Assertion (A): A cyclist always bends inwards while negotiating a curve
Reason (R): By bending he lowers his centre of gravity
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
A cyclist bends inwards to provide the necessary centripetal force and maintain rotational equilibrium by balancing torques. Thus, (A) is true.
Lowering the center of gravity is not the primary reason for bending, making (R) false.
Assertion (A): A wheel slides downward on frictionless inclined plane, without rolling.
Reason (R): In pure rolling work done against friction always zero.
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
On a frictionless inclined plane, there's no torque to cause rotation, so a wheel will only slide; thus (A) is true.
In pure rolling, the point of contact is instantaneously at rest, so static friction does no work *by* it. The statement 'work done against friction always zero' is not universally true, making (R) false.
Assertion (A): If a sphere starts pure rolling down a rough incline plane, work done by friction is zero.
Reason (R): Work done by friction for translational motion is negative and work done by friction for rotational motion is positive and equal in magnitude.
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 pure rolling, the point of contact is instantaneously at rest, so static friction does no work (A is true).
Friction opposes translational motion (negative work) and causes rotation (positive work); these works are equal in magnitude, leading to zero net work (R is true and correctly explains A).
Assertion (A): A solid copper and solid aluminium sphere of same masses are spinning about their axes with same angular velocities copper sphere has more angular momentum than aluminium.
Reason (R): Both copper and aluminium sphere have same radius.
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
Angular momentum \( L = I\omega = \frac{2}{5} MR^2 \omega \). If mass (M) and angular velocity (\( \omega \)) are the same, L depends on \( R^2 \). Copper is denser than aluminium (\( \rho_{Cu} > \rho_{Al} \)). For the same mass, \( V_{Cu} < V_{Al} \), implying \( R_{Cu} < R_{Al} \). Therefore, \( L_{Cu} < L_{Al} \), making (A) false.
Also, \( R_{Cu} < R_{Al} \), so (R) is false. Both (A) and (R) are false.
Assertion (A): A ladder is more likely to slip when a person is near the top than when he is near the bottom.
Reason (R): The friction between the ladder and floor decreases as he climbs up.
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; a ladder is more likely to slip when a person is higher up due to increased outward horizontal thrust. Reason (R) is false; the normal force at the base remains constant, so the maximum static friction available does not decrease.
A sphere is performing pure rolling on a rough horizontal surface with constant angular velocity.
Assertion (A): Frictional force acting on the sphere is zero.
Reason (R): Velocity of contact point is zero.
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; if a sphere rolls purely with constant angular velocity, no force (including friction) is needed to maintain its motion.
Reason (R) is true; for pure rolling, the contact point is instantaneously at rest.
However, R is a definition of pure rolling, not the explanation for zero friction in this specific case (due to zero acceleration).
Assertion (A): Moment of inertia about an axis passing through center of mass is maximum.
Reason (R): Theorem of parallel axis can be applied only for two dimensional body of negligible thickness.
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 because the moment of inertia about an axis passing through the center of mass is the minimum, not maximum, for parallel axes (\(I = I_{CM} + Md^2\)).
Reason (R) is false because the parallel axis theorem is applicable to any rigid body, not just 2D bodies of negligible thickness.
Assertion (A): If earth shrink (without change in mass) to half its present size, length of the day would become 6 hours.
Reason (R): As size of the earth changes its moment of inertia changes.
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; by conservation of angular momentum \(L = I\omega\), if the radius halves, moment of inertia (\(I \propto R^2\)) becomes one-fourth. Thus, angular velocity (\(omega\)) becomes four times, making day length \(24/4 = 6\text{ hours}\).
Reason (R) is true and explains A; moment of inertia depends on mass distribution and size.
Assertion (A): Speed of any point on a rigid body in pure rolling can be calculated by expression \(v = r\omega\), where \(r = \text{distance of points from instantaneous centre of rotation}\).
Reason (R): Pure rolling of rigid body can be considered as a pure rotation about instantaneous centre of rotation.
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 pure rolling, the point of contact is the instantaneous center of rotation. The velocity of any point on the rigid body is given by \(v = r\omega\) where \(r\) is its distance from the ICOR. Pure rolling is essentially rotation about the ICOR.
Assertion (A): A sphere rolls down a rough inclined plane without slipping. It gains rotational K.E due to friction.
Reason (R): In this situation, work done by static friction is negative.
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
For pure rolling, the point of contact is instantaneously at rest. Therefore, static friction does no work.
The static friction provides the necessary torque for angular acceleration and gain in rotational K.E.