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(D) The time taken for a body to roll down an inclined plane is given by $t = \sqrt{\frac{2s(1 + \beta)}{g \sin 	heta}}$,where $\beta = \frac{I}{MR^2

Vedclass · Accepted Answer

$1.5$

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(D) The time taken for a body to roll down an inclined plane is given by $t = \sqrt{\frac{2s(1 + \beta)}{g \sin 	heta}}$,where $\beta = \frac{I}{MR^2}$. For a thin spherical shell,$I = \frac{2}{3}MR^2$,so $\beta = 2/3$. However,for a shell with thickness $t$,the moment of inertia is $I = \frac{2}{3}M \frac{R_o^5 - R_i^5}{R_o^3 - R_i^3}$. Given $R_o = 20 \,cm$ and $t \ll R_o$,$I \approx \frac{2}{3}MR^2(1 + \frac{t}{R})$. Thus,$\beta \approx \frac{2}{3}(1 + \frac{t}{R})$. The time $t \propto \sqrt{1 + \beta}$. Since the time difference is $1 \%$,$\frac{\Delta t}{t} = \frac{1}{2} \frac{\Delta \beta}{1 + \beta} \approx 0.01$. Given $\beta \approx 2/3$,$1 + \beta \approx 5/3$. Then $\frac{1}{2} \frac{\Delta \beta}{5/3} = 0.01 \implies \Delta \beta = 0.033$. Since $\beta = \frac{2}{3}(1 + \frac{t}{R})$,$\Delta \beta = \frac{2}{3} \frac{\Delta t}{R}$. Substituting values: $0.033 = \frac{2}{3} \frac{\Delta t}{20} \implies \Delta t = 0.033 	imes 30 = 0.99 \,cm$. The total thickness is $0.5 + 0.99 = 1.49 \,cm$,which is closest to $1.5 \,cm$.

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