WBJEE · Physics · Rotational Motion
Three particles, each of mass ' \(m\) ' grams situated at the vertices of an equilateral \(\triangle A B C\) of side ' \(a\) ' \(\mathrm{cm}\) (as shown in the figure). The moment of inertia of the system about a line \(\mathrm{AX}\) perpendicular to \(\mathrm{AB}\) and in the plane of \(\mathrm{ABC}\) in \(\mathrm{g}-\mathrm{cm}^2\) units will be
- A \(2 \mathrm{ma}^2\)
- B \(\frac{3}{2} \mathrm{ma}^2\)
- C \(\frac{3}{4} \mathrm{ma}^2\)
- D \(\frac{5}{4} \mathrm{ma}^2\)
Answer & Solution
Correct Answer
(D) \(\frac{5}{4} \mathrm{ma}^2\)
Step-by-step Solution
Detailed explanation
\(\mathrm{I}=\mathrm{~ma}^2+\mathrm{m}\left(\mathrm{a} \cos 60^{\circ}\right)^2=m \mathrm{a}^2+\frac{\mathrm{ma}^2}{4}=\frac{5 \mathrm{ma}^2}{4}\)
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