Rotational Motion

JEE Main level, single correct, medium

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Question 16 of 280

A solid sphere of mass 5 kg and radius 10 cm is kept in contact with another solid sphere of mass 10 kg and radius 20 cm. The moment of inertia of this pair of spheres about the tangent passing through the point of contact is ____\_\_\_\_ kg.m2\mathrm{kg}. \mathrm{m}^{2}.

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Answer: (D) 0.63

For two solid spheres in contact with the tangent line at the contact point.
Sphere 1: m1=5m_1 = 5 kg, r1=0.1r_1 = 0.1 m
Sphere 2: m2=10m_2 = 10 kg, r2=0.2r_2 = 0.2 m
For a solid sphere, Icm=25mr2I_{cm} = \frac{2}{5}mr^2
The tangent at contact is perpendicular to the line joining centers. The perpendicular distance from each sphere's center to this tangent line equals its radius.
Using parallel axis theorem: Itotal=Icm+md2I_{total} = I_{cm} + md^2
For sphere 1: I1=25(5)(0.1)2+5(0.1)2=0.02+0.05=0.07I_1 = \frac{2}{5}(5)(0.1)^2 + 5(0.1)^2 = 0.02 + 0.05 = 0.07 kg·m²
For sphere 2: I2=25(10)(0.2)2+10(0.2)2=0.16+0.4=0.56I_2 = \frac{2}{5}(10)(0.2)^2 + 10(0.2)^2 = 0.16 + 0.4 = 0.56 kg·m²
Total: I=0.07+0.56=0.63I = 0.07 + 0.56 = 0.63 kg·m²

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