Circular Motion - NEET Physics Questions
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Circular Motion

Question 21: easy

A cyclist paddling at a speed of 10 m/s on a level road takes a sharp circular turn of radius 10 m without reducing the speed. The angle made by cyclist with vertical is

1. π/4
2. π/3
3. π/6
4. π/2
View Answer

Using the formula

tan⁔θ=v2rg\tan \theta = \frac{v^2}{r g}

and given values

v=10v = 10

m/s,

r=10r = 10

m, and

g=10g = 10

m/s², we get:

 

tan⁔θ=10210Ɨ10=1\tan \theta = \frac{10^2}{10 \times 10} = 1

 

Thus,

Īø=tanā”āˆ’1(1)=45∘\theta = \tan^{-1}(1) = 45^\circ

. The cyclist leans at 45° or Ļ€/4 with the vertical.

Question 22:

A particle is executing uniform circular motion with velocity \(\vec{v}\) and acceleration \(\vec{a}\). Which of the following is true?

1. \(\vec{v}\) is a constant; \(\vec{a}\) is a constant
2. \(\vec{v}\) is not a constant; \(\vec{a}\) is a constant
3. \(\vec{v}\) is a constant; \(\vec{a}\) is not a constant
4. \(\vec{v}\) is not a constant; \(\vec{a}\) is not a constant
View Answer

In uniform circular motion, the magnitudes of velocity and centripetal acceleration are constant, but their directions continuously change as the particle moves along the circle. Thus, both vector quantities are non-constant.

Question 23: easy

A particle moves in a circular path so that its distance travel varies with time \(t\) as \(s = 3t^2 + 6t\). Then its acceleration at \(t = 1\text{ sec.}\) is (radius of path is \(12\text{ m}\)) –

1. \(6\sqrt{5}\text{ m/s}^2\)
2. \(6\text{ m/s}^2\)
3. \(12\text{ m/s}^2\)
4. \(12\sqrt{3}\text{ m/s}^2\)
View Answer

Speed is \(v = \frac{ds}{dt} = 6t + 6\). At \(t = 1\text{ s}\), \(v = 12\text{ m/s}\). Tangential acceleration is \(a_t = \frac{dv}{dt} = 6\text{ m/s}^2\). Centripetal acceleration is \(a_c = \frac{v^2}{R} = \frac{12^2}{12} = 12\text{ m/s}^2\). Total acceleration is \(a = \sqrt{a_t^2 + a_c^2} = \sqrt{6^2 + 12^2} = 6\sqrt{5}\text{ m/s}^2\).

Question 24: moderate

The kinetic energy \((K)\) of particle moving along a circle of radius \(R\) depends upon the distance covered \(S\) and is given by \(K = aS\) where \(a\) is a constant. Then the centripetal force acting on the particle is:

1. \(\frac{aS}{R}\)
2. \(\frac{2(aS)^2}{R}\)
3. \(\frac{aS^2}{R}\)
4. \(\frac{2aS}{R}\)
View Answer

Kinetic energy is \(K = \frac{1}{2}mv^2 = aS\). Centripetal force is \(F_c = \frac{mv^2}{R}\) ( Since \(mv^2 = 2aS\), we get \(F_c = \frac{2aS}{R}\).

Question 25: moderate

A particle of mass \(m\) is tied to a string of length \(L\) and rotated in vertical circle about other end with critical speed so that it is just able to complete the vertical loop. Then tension in string, when string is at horizontal position will be:

1. \(2mg\)
2. \(3mg\)
3. \(4mg\)
4. \(5mg\)
View Answer

To just complete the vertical loop, the velocity at the bottom is \(\sqrt{5gL}\). By energy conservation, the velocity at the horizontal position is \(v = \sqrt{3gL}\). The tension at this point is \(T = \frac{mv^2}{L} = 3mg\).

Question 26: easy

A particle start revolving on a circular path with constant angular acceleration \(\frac{\pi}{2}\text{ rad/sec}^2\). Then find number of cycles it will complete in first 12 seconds:

1. \(12\text{ cycles}\)
2. \(18\text{ cycles}\)
3. \(36\text{ cycles}\)
4. \(72\text{ cycles}\)
View Answer

Angular displacement is \(\theta = \frac{1}{2}\alpha t^2 = \frac{1}{2} \left(\frac{\pi}{2}\right) (12)^2 = 36\pi\text{ rad}\). Number of cycles \(N = \frac{\theta}{2\pi} = \frac{36\pi}{2\pi} = 18\).

Question 27: easy

Assertion: In uniform circular motion tangential acceleration of particle is zero.


Reason: In uniform circular motion net force on particle is always directed towards centre of circular path.


 

1. Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
2. Both Assertion and Reason are true but Reason is not correct explanation of Assertion.
3. Assertion is true but Reason is false.
4. Assertion and Reason are false.
View Answer

Tangential acceleration is zero because speed is constant. The net force is centripetal, which is directed towards the center. Thus, both are true but the Reason is not the explanation of the Assertion.

Question 28: easy

A road of width \(20\text{ m}\) forms an arc of radius \(15\text{ m}\), its outer edge is \(2\text{ m}\) higher than its inner edge. For what speed the road is banked?

1. \(\sqrt{10}\text{ m/s}\)
2. \(\sqrt{14.7}\text{ m/s}\)
3. \(\sqrt{9.8}\text{ m/s}\)
4. None of these
View Answer

The angle of banking is given by \(sin\theta \approx \tan\theta = \frac{h}{w} = \frac{2}{20} = 0.1\). Also, \(\tan\theta = \frac{v^2}{Rg}\). Equating these, \(frac{v^2}{15 \times 9.8} = 0.1 ⇒ v^2 = 14.7 ⇒ v = \sqrt{14.7}\text{ m/s}\).

Question 29: easy

A particle moves in a circle of radius \(10\text{ cm}\) with constant speed and time period \(\pi\text{ s}\). The acceleration of the particle is :

1. \(10\text{ cm/s}^2\)
2. \(20\text{ cm/s}^2\)
3. \(40\text{ cm/s}^2\)
4. \(50\text{ cm/s}^2\)
View Answer

The angular frequency is \(\omega = \frac{2\pi}{T} = \frac{2\pi}{\pi} = 2\text{ rad/s}\). The centripetal acceleration is \(a = \omega^2 R = 2^2 \times 10 = 40\text{ cm/s}^2\).

Question 30: easy

Assertion (A): A particle moving at constant speed and constant magnitude of radial acceleration must be undergoing uniform circular motion.


Reason (R): In uniform circular motion speed cannot change as there is no tangential acceleration.


 

1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer

Assertion (A): Constant speed and constant magnitude of radial acceleration ((v^2/r)) imply constant radius ((r)), which defines uniform circular motion. So (A) is True.
Reason (R): In uniform circular motion, acceleration is purely centripetal (radial), with no component tangential to the path. Thus, speed remains constant. So (R) is True.
Reason (R) correctly explains why constant speed and constant radial acceleration magnitude lead to uniform circular motion by implying constant radius and absence of tangential acceleration.