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NEET Physics
QUESTION #1057
Question 1
A sphere of radius \(R\) is cut from a larger sphere of radius \(2R\). Ratio of moment of inertia of smaller sphere to rest part about Y-axis:
Correct Answer Explanation
Let density \(=\rho\). Mass of large sphere: \(M=\frac{4}{3}\pi(2R)^3\rho=\frac{32\pi R^3\rho}{3}\). Mass of small sphere: \(m=\frac{4}{3}\pi R^3\rho=\frac{M}{8}\). \(I_{small}=\frac{2}{5}mR^2\). Small sphere center is at \(R\) from Y-axis (it's cut from side). \(I_{small,Y}=\frac{2}{5}mR^2+mR^2=\frac{7mR^2}{5}\). \(I_{large,Y}=\frac{2}{5}M(2R)^2=\frac{8MR^2}{5}\). \(I_{rest}=I_{large}-I_{small,Y}=\frac{8MR^2}{5}-\frac{7mR^2}{5}=\frac{8MR^2-7(M/8)R^2}{5}=\frac{MR^2(8-7/8)}{5}=\frac{57MR^2}{40}\). Ratio \(=\frac{7mR^2/5}{57MR^2/40}=\frac{7M/5\times40}{8\times57M}=\frac{7\times40}{5\times8\times57}=\frac{280}{2280}=\frac{7}{57}\).
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