JEE MAIN - Physics (2003 - No. 29)
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
$${I_Y} = 32{I_X}$$
$${I_Y} = 16{I_X}$$
$${I_Y} = {I_X}$$
$${I_Y} = 64{I_X}$$
Explanation
We know that density $$\left( d \right) = {{mass\left( M \right)} \over {volume\left( V \right)}}$$
$$\therefore$$ Mass of disc $$M = d \times V = d \times \left( {\pi {R^2} \times t} \right).$$
The moment of inertia of any disc is $$I = {1 \over 2}M{R^2}$$
$$\therefore$$ $$I = {1 \over 2}\left( {d \times \pi {R^2} \times t} \right){R^2} = {{\pi d} \over 2}t \times {R^4}$$
$$\therefore$$ $${{{I_X}} \over {{I_Y}}} = {{{t_X}R_X^4} \over {{t_Y}R_Y^4}}$$
$$ = {{t \times {R^4}} \over {{t \over 4} \times {{\left( {4R} \right)}^4}}}$$
$$ = {1 \over {64}}$$
$$\therefore$$ Mass of disc $$M = d \times V = d \times \left( {\pi {R^2} \times t} \right).$$
The moment of inertia of any disc is $$I = {1 \over 2}M{R^2}$$
$$\therefore$$ $$I = {1 \over 2}\left( {d \times \pi {R^2} \times t} \right){R^2} = {{\pi d} \over 2}t \times {R^4}$$
$$\therefore$$ $${{{I_X}} \over {{I_Y}}} = {{{t_X}R_X^4} \over {{t_Y}R_Y^4}}$$
$$ = {{t \times {R^4}} \over {{t \over 4} \times {{\left( {4R} \right)}^4}}}$$
$$ = {1 \over {64}}$$
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