MHT CET · Physics · Rotational Motion
The moment of inertia of a thin uniform rod of length \(L\) and mass \(M\) about an axis passing through a point at a distance of \(1 / 3\) from one of its ends and perpendicular to the rod is
- A \(\frac{M L^{2}}{12}\)
- B \(\frac{M L^{2}}{9}\)
- C \(\frac{7 M L^{2}}{48}\)
- D \(\frac{M L^{2}}{48}\)
Answer & Solution
Correct Answer
(B) \(\frac{M L^{2}}{9}\)
Step-by-step Solution
Detailed explanation
\(I_{\mathrm{CM}}=\frac{M L^{2}}{12}\)
(about middle point)
\(\therefore\)
\(I=I_{\mathrm{CM}}+M x^{2}\)
\(\begin{array}{l}=\frac{M L^{2}}{12}+M\left(\frac{L}{6}\right)^{2} \\I=\frac{M L^{2}}{9}\end{array}\)
(about middle point)
\(\therefore\)
\(I=I_{\mathrm{CM}}+M x^{2}\)
\(\begin{array}{l}=\frac{M L^{2}}{12}+M\left(\frac{L}{6}\right)^{2} \\I=\frac{M L^{2}}{9}\end{array}\)
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