JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2015
- Hard
- MCQ
From a solid sphere of mass \(M\) and radius \(R\) a cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its center and perpendicular to one of its faces is
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(\frac{{4M{R^2}}}{{9\sqrt 3 \pi }}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}
Here\,a = \frac{2}{{\sqrt 3 }}R\\
Now,\frac{M}{{M'}} = \frac{{\frac{4}{3}\pi {R^3}}}{{{a^3}}}\\
= \frac{{\frac{4}{3}\pi {R^3}}}{{{{\left( {\frac{2}{{\sqrt 3 }}R} \right)}^3}}} = \frac{{\sqrt 3 }}{2}\pi .\,M' = \frac{{2M}}{{\sqrt 3 \pi }}
\end{array}\)
Here\,a = \frac{2}{{\sqrt 3 }}R\\
Now,\frac{M}{{M'}} = \frac{{\frac{4}{3}\pi {R^3}}}{{{a^3}}}\\
= \frac{{\frac{4}{3}\pi {R^3}}}{{{{\left( {\frac{2}{{\sqrt 3 }}R} \right)}^3}}} = \frac{{\sqrt 3 }}{2}\pi .\,M' = \frac{{2M}}{{\sqrt 3 \pi }}
\end{array}\)
Moment of inertia of the cube about the given axis,
\(I = \frac{{M'{a^2}}}{6} = \frac{{\frac{{2M}}{{\sqrt 3 \pi }} \times {{\left( {\frac{2}{{\sqrt R }}R} \right)}^2}}}{6} = \frac{{4M{R^2}}}{{9\sqrt 3 \pi }}\)
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