JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A uniform thin bar of mass \(6 \,kg\) and length \(2.4\, meter\) is bent to make an equilateral hexagon. The moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of hexagon is ...... \(\times \,10^{-1} \,kg m ^{2}\)
- A \(16\)
- B \(8\)
- C \(24\)
- D \(4\)
Answer & Solution
Correct Answer
(B) \(8\)
Step-by-step Solution
Detailed explanation
\(6 \ell=2.4 \quad \ell=0.4 \,m\) \(\sin 60^{\circ}=\frac{ r }{\ell}\) \(r =\ell \sin 60^{\circ}=\frac{\ell \sqrt{3}}{2}\) MOI, \(\quad I=\left[\frac{m \ell^{2}}{12}+\operatorname{mr}^{2}\right] 6\)…
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