Circular Motion - NEET Physics Questions
← All Chapters

Circular Motion

Question 11: easy

A particle moves in a circle of radius 5 m with constant speed and time period 2π s. The acceleration of the particle is :

1. 15 m/s²
2. 25 m/s²
3. 36 m/s²
4. 5 m/s²
View Answer

As speed of the particle is constant only centripetal acceleration will act on the object.

Centripetal Acceleration = ω²R

a c = (2π/2π)× 5= 5 m/s² 

Question 12: easy

The maximum speed of car with which it can go around a level road of radius 10 m is (coefficient of friction between the road and tyre is 0.5)
(g = 9.8 m/s²)

1. √19 m/s
2. √29 m/s
3. √39 m/s
4. √49 m/s
View Answer

The maximum speed on a level road is limited by static friction: u=v²/rg

Given μs​=0.5, g=9.8m/s², and r=10m.

Substituting these values: V²max​=(0.5)(9.8)(10)​= 49 ​= 7m/s.

Therefore, the maximum speed is 7 m/s.

Question 13: 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

tanθ=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:

 

tanθ=10210×10=1\tan \theta = \frac{10^2}{10 \times 10} = 1

 

Thus,

θ=tan1(1)=45\theta = \tan^{-1}(1) = 45^\circ

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

Question 14: 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 15: 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 16: 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.