The ratio of kinetic energy of two bodies is $2 : 1$ and their angular momentum are in the ratio of $1 : 2$ . Then the ratio of their moment of inertia will be (assume pure rotational motion)

1 : 4

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4 : 1

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8 : 1

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1 : 8

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Solution

The correct option is D $1 : 8$
Let, kinetic energy of the first body is $K E_{1}$ and second body is $K E_{2}$
$L_{1} & L_{2}$ and $I_{1} & I_{2}$ are angular momentum and moment of Inertia of the first body and second body respectively.
Given, $\frac{K E_{1}}{K E_{2}} = \frac{2}{1}, \frac{L_{1}}{L_{2}} = \frac{1}{2}$
Now,
$\frac{K E_{1}}{K E_{2}} = \frac{\frac{1}{2} I_{1} ω_{1}^{2}}{\frac{1}{2} I_{2} ω^{2}} = \frac{\frac{1}{2} \frac{(I_{1} ω_{1})^{2}}{I_{1}}}{\frac{1}{2} \frac{(I_{2} ω_{2})^{2}}{I_{2}}}$
$\Rightarrow \frac{K E_{1}}{K E_{2}} = \frac{\frac{(L_{1})^{2}}{I_{1}}}{\frac{(L_{2})^{2}}{I_{2}}}$ as $(I_{1} ω_{1} = L_{1} and I_{2} ω_{2} = L_{2})$
$\Rightarrow 2 = \frac{1}{4} \frac{I_{2}}{I_{1}}$
$\Rightarrow \frac{I_{2}}{I_{1}} = \frac{8}{1} \Rightarrow \frac{I_{1}}{I_{2}} = \frac{1}{8}$

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