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: A motorcyclist rides inside a vertical "globe of death" of diameter 10\text{ m}. The minimum speed he must have at the topmost point is (g = 10\text{ m/s}^2):

7.07\text{ m/s}. Radius r = 5\text{ m}. At the top, gravity alone supplies the centripetal force in the limiting case: mg = \dfrac{mv^2}{r} \Rightarrow v_{\text{top}} = \sqrt{gr} = \sqrt{50} \approx 7.07\text{ m/s}.

Q: A body of mass 2\text{ kg} tied to a string of length 1\text{ m} is whirled in a vertical circle. The minimum speed at the lowest point needed to complete the circle is (g = 10\text{ m/s}^2):

7.07\text{ m/s}. For a just-completed vertical circle, v_{\text{lowest}} = \sqrt{5gr} = \sqrt{5 \times 10 \times 1} = \sqrt{50} \approx 7.07\text{ m/s}. Note the answer does not depend on the mass.

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 travels on a road banked at angle \theta with coefficient of friction \mu. The maximum safe speed without skidding outward is:

\sqrt{rg\left(\dfrac{\tan\theta + \mu}{1 - \mu\tan\theta}\right)}. At maximum speed the vehicle tends to slide up the bank, so friction acts down the incline. Resolving along and perpendicular to the road gives v_{\max} = \sqrt{rg\left(\dfrac{\tan\theta + \mu}{1 - \mu\tan\theta}\right)}.

Q: On a rough banked curve of radius r, banking angle \theta and coefficient of friction \mu, the minimum speed needed to prevent the vehicle from sliding down is:

\sqrt{rg\left(\dfrac{\tan\theta - \mu}{1 + \mu\tan\theta}\right)}. Below the design speed the vehicle tends to slip down, so friction acts up the bank. Solving N\sin\theta - f\cos\theta = \dfrac{mv^2}{r} with N\cos\theta + f\sin\theta = mg gives v_{\min} = \sqrt{rg\left(\dfrac{\tan\theta - \mu}{1 + \mu\tan\theta}\right)}.

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 motorcyclist rides inside a vertical "globe of death" of diameter 10	ext{ m}. The minimum speed he must have at the topmost point is (g = 10	ext{ m/s}^2):

Accepted Answer

7.07	ext{ m/s}. Radius r = 5	ext{ m}. At the top, gravity alone supplies the centripetal force in the limiting case: mg = \dfrac{mv^2}{r} \Rightarrow v_{	ext{top}} = \sqrt{gr} = \sqrt{50} \approx 7.07	ext{ m/s}.

Question 5

A body of mass 2	ext{ kg} tied to a string of length 1	ext{ m} is whirled in a vertical circle. The minimum speed at the lowest point needed to complete the circle is (g = 10	ext{ m/s}^2):

Accepted Answer

7.07	ext{ m/s}. For a just-completed vertical circle, v_{	ext{lowest}} = \sqrt{5gr} = \sqrt{5 	imes 10 	imes 1} = \sqrt{50} \approx 7.07	ext{ m/s}. Note the answer does not depend on the mass.

Question 6

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 7

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 8

A car travels on a road banked at angle 	heta with coefficient of friction \mu. The maximum safe speed without skidding outward is:

Accepted Answer

\sqrt{rg\left(\dfrac{	an	heta + \mu}{1 - \mu	an	heta}ight)}. At maximum speed the vehicle tends to slide up the bank, so friction acts down the incline. Resolving along and perpendicular to the road gives v_{\max} = \sqrt{rg\left(\dfrac{	an	heta + \mu}{1 - \mu	an	heta}ight)}.

Question 9

On a rough banked curve of radius r, banking angle 	heta and coefficient of friction \mu, the minimum speed needed to prevent the vehicle from sliding down is:

Accepted Answer

\sqrt{rg\left(\dfrac{	an	heta - \mu}{1 + \mu	an	heta}ight)}. Below the design speed the vehicle tends to slip down, so friction acts up the bank. Solving N\sin	heta - f\cos	heta = \dfrac{mv^2}{r} with N\cos	heta + f\sin	heta = mg gives v_{\min} = \sqrt{rg\left(\dfrac{	an	heta - \mu}{1 + \mu	an	heta}ight)}.

Question 10

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 11

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 12

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 13

A body of mass m just completes a vertical circle of radius r. The tension in the string at the lowest point is:

Accepted Answer

6mg. For a just-completed circle v_{	ext{low}}^2 = 5gr. At the lowest point T - mg = \dfrac{mv^2}{r} \Rightarrow T = mg + 5mg = 6mg.

Question 14

For a body moving in a vertical circle of radius r, the difference between the tension at the lowest point and that at the highest point is:

Accepted Answer

6mg. At the lowest point T_L = \dfrac{mv_L^2}{r} + mg; at the top T_H = \dfrac{mv_H^2}{r} - mg. Energy conservation gives v_L^2 - v_H^2 = 4gr, so T_L - T_H = \dfrac{m(v_L^2 - v_H^2)}{r} + 2mg = 4mg + 2mg = 6mg, independent of speed.

Question 15

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 16

The period of revolution of a conical pendulum of string length L making an angle 	heta with the vertical is:

Accepted Answer

2\pi\sqrt{\dfrac{L\cos	heta}{g}}. With r = L\sin	heta and T\cos	heta = mg, T\sin	heta = m\omega^2 r gives \omega^2 = \dfrac{g}{L\cos	heta}. Hence the period is \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{L\cos	heta}{g}}.

Question 17

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 18

A car negotiates a level curve of radius 100	ext{ m} with coefficient of friction 0.4. The maximum speed without skidding is (g = 10	ext{ m/s}^2):

Accepted Answer

20	ext{ m/s}. v_{\max} = \sqrt{\mu r g} = \sqrt{0.4 	imes 100 	imes 10} = \sqrt{400} = 20	ext{ m/s}.

Question 19

A road of radius 80	ext{ m} is banked at an angle 	heta such that 	an	heta = 0.5. The design speed for which the banking is ideal is (g = 10	ext{ m/s}^2):

Accepted Answer

20	ext{ m/s}. For ideal banking 	an	heta = \dfrac{v^2}{rg}, so v = \sqrt{rg	an	heta} = \sqrt{80 	imes 10 	imes 0.5} = \sqrt{400} = 20	ext{ m/s}.

Question 20

A conical pendulum of string length 1	ext{ m} makes an angle of 60^\circ with the vertical. Its period of revolution is (g = 10	ext{ m/s}^2):

Accepted Answer

1.40	ext{ s}. T = 2\pi\sqrt{\dfrac{L\cos	heta}{g}} = 2\pi\sqrt{\dfrac{1 	imes 0.5}{10}} = 2\pi\sqrt{0.05} = 2\pi(0.2236) \approx 1.40	ext{ s}.

Circular Motion & Banking of Roads — MH-CET Physics MCQs with Solutions

Questions with solutions