AP EAMCET · PHYSICS · Oscillations
A thin magnetic iron rod of length \(30 \mathrm{~cm}\) is suspended in a uniform magnetic field. Its time period of oscillation is \(4 \mathrm{~s}\). It is broken into three equal parts. The time period in seconds of oscillation of one part when suspended in the same magnetic field is
- A \(\frac{1}{\sqrt{3}}\)
- B \(\frac{2}{\sqrt{3}}\)
- C \(\sqrt{3}\)
- D \(\frac{4}{\sqrt{3}}\)
Answer & Solution
Correct Answer
(D) \(\frac{4}{\sqrt{3}}\)
Step-by-step Solution
Detailed explanation
Time period of magnet \[ T=2 \pi \sqrt{\frac{I}{M H}} \] \[ M=\text { magnetic moment }=m \times l \] \[ I=\text { moment of inertia }=\frac{m l^2}{12} \] When magnet is broken into three equal parts, Then, \(M^{\prime}=m \times \frac{l}{3}=\frac{M}{3}\)…
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