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(C) The moment of inertia of a circular disc about an axis passing through its centre and perpendicular to its plane is given by $I = \frac{1}{2} M R^

Vedclass · Accepted Answer

$0.022$

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(C) The moment of inertia of a circular disc about an axis passing through its centre and perpendicular to its plane is given by $I = \frac{1}{2} M R^2$.Since the mass $M$ remains constant, the change in moment of inertia depends on the change in radius $R$.For a small change in temperature $\Delta t$, the change in radius is given by $\Delta R = R \alpha \Delta t$.The new radius is $R' = R(1 + \alpha \Delta t)$.The new moment of inertia is $I' = \frac{1}{2} M (R')^2 = \frac{1}{2} M R^2 (1 + \alpha \Delta t)^2$.Using the binomial approximation $(1 + x)^n \approx 1 + nx$ for small $x$, we get $I' \approx \frac{1}{2} M R^2 (1 + 2 \alpha \Delta t) = I(1 + 2 \alpha \Delta t)$.The fractional change in moment of inertia is $\frac{\Delta I}{I} = \frac{I' - I}{I} = 2 \alpha \Delta t$.Substituting the given values: $\alpha = 11 	imes 10^{-6} /{ }^{\circ} C$ and $\Delta t = 10^{\circ} C$.$\frac{\Delta I}{I} = 2 	imes (11 	imes 10^{-6}) 	imes 10 = 220 	imes 10^{-6} = 0.00022$.The percentage increase is $0.00022 	imes 100 = 0.022 \%$.

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