The ratio of moment of inertia of a solid sphere to that of a hollow sphere of same mass and radius is (a) 5/3 (b) 4/5 (c) 3/5 (d) 2/5
The Correct Answer and Explanation is:
Correct Answer: (a) 5/3
Explanation:
The moment of inertia (I) of an object depends on how its mass is distributed relative to the axis of rotation. For spheres, the moment of inertia about an axis through the center is different for a solid and a hollow (thin-walled) sphere, even if they have the same mass (M) and radius (R).
Let us recall the standard formulas:
- Solid Sphere:
Isolid=25MR2I_{solid} = \frac{2}{5}MR^2Isolid=52MR2 - Hollow Sphere (thin spherical shell):
Ihollow=23MR2I_{hollow} = \frac{2}{3}MR^2Ihollow=32MR2
We are asked to find the ratio of the moment of inertia of the solid sphere to that of the hollow sphere.
Let us compute the ratio: IsolidIhollow=25MR223MR2\frac{I_{solid}}{I_{hollow}} = \frac{\frac{2}{5}MR^2}{\frac{2}{3}MR^2}IhollowIsolid=32MR252MR2
Canceling out MR2MR^2MR2 from both the numerator and denominator: =2/52/3=25×32=610=35= \frac{2/5}{2/3} = \frac{2}{5} \times \frac{3}{2} = \frac{6}{10} = \frac{3}{5}=2/32/5=52×23=106=53
Wait — this result matches (c) 3/5. So this indicates the original assumption must be reconsidered. Let’s carefully distinguish between thin-walled hollow sphere and uniform spherical shell.
In many physics problems:
- A hollow sphere or thin spherical shell has:
I=23MR2I = \frac{2}{3}MR^2I=32MR2 - A solid sphere has:
I=25MR2I = \frac{2}{5}MR^2I=52MR2
So the ratio becomes: IsolidIhollow=2/52/3=35\frac{I_{solid}}{I_{hollow}} = \frac{2/5}{2/3} = \frac{3}{5}IhollowIsolid=2/32/5=53
Thus, the correct answer is (c) 3/5
Final Answer: (c) 3/5
Conceptual Insight:
The hollow sphere has more mass distributed farther from the axis, increasing its moment of inertia compared to the solid sphere. So, even though both have the same mass and radius, the solid sphere resists rotational motion less, which is why its moment of inertia is lower.
