MHT CET · Physics · Rotational Motion
Three solid spheres each of mass ' \(M\) ' and radius ' \(R\) ' are arranged as shown in the figure. The moment of inertia of the system about YY' will be
- A \(\frac{16}{5} \mathrm{MR}^2\)
- B \(\frac{21}{5} \mathrm{MR}^2\)
- C \(\frac{7}{5} \mathrm{MR}^2\)
- D \(\frac{11}{5} \mathrm{MR}^2\)
Answer & Solution
Correct Answer
(A) \(\frac{16}{5} \mathrm{MR}^2\)
Step-by-step Solution
Detailed explanation
Moment of inertia of the upper sphere \(=\frac{2}{5} \mathrm{MR}^2\)
For each lower sphere M.I. \(=\frac{2}{5} \mathrm{MR}^2+\mathrm{MR}^2=\frac{7}{5} \mathrm{MR}^2\)
\(\therefore\) Total moment of inertia \(\mathrm{I}=\frac{2}{5} \mathrm{MR}^2+2 \times \frac{7}{5} \mathrm{MR}^2=\frac{16}{5} \mathrm{MR}^2\)
For each lower sphere M.I. \(=\frac{2}{5} \mathrm{MR}^2+\mathrm{MR}^2=\frac{7}{5} \mathrm{MR}^2\)
\(\therefore\) Total moment of inertia \(\mathrm{I}=\frac{2}{5} \mathrm{MR}^2+2 \times \frac{7}{5} \mathrm{MR}^2=\frac{16}{5} \mathrm{MR}^2\)
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