1
AIEEE 2003
MCQ (Single Correct Answer)
+4
-1
A particle performing uniform circular motion has angular frequency is doubled & its kinetic energy halved, then the new angular momentum is
A
$${L \over 4}$$
B
$$2L$$
C
$$4L$$
D
$${L \over 2}$$
2
AIEEE 2003
MCQ (Single Correct Answer)
+4
-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
A
$${I_Y} = 32{I_X}$$
B
$${I_Y} = 16{I_X}$$
C
$${I_Y} = {I_X}$$
D
$${I_Y} = 64{I_X}$$
3
AIEEE 2003
MCQ (Single Correct Answer)
+4
-1
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
A
$$\overrightarrow {r.} \overrightarrow \tau = 0\,\,$$ and $$\overrightarrow {F.} \overrightarrow \tau \ne 0\,\,$$
B
$$\overrightarrow {r.} \vec \tau \ne 0{\mkern 1mu} {\mkern 1mu} $$ and $$\overrightarrow {F.} \overrightarrow \tau = 0\,\,$$
C
$$\overrightarrow {r.} \vec \tau \ne 0{\mkern 1mu} $$ and $$\overrightarrow {F.} \overrightarrow \tau \ne 0$$
D
$$\overrightarrow {r.} \vec \tau = 0{\mkern 1mu} $$ and $$\overrightarrow {F.} \overrightarrow \tau = 0\,\,$$
4
AIEEE 2003
MCQ (Single Correct Answer)
+4
-1
The time period of satellite of earth is $$5$$ hours. If the separation between the earth and the satellite is increased to $$4$$ times the previous value, the new time period will become
A
$$10$$ hours
B
$$80$$ hours
C
$$40$$ hours
D
$$20$$ hours
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