question_answer

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 distance 'L/2' from its centre on both sides, it reduces the oscillation frequency by \[20%\] The value of ratio m/M is close to: [JEE Main Online Paper (Held On 09-Jan-2019 Evening]

A) 0.37     

B) 0.57

C) 0.77                 

D) 0.17

Correct Answer: A

Solution :

\[{{T}^{1}}=2\pi \,\,\sqrt{\frac{{{I}^{1}}}{C}}\] \[\frac{T}{{{T}^{1}}}\,\,=\,\,\frac{I}{{{I}^{1}}}\,\,=\,\,\frac{{{f}^{1}}}{f}\,\,=\,\,\frac{0.8f}{f}\,\,=\,\,0.8\] \[\frac{I}{{{I}^{1}}}\,\,=\,\,0.64\] \[\frac{M{{(2L)}^{2}}}{12}\,\,=\,\,0.64\,\left[ \frac{M{{(2L)}^{2}}}{12}+2m{{\left( \frac{L}{2} \right)}^{2}} \right]\] \[\frac{M{{L}^{2}}}{3\,\,\times \,\,0.64}\,\,=\,\,\frac{M{{L}^{2}}}{3}\,+\,\frac{M{{L}^{2}}}{2}\] \[\frac{M}{1.92}\,\,-\,\,\frac{M}{3}\,\,=\,\,\frac{m}{2}\] \[\frac{1.08}{3\times 1.92}\,M\,\,=\,\,\frac{m}{2}\] \[\Rightarrow \,\,\frac{m}{M}\,\,=\,\,\frac{1.08\,\,\times \,\,2}{3\,\times \,1.92}\,=\,\frac{2.16}{5.76}\,\,\approx \,\,0.37\]