	ext{ m} is banked for a design speed of 72	ext{ km/h}. Taking g = 10	ext{ m/s}^2, the optimum angle of banking is:

Accepted Answer

an^{-1}(0.4). v = 72	ext{ km/h} = 20	ext{ m/s}. For an ideally banked road, 	an	heta = \dfrac{v^2}{rg} = \dfrac{400}{100 	imes 10} = 0.4, so 	heta = 	an^{-1}(0.4).

Question 2

A vehicle negotiates an unbanked circular curve of radius 50	ext{ m} on a level road. If the coefficient of friction is \mu = 0.2 and g = 9.8	ext{ m/s}^2, the maximum speed without skidding is:

Accepted Answer

9.9	ext{ m/s}. On a level road friction alone supplies the centripetal force: \mu mg = \dfrac{mv^2}{r}, so v_{\max} = \sqrt{\mu r g} = \sqrt{0.2 	imes 50 	imes 9.8} = \sqrt{98} \approx 9.9	ext{ m/s}.

Question 3

In a conical pendulum, a bob of mass m revolves in a horizontal circle while the string makes an angle 	heta with the vertical. The tension in the string is:

Accepted Answer

\dfrac{mg}{\cos	heta}. The bob has no vertical acceleration, so the vertical component of tension balances weight: T\cos	heta = mg \Rightarrow T = \dfrac{mg}{\cos	heta}. The horizontal component T\sin	heta supplies the centripetal force.

Question 4

A particle of mass m moves in a circle of radius r with uniform speed v. The centripetal force acting on it is:

Accepted Answer

\dfrac{mv^2}{r}. Centripetal acceleration is a_c = \dfrac{v^2}{r}, so the required force directed towards the centre is F = ma_c = \dfrac{mv^2}{r}.

Question 5

The optimum angle of banking of a curved road is independent of:

Accepted Answer

The mass of the vehicle. From 	an	heta = \dfrac{v^2}{rg}, the banking angle depends on speed, radius and g only. The mass cancels out, which is why one banking angle serves cars and trucks alike.

Question 6

A car takes a turn of radius 20	ext{ m} on a level road with coefficient of friction 0.5. The maximum speed with which it can turn is (g = 10	ext{ m/s}^2):

Accepted Answer

10	ext{ m/s}. v_{\max} = \sqrt{\mu r g} = \sqrt{0.5 	imes 20 	imes 10} = \sqrt{100} = 10	ext{ m/s}.

Question 7

The work done by the centripetal force on a particle in uniform circular motion during one complete revolution is:

Accepted Answer

Zero. The centripetal force is always directed towards the centre, perpendicular to the instantaneous displacement. Since W = \vec{F}\cdot\vec{s} and the two are always at 90^\circ, the work done is zero.

Question 8

In a conical pendulum the centripetal force required for the circular motion is provided by:

Accepted Answer

The horizontal component of the tension. Tension resolves into T\cos	heta (vertical, balancing mg) and T\sin	heta (horizontal, pointing at the axis). It is the horizontal component T\sin	heta that supplies \dfrac{mv^2}{r}.

Question 9

A stone tied to a string is whirled in a horizontal circle. If the string suddenly breaks, the stone flies off:

Accepted Answer

Along the tangent to the circle at that point. Once the tension vanishes there is no centripetal force, so by Newton's first law the stone continues with the velocity it had at that instant — which is tangential to the circle.

Question 10

Centrifugal force is described as a pseudo force because it:

Accepted Answer

Appears only in a rotating (non-inertial) frame of reference. Centrifugal force has no physical agent exerting it. It is introduced only so that Newton's laws can be applied inside a rotating, non-inertial frame; in the ground (inertial) frame it does not exist.

Question 11

A wheel starts from rest and rotates with a constant angular acceleration of 2	ext{ rad/s}^2. The total angle turned in the first 5	ext{ seconds} is:

Accepted Answer

25	ext{ rad}. 	heta = \omega_0 t + 	frac{1}{2}\alpha t^2 = 0 + 	frac{1}{2}(2)(5)^2 = 25	ext{ rad}.

Question 12

The angular velocity of a rotating body changes from 4	ext{ rad/s} to 12	ext{ rad/s} in 2	ext{ seconds}. Its angular acceleration is:

Accepted Answer

4	ext{ rad/s}^2. \alpha = \dfrac{\Delta\omega}{\Delta t} = \dfrac{12 - 4}{2} = 4	ext{ rad/s}^2.

Question 13

A particle moves along a circle of radius 0.5	ext{ m} with angular velocity 10	ext{ rad/s}. Its linear (tangential) speed is:

Accepted Answer

5	ext{ m/s}. v = \omega r = 10 	imes 0.5 = 5	ext{ m/s}.

Question 14

A body moves in a circle of radius 4	ext{ m} with a constant speed of 8	ext{ m/s}. Its centripetal acceleration is:

Accepted Answer

16	ext{ m/s}^2. a_c = \dfrac{v^2}{r} = \dfrac{8^2}{4} = \dfrac{64}{4} = 16	ext{ m/s}^2.

Question 15

A ceiling fan rotates at 300	ext{ rpm}. Its angular velocity in rad/s is:

Accepted Answer

10\pi. \omega = \dfrac{2\pi N}{60} = \dfrac{2\pi 	imes 300}{60} = 10\pi \approx 31.4	ext{ rad/s}.

Question 16

The tangential acceleration a_t of a particle moving on a circle of radius r with angular acceleration \alpha is:

Accepted Answer

\alpha r. Differentiating v = \omega r with r constant gives a_t = \dfrac{dv}{dt} = r\dfrac{d\omega}{dt} = \alpha r. (\omega^2 r is the centripetal, not tangential, acceleration.)

Question 17

A particle on a circle of radius 2	ext{ m} speeds up uniformly from rest to 10	ext{ m/s} in 5	ext{ s}. Its tangential acceleration is:

Accepted Answer

2	ext{ m/s}^2. Tangential acceleration is the rate of change of speed: a_t = \dfrac{10 - 0}{5} = 2	ext{ m/s}^2.

Question 18

A particle moves on a circle of radius 2	ext{ m}. At the instant its speed is 10	ext{ m/s}, its centripetal acceleration is:

Accepted Answer

50	ext{ m/s}^2. a_c = \dfrac{v^2}{r} = \dfrac{100}{2} = 50	ext{ m/s}^2, directed towards the centre.

Question 19

The angular velocity \omega of a body making n revolutions per second is:

Accepted Answer

2\pi n. One revolution corresponds to 2\pi radian, so n revolutions per second means \omega = 2\pi n	ext{ rad/s}.

Question 20

For a particle in uniform circular motion, which of the following remains constant?

Accepted Answer

The magnitude of its velocity. In uniform circular motion the speed is constant, but the direction of motion changes continuously. Hence velocity, acceleration and momentum are all changing vectors, while the magnitude of velocity (the speed) stays fixed.

Rotational Dynamics — MH-CET Physics MCQs with Solutions

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