MHT CET 2026 13th April Morning Shift | Rotational Motion Question 29 | Physics

A body slides down a smooth inclined plane of inclination $\theta$ and reaches the bottom with velocity 'V'. If the same body is a ring which rolls down the same inclined plane then linear velocity at the bottom of the plane is

$\dfrac{V}{\sqrt{2}}$

$\dfrac{V}{2}$

$V$

$2V$

MHT CET 2026 13th April Morning Shift

MCQ (Single Correct Answer)

-0

A disc of radius 0.4 m and mass 1 kg rotates about an axis passing through its centre and perpendicular to its plane. The angular acceleration is 10 rad/s$^2$. The tangential force applied to the rim of the disc is

$1$ N

$2$ N

$3$ N

$4$ N

MHT CET 2026 11th April Evening Shift

MCQ (Single Correct Answer)

-0

The moment of inertia of a ring about an axis passing through its centre and perpendicular to its plane is I. It is rotating with angular velocity $\omega$. Another identical ring is gently placed on it so that their centres coincide. If both rings are rotating about the same axis then loss in kinetic energy is

$I\omega^2$

$\dfrac{I\omega^2}{2}$

$\dfrac{I\omega^2}{4}$

$\dfrac{I\omega^2}{8}$

MHT CET 2026 11th April Evening Shift

MCQ (Single Correct Answer)

-0

Let M and L be the mass and length of thin uniform rod respectively. In first case, axis of rotation is passing through centre and perpendicular to length of rod. In second case axis of rotation is passing through one end and perpendicular to length of rod. The ratio of radius of gyration in first case to second case is

$\dfrac{1}{4}$

$\dfrac{1}{2}$

$\dfrac{1}{8}$

$\dfrac{1}{6}$

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