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JEE MAIN 2019
09-01-19 S2
Question
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:
Select the correct option:
A
0.77
B
0.17
C
0.37
D
0.57
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$$ \begin{aligned} & \mathrm{I}_1=\frac{\mathrm{M}(2 \mathrm{~L})^2}{12}=\frac{\mathrm{ML}^2}{3} \\ & \mathrm{I}_2=\mathrm{I}_1+2 \frac{\mathrm{~mL}^2}{4}=\frac{\mathrm{ML}^2}{3}+\frac{\mathrm{mL}^2}{2} \\ & \omega \propto \frac{1}{\sqrt{ } \mathrm{I}} \end{aligned} $$ $$ \frac{\omega_1}{\omega_2}=\frac{1}{0.8}=\sqrt{\frac{\frac{M}{3}+\frac{m}{2}}{\frac{M}{3}}} \quad \Rightarrow \frac{m}{M}=0.375 $$
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