JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
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
- A \(\frac{{M{R^2}}}{{16\sqrt 2 \pi }}\)
- B \(\frac{{4M{R^2}}}{{9\sqrt 3 \pi }}\)
- C \(\;\frac{{4M{R^2}}}{{3\sqrt 3 \pi }}\)
- D \(\;\frac{{M{R^2}}}{{32\sqrt 2 \pi }}\)
Answer & Solution
Correct Answer
(B) \(\frac{{4M{R^2}}}{{9\sqrt 3 \pi }}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l} Here\,a = \frac{2}{{\sqrt 3 }}R\\ Now,\frac{M}{{M'}} = \frac{{\frac{4}{3}\pi {R^3}}}{{{a^3}}}\\ = \frac{{\frac{4}{3}\pi {R^3}}}{{{{\left( {\frac{2}{{\sqrt 3 }}R} \right)}^3}}} = \frac{{\sqrt 3 }}{2}\pi .\,M' = \frac{{2M}}{{\sqrt 3 \pi }} \end{array}\) Moment…
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