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(C) The frequency of torsional oscillations is given by $f = \frac{1}{2 \pi} \sqrt{\frac{C}{I}}$,where $C$ is the torsional constant and $I$ is the mo

Vedclass · Accepted Answer

$0.37$

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(C) The frequency of torsional oscillations is given by $f = \frac{1}{2 \pi} \sqrt{\frac{C}{I}}$,where $C$ is the torsional constant and $I$ is the moment of inertia.For the rod of mass $M$ and length $2L$ suspended at its center,the moment of inertia is $I_1 = \frac{M(2L)^2}{12} = \frac{ML^2}{3}$.Thus,$f = \frac{1}{2 \pi} \sqrt{\frac{C}{ML^2/3}}$.When two masses $m$ are attached at distance $L/2$ from the center,the new moment of inertia is $I_2 = I_1 + 2 	imes m(L/2)^2 = \frac{ML^2}{3} + \frac{mL^2}{2}$.The new frequency is $f' = f - 0.20f = 0.8f$.So,$0.8f = \frac{1}{2 \pi} \sqrt{\frac{C}{I_2}}$.Dividing the two equations: $\frac{f}{0.8f} = \sqrt{\frac{I_2}{I_1}} \Rightarrow \frac{1}{0.8} = \sqrt{\frac{ML^2/3 + mL^2/2}{ML^2/3}}$.Squaring both sides: $\frac{1}{0.64} = 1 + \frac{3m}{2M} \Rightarrow 1.5625 = 1 + \frac{3m}{2M}$.$\frac{3m}{2M} = 0.5625 \Rightarrow \frac{m}{M} = 0.5625 	imes \frac{2}{3} = 0.375$.Thus,the ratio $m/M$ is close to $0.37$.

$A$ rod of mass $M$ and length $2L$ is suspended at its middle by a wire. It exhibits torsional oscillations. If two masses each of $m$ are attached at a distance $L/2$ from its centre on both sides,the oscillation frequency reduces by $20\%$. The value of the ratio $m/M$ is close to:

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