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JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion

Two identical solid spheres each of mass \(2\,kg\) and radii \(10\,cm\) are fixed at the ends of a light rod. The separation between the centres of the spheres is \(40\,cm\). The moment of inertia of the system about an axis perpendicular to the rod passing through its middle point is \(........... \times 10^{-3}\,kg - m ^2\)

  1. A \(177\)
  2. B \(178\)
  3. C \(198\)
  4. D \(176\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(176\)

Step-by-step Solution

Detailed explanation

\(I =2\left( I _{ cm }+ md ^2\right)\) \(=2\left(\frac{2}{5} mr ^2+ md ^2\right)\) \(=\frac{4}{5} \times 2 \times(0.1)^2+2(2)(0.20)^2\) \(=\frac{8}{5} \times 10^{-2}+16 \times 10^{-2}\) \(=(1.6+16) \times 10^{-2}\) \(=17.6 \times 10^{-2}\) \(I =176 \times 10^{-3}\,kg\,m ^2\)
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