MHT CET · Physics · Rotational Motion
A thin metal rod of mass \(M\) and length \(L\) is into four equal parts by cutting it perpendicular to the length. If moment of inertia of the rod about an axis passing through its centre and perpendicular to its axis it \(I\), then moment of inertia of each part about the similar axis is
- A \(\frac{I}{64}\)
- B \(\frac{I}{8}\)
- C \(\frac{I}{16}\)
- D \(\frac{I}{32}\)
Answer & Solution
Correct Answer
(A) \(\frac{I}{64}\)
Step-by-step Solution
Detailed explanation
We know, that moment of inertia has following functional dependence on mass and length of the rod:
\(I \propto M L^2\)
If thin rod is split into four equal parts.
\(\begin{aligned} & M^{\prime}=\frac{M}{4} \\ & L^{\prime}=\frac{L}{4} \\ & \therefore I^{\prime} \propto \frac{M}{4}\left(\frac{L^2}{16}\right) \\ & \Rightarrow I^{\prime} \propto \frac{M L^2}{64} \\ & \Rightarrow I^{\prime}=\frac{I}{64}\end{aligned}\)
\(I \propto M L^2\)
If thin rod is split into four equal parts.
\(\begin{aligned} & M^{\prime}=\frac{M}{4} \\ & L^{\prime}=\frac{L}{4} \\ & \therefore I^{\prime} \propto \frac{M}{4}\left(\frac{L^2}{16}\right) \\ & \Rightarrow I^{\prime} \propto \frac{M L^2}{64} \\ & \Rightarrow I^{\prime}=\frac{I}{64}\end{aligned}\)
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