MHT CET · Physics · Rotational Motion
A thin metal rod of mass 'M' and length 'L' is cut into 4 equal parts by cutting it perpendicular to its length, If moment of inertia of the rod about an axis passing through its centre and perpendicular to its axis is 'I' then moment of inertia of each part about the similar axis is
- A \(\frac{\mathrm{I}}{16}\)
- B \(\frac{\mathrm{I}}{32}\)
- C \(\frac{\mathrm{I}}{128}\)
- D \(\frac{\mathrm{I}}{64}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{I}}{64}\)
Step-by-step Solution
Detailed explanation
\(I=\frac{M L^{2}}{12}\)
\(\therefore \frac{I^{\prime}}{I}=\frac{\frac{M}{4} \cdot\left(\frac{L}{4}\right)^{2}}{M L^{2}}=\frac{1}{64}\)
\(\therefore \frac{I^{\prime}}{I}=\frac{\frac{M}{4} \cdot\left(\frac{L}{4}\right)^{2}}{M L^{2}}=\frac{1}{64}\)
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