AIPMT 2010 Prelims | Rotational Motion Question 57 | Physics

AIPMT 2010 Prelims

MCQ (Single Correct Answer)

-1

A circular disk of moment of inertia $${I_t}$$ is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed $${\omega _i}$$. Another disk of moment of inertia $${I_b}$$ is dropped coaxially onto the rotating disk. Initially the second disk has zero angular speed. Eventually both the disks rotate with a constant angular speed $$\omega $$. The energy lost by the initially rotating disc to friction is

$${1 \over 2}{{I_b^2} \over {\left( {{I_t} + {I_b}} \right)}}\omega _i^2$$

$${1 \over 2}{{I_t^2} \over {\left( {{I_t} + {I_b}} \right)}}\omega _i^2$$

$${{{I_b} - {I_t}} \over {\left( {{I_t} + {I_b}} \right)}}\omega _i^2$$

$${1 \over 2}{{{I_b}{I_t}} \over {\left( {{I_t} + {I_b}} \right)}}\omega _i^2$$

AIPMT 2009

MCQ (Single Correct Answer)

-1

Four identical thin rods each of mass M and length $$l$$, form a square frame. Moment of inertia of this frame about an axis through the centre of the square and perpendicular to its plane is

$${2 \over 3}M{l^2}$$

$${{13} \over 3}M{l^2}$$

$${1 \over 3}M{l^2}$$

$${4 \over 3}M{l^2}$$

AIPMT 2009

MCQ (Single Correct Answer)

-1

If $$\overrightarrow F $$ is the force acting on a particle having position vector $$\overrightarrow r $$ and $$\overrightarrow \tau $$ be the torque of this force about the origin, then

$$\overrightarrow r .\overrightarrow \tau > 0$$ and $$\overrightarrow F .\overrightarrow \tau < 0$$

$$\overrightarrow r .\overrightarrow \tau = 0$$ and $$\overrightarrow F .\overrightarrow \tau = 0$$

$$\overrightarrow r .\overrightarrow \tau = 0$$ and $$\overrightarrow F .\overrightarrow \tau \ne 0$$

$$\vec r.\vec \tau \ne 0$$ and $$\overrightarrow F .\overrightarrow \tau = 0$$

AIPMT 2009

MCQ (Single Correct Answer)

-1

A thin circular ring of mass M and radius R is rotating in a horizontal plane about an axis vertical to its plane with a constant angular velocity $$\omega $$. If two objects each of mass m be attached gently to the opposite ends of a diameter of the ring, the ring will then rotate with an angular velocity

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

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

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

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

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