JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
Moment of inertia of a square plate of side \(l\) about the axis passing through one of the corner and perpendicular to the plane of square plate is given by
- A \(\frac{{M} l^{2}}{6}\)
- B \({M} l^{2}\)
- C \(\frac{{M} l^{2}}{12}\)
- D \(\frac{2}{3} {M} l^{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{2}{3} {M} l^{2}\)
Step-by-step Solution
Detailed explanation
According to perpendicular Axis theorem. \({I}_{{x}}+{I}_{{y}}={I}_{z}\) \({I}_{z} \Rightarrow \frac{{m} \ell^{2}}{3}+\frac{{m} \ell^{2}}{3}\) \(=\frac{2 {m} \ell^{2}}{3}\)
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