JEE MAIN - Physics (2004 - No. 47)
One solid sphere $$A$$ and another hollow sphere $$B$$ are of same mass and same outer radii. Their moment of inertia about their diameters are respectively $${I_A}$$ and $${I_B}$$ such that
$${I_A} < {I_B}$$
$${I_A} > {I_B}$$
$${I_A} = {I_B}$$
$${{{I_A}} \over {{I_B}}} = {{{d_A}} \over {{d_B}}}$$
where $${d_A}$$ and $${d_B}$$ are their densities.
Explanation
For solid sphere the moment of inertia of $$A$$ about its diameter
$${I_A} = {2 \over 5}M{R^2}.$$
The moment of inertia of a hollow sphere $$B$$ about its diameter
$${I_B} = {2 \over 3}M{R^2}.$$
$$\therefore$$ $${I_A} < {I_B}$$
$${I_A} = {2 \over 5}M{R^2}.$$
The moment of inertia of a hollow sphere $$B$$ about its diameter
$${I_B} = {2 \over 3}M{R^2}.$$
$$\therefore$$ $${I_A} < {I_B}$$
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