Question

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

• A. 12
• B. 14
• C. 16
• D. 18

Answer 1

The correct option is C 16

$f_0=\frac{1}{2\pi }\sqrt{\frac{c}{l_o}}$

$f_0=\frac{1}{2\pi }\sqrt{\frac{3c}{M{L}^2}}$ ..........(1)

$f^{\prime }=\frac{M{L}^2}{3}+2\frac{m{L}^2}{4}$

$f^{\prime }=\frac{1}{2\pi }\sqrt{\frac{c}{L^2(\frac{M}{a}+\frac{m}{2})}}=0.2f_0$ .......... (2)

from equation (1) and (2)

$\frac{6}{2m+3m}=\frac{0.04}{M}\times 3$

$\frac{m}{M}=\frac{96}{6}=16$