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# Effect of moment of inertia of a rigid body due to thermal expansion is  $\frac{\Delta I}{I}=x\alpha \Delta t$ where  $x=$

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1
b
2
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detailed solution

Correct option is B

moment of inertia I=mK2 m is mass; K is radius of gyration on differentiating above equation and dividing same with above equation ∆II=2∆KK fractional change in moment of inertia is ∆II=2α∆t α is coefficient of linear expansion of solid ∆t is change in temperature

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The coefficient of linear expansion of brass and steel are ${\mathrm{\alpha }}_{1}$ and ${\alpha }_{2}$. If we take a brass rod of length ${l}_{1}$ and steel rod of length ${l}_{\mathit{2}}$ at 0oC, their difference in length $\left({l}_{2}-{l}_{1}\right)$will remain the same at all temperatures if

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