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 __________ × 10⁻³ kg m² Numerical answer type. Work it out, then open the answer.

Answer: 176. Each sphere: I = 2/5mr² + md² with r = 0.1 m, d = 0.2 m. I₁ = 2/5(2)(0.01) + 2(0.04) = 0.008 + 0.08 = 0.088 kg m² I = 2I₁ = 0.176 kg m² = 176 × 10⁻³ kg m² So the answer is 176.

Two identical solid spheres each of mass 2 kg and radii 10 cm are… | JEE Main 2023 (6 Apr, Shift 1)

Q95JEE Main 2023 · 6 APR · SHIFT 1Medium

Moment of Inertia & Radius of Gyration

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

Numerical answer type. Work it out, then open the answer.

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Answer: 176

Each sphere: $I = \frac{2}{5}mr^2 + md^2$ with $r = 0.1\text{ m}$ , $d = 0.2\text{ m}$ .
$I_1 = \frac{2}{5}(2)(0.01) + 2(0.04) = 0.008 + 0.08 = 0.088\text{ kg m}^2$
$I = 2I_1 = 0.176\text{ kg m}^2 = 176 \times 10^{-3}\text{ kg m}^2$
So the answer is $176$ .

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