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.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
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.
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 \(v^2/r\).
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
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.
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.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
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.
Assertion (A): In uniform circular motion, angular acceleration is zero.
Reason (R): In uniform circular motion, acceleration is constant.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
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.
Assertion (A): Infinitesimally small angular displacement is a vector quantity.
Reason (R): Angular velocity doesn’t depend upon reference frame.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
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.
Assertion (A): Average angular velocity is a scalar quantity.
Reason (R): Large angular displacements \( (\Delta \theta) \) is a scalar.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
Instantaneous angular velocity \( \vec{\omega} \) is a vector. However, finite angular displacement \( \Delta \theta \) is not a vector, but a scalar, as stated in Reason (R). Therefore, if average angular velocity is defined as the scalar \( \Delta \theta / \Delta t \), then Assertion (A) is considered true. In this context, both (A) and (R) are true, and (R) provides the explanation for (A).