JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
Two identical spherical balls of mass \(M\) and radius \(R\) each are stuck on two ends of a rod of length \(2R\) and mass \(M\) (see figure). The moment of inertia of the system about the axis passing perpendicularly through the centre of the rod is
- A \(\frac{{137}}{{15}}M{R^2}\)
- B \(\frac{{17}}{{15}}M{R^2}\)
- C \(\frac{{209}}{{15}}M{R^2}\)
- D \(\frac{{152}}{{15}}M{R^2}\)
Answer & Solution
Correct Answer
(A) \(\frac{{137}}{{15}}M{R^2}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l} l = 2\left[ {\frac{2}{5}M{R^2} + M4{R^2}} \right] + M\frac{{4{R^2}}}{{12}}\\ \,\,\, = M{R^2}\left[ {\frac{1}{3} + \frac{4}{5} + 8} \right] = \frac{{137}}{{15}}M{R^2} \end{array}\)
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