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Q.

A circular disc A of radius r is made from an iron plate of thickness t and another circular disc B of radius 4r is made from an iron plate of thickness t/4.The relation between moments of inertia $l_{A}$ and $l_{B}$ is

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a

$l_{A} < l_{B}$

b

$l_{A} > l_{B}$

c

Depends on the actual value of t and r

d

$l_{A} = l_{B}$

answer is C.

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Detailed Solution

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$\begin{array}{l} l = \frac{m r^{2}}{2} = ρ (π r^{2} t) \frac{r^{2}}{2} \\ I \infty r^{4} t \\ \frac{I_{A}}{I_{B}} = {[\frac{r_{A}}{r_{B}}]}^{4} . \frac{t_{A}}{t_{B}} = {[\frac{1}{4}]}^{4} . \frac{4}{1} = {[\frac{1}{4}]}^{3} \\ H e n c e, I_{A} = \frac{I_{B}}{4^{3}} o r I_{A} < I_{B} \end{array}$

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A circular disc A of radius r is made from an iron plate of thickness t and another circular disc B of radius 4r is made from an iron plate of thickness t/4.The relation between moments of inertia lA and lB is