ExamBro
ExamBro
JEE Mains · Physics · STD 11 - 13. oscillations

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

  1. A \(0.77\)
  2. B \(0.57\)
  3. C \(0.37\)
  4. D \(0.17\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(0.37\)

Step-by-step Solution

Detailed explanation

\(\mathrm{f}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{C}}{\left(\frac{\mathrm{ML}^{2}}{3}\right)}} \& 0.8 \mathrm{f}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{C}}{\left(\frac{\mathrm{ML}^{2}}{3}+\frac{\mathrm{mL}^{2}}{2}\right)}}\)…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app