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NEET 2025
Exam
NEET
Year
2025
Subject
Physics
Chapter
Rotational Motion

NEET 2025 · Physics · Rotational Motion

A sphere of radius R is cut from a larger solid sphere of radius 2...

A sphere of radius RR is cut from a larger solid sphere of radius 2R2 R as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the Y-axis is: A sphere of radius R is cut from a larger solid sphere of radius 2 R as shown in the figure. The ratio of the moment of

Options

  1. A78\frac{7}{8}
  2. B740\frac{7}{40}
  3. C757\frac{7}{57}
  4. D764\frac{7}{64}

Correct answer

C. 757\frac{7}{57}

Step-by-step solution

A sphere of radius R is cut from a larger solid sphere of radius 2 R as shown in the figure. The ratio of the moment of
I1=25M(2R)2I2=25mr2+md2=25(M8)R2+M8(R)2=7MR240I=I1−I2=8MR25−7MR240=5740MR2\begin{aligned} & \mathrm{I}_1=\frac{2}{5} \mathrm{M}(2 \mathrm{R})^2 \\ & \mathrm{I}_2=\frac{2}{5} \mathrm{mr}^2+\mathrm{md}^2 \\ & =\frac{2}{5}\left(\frac{\mathrm{M}}{8}\right) \mathrm{R}^2+\frac{\mathrm{M}}{8}(\mathrm{R})^2 \\ & =\frac{7 \mathrm{MR}^2}{40} \\ & \mathrm{I}=\mathrm{I}_1-\mathrm{I}_2=\frac{8 \mathrm{MR}^2}{5}-\frac{7 \mathrm{MR}^2}{40}=\frac{57}{40} \mathrm{MR}^2\end{aligned}\frac{I\mathrm{I}_2}{I\mathrm{I}_1}=\frac{7}{57}$
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