AIEEE 2004 | Rotational Motion Question 202 | Physics

Let $$\overrightarrow F $$ be 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 $$\overrightarrow {F.} \overrightarrow \tau \ne 0\,\,$$

$$\overrightarrow {r.} \vec \tau \ne 0{\mkern 1mu} {\mkern 1mu} $$ and $$\overrightarrow {F.} \overrightarrow \tau = 0\,\,$$

$$\overrightarrow {r.} \vec \tau \ne 0{\mkern 1mu} $$ and $$\overrightarrow {F.} \overrightarrow \tau \ne 0$$

$$\overrightarrow {r.} \vec \tau = 0{\mkern 1mu} $$ and $$\overrightarrow {F.} \overrightarrow \tau = 0\,\,$$

AIEEE 2003

MCQ (Single Correct Answer)

-1

A particle performing uniform circular motion has angular frequency is doubled & its kinetic energy halved, then the new angular momentum is

$${L \over 4}$$

$$2L$$

$$4L$$

$${L \over 2}$$

AIEEE 2003

MCQ (Single Correct Answer)

-1

A circular disc $$X$$ of radius $$R$$ is made from an iron plate of thickness $$t,$$ and another disc $$Y$$ of radius $$4$$ $$R$$ is made from an iron plate of thickness $${t \over 4}.$$ Then the relation between the moment of inertia $${I_X}$$ and $${I_Y}$$ is

$${I_Y} = 32{I_X}$$

$${I_Y} = 16{I_X}$$

$${I_Y} = {I_X}$$

$${I_Y} = 64{I_X}$$

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