AIPMT 2011 Prelims | Rotational Motion Question 51 | Physics

The instantaneous angular position of a point on a rotating wheel is given by the equation $$\theta \left( t \right) = 2{t^3} - 6{t^2}$$

The torque on the wheel becomes zero at

t = 1 s

t = 0.5 s

t = 0.25 s

t = 2 s

AIPMT 2011 Prelims

MCQ (Single Correct Answer)

-1

The moment of inertia of a thin uniform rod of mass M and length L about an axis passing through its midpoint and perpendicular to its length is $$I$$₀. Its moment of inertia about an axis passing through one of its ends and perpendicular to its length is

$${I_0} + M{L^2}/2$$

$${I_0} + M{L^2}/4$$

$${I_0} + 2M{L^2}$$

$${I_0} + M{L^2}$$

AIPMT 2010 Mains

MCQ (Single Correct Answer)

-1

From a circular disc of radius R and mass 9M, a small disc of mass M and radius $${R \over 3}$$ is removed concentrically. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre is

$${{40} \over 9}$$ MR²

MR²

4MR²

$${4 \over 9}$$ MR²

AIPMT 2010 Mains

MCQ (Single Correct Answer)

-1

A thin circular ring of mass M and radius r is rotating about its axis with constant angular velocity $$\omega $$. Two objects each of msass m are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with angular velocity given by

$${{\left( {M + 2m} \right)\omega } \over {2m}}$$

$${{2M\omega } \over {M + 2m}}$$

$${{\left( {M + 2m} \right)\omega } \over M}$$

$${{M\omega } \over {M + 2m}}$$

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