AIEEE 2007 | Rotational Motion Question 193 | Physics

A round uniform body of radius $$R,$$ mass $$M$$ and moment of inertia $$I$$ rolls down (without slipping) an inclined plane making an angle $$\theta $$ with the horizontal. Then its acceleration is

$${{g\,\sin \theta } \over {1 - M{R^2}/I}}$$

$${{g\,\sin \theta } \over {1 + I/M{R^2}}}$$

$${{g\,\sin \theta } \over {1 + M{R^2}/I}}$$

$${{g\,\sin \theta } \over {1 - I/M{R^2}}}$$

AIEEE 2007

MCQ (Single Correct Answer)

-1

Angular momentum of the particle rotating with a central force is constant due to

constant torque

constant force

constant linear momentum

zero torque

AIEEE 2007

MCQ (Single Correct Answer)

-1

For the given uniform square lamina $$ABCD$$, whose center is $$O,$$ AIEEE 2007 Physics - Rotational Motion Question 185 English

$${I_{AC}} = \sqrt 2 \,\,{I_{EF}}$$

$$\sqrt 2 {I_{AC}} = {I_{EF}}$$

$${I_{AD}} = 3{I_{EF}}$$

$${I_{AC}} = {I_{EF}}$$

AIEEE 2006

MCQ (Single Correct Answer)

-1

Four point masses, each of value $$m,$$ are placed at the corners of a square $$ABCD$$ of side $$l$$. The moment of inertia of this system about an axis passing through $$A$$ and parallel to $$BD$$ is

$$2m{l^2}$$

$$\sqrt 3 m{l^2}$$

$$3m{l^2}$$

$$m{l^2}$$

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