Two small spheres of masses 10kg and 30 kg are joined by a rod of length 0.5 m and of negligible mass. The M.I of the system about a normal axis through centre of mass of the system is

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1.875 kgm²

2.45 kgm²

0.75 kgm²

1.75 kgm²

answer is A.

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Detailed Solution

When two particles of masses m₁ and m₂ are at a distance d apart, M.I above centre of mass is
$I = \frac{m_{1} m_{2}}{m_{1} + m_{2}} d^{2} = \frac{10 x 30}{10 + 30} {(\frac{1}{2})}^{2} = 1.875 kg m^{2}$

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