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 09-Jan-2019 Evening]

A) 0.37           

B)       0.57

C)  0.77                

D)       0.17

Correct Answer: A

Solution :

[a]

\[{{T}^{1}}=2\pi \,\,\sqrt{\frac{{{I}^{1}}}{C}}\]

\[\frac{T}{{{T}^{1}}}\,\,=\,\,\frac{I}{{{I}^{1}}}\,\,=\,\,\frac{{{f}^{1}}}{f}\,\,=\,\,\frac{0.8f}{f}\,\,=\,\,0.8\]

\[\frac{I}{{{I}^{1}}}\,\,=\,\,0.64\]

\[\frac{M{{(2L)}^{2}}}{12}\,\,=\,\,0.64\,\left[ \frac{M{{(2L)}^{2}}}{12}+2m{{\left( \frac{L}{2} \right)}^{2}} \right]\]

\[\frac{M{{L}^{2}}}{3\,\,\times \,\,0.64}\,\,=\,\,\frac{M{{L}^{2}}}{3}\,+\,\frac{M{{L}^{2}}}{2}\]

\[\frac{M}{1.92}\,\,-\,\,\frac{M}{3}\,\,=\,\,\frac{m}{2}\]

\[\frac{1.08}{3\times 1.92}\,M\,\,=\,\,\frac{m}{2}\]

\[\Rightarrow \,\,\frac{m}{M}\,\,=\,\,\frac{1.08\,\,\times \,\,2}{3\,\times \,1.92}\,=\,\frac{2.16}{5.76}\,\,\approx \,\,0.37\]