MHT CET 2020 16th October Evening Shift | Rotational Motion Question 53 | Physics

A body slides down a smooth inclined plane having angle $$\theta$$ and reaches the bottom with velocity $$v$$. If a body is a sphere, then its linear velocity at the bottom of the plane is

$$\sqrt{\frac{9}{7}} v$$

$$\sqrt{\frac{5}{7}} v$$

$$\sqrt{\frac{2}{7}} v$$

$$\sqrt{\frac{3}{7}} v$$

MHT CET 2020 16th October Evening Shift

MCQ (Single Correct Answer)

-0

A thin uniform rod has mass $$M$$ and length $$L$$ The moment of inertia about an axis perpendicular to it and passing through the point at a distance $$\frac{L}{3}$$ from one of its ends, will be

$$\frac{M L^2}{12}$$

$$\frac{7}{8} M L^2$$

$$\frac{M L^2}{9}$$

$$\frac{M L^2}{3}$$

MHT CET 2020 16th October Evening Shift

MCQ (Single Correct Answer)

-0

The ratio of radii of gyration of a ring to a disc (both circular) of same radii and mass, about a tangential axis perpendicular to the plane is

$$\frac{\sqrt{3}}{\sqrt{2}}$$

$$\frac{2}{\sqrt{5}}$$

$$\frac{\sqrt{2}}{1}$$

$$\frac{2}{\sqrt{3}}$$

MHT CET 2020 16th October Morning Shift

MCQ (Single Correct Answer)

-0

If there is a change of angular momentum from $$1 \mathrm{j}$$-$$\mathrm{s}$$ to $$4 \mathrm{j}$$-$$\mathrm{s}$$ in $$4 \mathrm{~s}$$, then the torque

$$\left(\frac{5}{4}\right) \mathrm{J}$$

$$\left(\frac{3}{4}\right) \mathrm{J}$$

$$1 \mathrm{~J}$$

$$\left(\frac{4}{3}\right) \mathrm{J}$$

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