Kinematics of Circular Motion - NEET Physics Chapterwise MCQs & PYQs

NEET Kinematics of Circular Motion MCQs & PYQs

Kinematics of circular motion is important for preparation of NEET, JEE Mains, JEE Advance , CUET and other exams in which Physics is asked

Question 1:

easy

A stone is moved round a horizontal circle with a 20 cm long string tied to If centripetal acceleration is 9.8 m/s2, then its angular velocity will be :

Centripetal Acceleration is given by a= ω²R

⇒ 9.8 = ω²× 1/5

⇒ ω² =49

⇒ ω = 7 rad/sec

Question 2:

easy

A wheel having diameter of 3 m starts from rest and accelerates uniformly to an angular velocity of 210 rpm in 5 seconds. Angular acceleration of the wheel is :

ω = ω + α . t 

⇒(210 × 2π)/60 = 0 + α × 5 

⇒ α = 1.4 π rad/sec ²

Question 3:

easy

The angular velocity of earth about its axis of rotation is :

Angular speed ω = 2π / T = 2π / (60×60×24) rad /sec 

Question 4:

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 \(12\text{ m}\)) –

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 5:

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:

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 6:

easy

Assertion (A): In non-uniform circular motion, velocity vector and acceleration vector are not perpendicular to each other.


Reason (R): In non-uniform circular motion, particle has normal as well as tangential acceleration.


 

In non-uniform circular motion, there is both tangential and centripetal acceleration. The tangential acceleration is parallel to velocity, so the resultant acceleration is not perpendicular to velocity. Reason (R) correctly identifies the components of acceleration, explaining why (A) is true.

Question 7:

easy

Assertion (A): If a body is in state of uniform circular motion then its velocity and acceleration both are varying.


Reason (R): If magnitude of velocity is \(v\) and radius of uniform circular motion is \(r\) then magnitude of acceleration is \(v^2/r\).


 

In uniform circular motion, speed is constant, but velocity (direction) and acceleration (direction) vary, making (A) true. Reason (R) gives the correct magnitude of centripetal acceleration \(a = v^2/r\), so (R) is true. However, (R) describes the magnitude, not why the vectors are varying, so it's not the correct explanation.

Question 8:

easy

Assertion (A): The equation of motion can be applied only if the acceleration is along the direction of velocity and is constant.


Reason (R): In circular motion, if velocity is constant then its motion is called uniform circular motion.


 

Assertion (A) is false; kinematic equations apply for constant acceleration (vector), not necessarily along velocity. Reason (R) is false; if velocity (vector) is constant, it's rectilinear motion, not circular motion. In uniform circular motion, *speed* is constant, but velocity changes direction.

Question 9:

easy

Assertion (A): In uniform circular motion, angular acceleration is zero.


Reason (R): In uniform circular motion, acceleration is constant.


 

Assertion (A) is true because angular speed \(omega\) is constant, thus \(alpha = domega/dt = 0\). Reason (R) is false; in uniform circular motion, the *magnitude* of acceleration is constant, but its *direction* continuously changes, so the acceleration vector is not constant.

Question 10:

easy

Assertion (A): Infinitesimally small angular displacement is a vector quantity.


Reason (R): Angular velocity doesn’t depend upon reference frame.


 

Infinitesimally small angular displacement \( d\vec{\theta} \) is a vector because it obeys the commutative law of vector addition. Thus (A) is true.


Angular velocity \( \vec{\omega} \) is a vector quantity, and its value depends on the chosen reference frame. Hence (R) is false.