Question

A rod of mass $M$ and length $2L$ is performing SHM as torsional pendulum in the horizontal plane. Two blocks, each of mass $m$ , are put at distance $\frac{L}{2}$ from the centre. The frequency after putting blocks of mass $m$ is $20\%$ of initial frequency. Then, the ratio of $\frac{m}{M}$ will be

• A. $12$
• B. $14$
• C. $16$
• D. $18$

Answer 1

Answer: (C)

$T=2\pi \, \sqrt{\frac{I}{C}}$
$f=\frac{1}{2 \pi } \, \sqrt{\frac{C}{I}}$
$f_{i n i t i a l}=\frac{1}{2 \pi } \, \sqrt{\frac{12 C}{M \left(2 L\right)^{2}}}$
$f_{f i n a l}=\frac{1}{2 \pi }\sqrt{\frac{C}{\left(\frac{M L^{2}}{3} + 2 m \frac{L^{2}}{4}\right)}}$
Since, $f_{final}=0.2f_{initial}$
$0.2\times \frac{1}{2 \pi } \, \sqrt{\frac{3 C}{M L^{2}}}=\frac{1}{2 \pi }\sqrt{\frac{C}{\frac{M L^{2}}{3} + \frac{2 m L^{2}}{4}}} \, $
$\Rightarrow 0.04\times \frac{3}{M}=\frac{12}{\left(4 M + 6 m\right)}$
$\Rightarrow M=0.04M+0.06m$
$\Rightarrow \frac{m}{M}=\frac{0 . 96}{0.06}=16$