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6. system of particles and rotational motion — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysics6. system of particles and rotational motion1 Mark
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
  • A
    $\frac{{M{R^2}}}{{16\sqrt 2 \pi }}$
  • ✓
    $\frac{{4M{R^2}}}{{9\sqrt 3 \pi }}$
  • C
    $\;\frac{{4M{R^2}}}{{3\sqrt 3 \pi }}$
  • D
    $\;\frac{{M{R^2}}}{{32\sqrt 2 \pi }}$
✓

Answer

Correct option: B.
$\frac{{4M{R^2}}}{{9\sqrt 3 \pi }}$
b
$\begin{array}{l}
Here\,a = \frac{2}{{\sqrt 3 }}R\\
Now,\frac{M}{{M'}} = \frac{{\frac{4}{3}\pi {R^3}}}{{{a^3}}}\\
 = \frac{{\frac{4}{3}\pi {R^3}}}{{{{\left( {\frac{2}{{\sqrt 3 }}R} \right)}^3}}} = \frac{{\sqrt 3 }}{2}\pi .\,M' = \frac{{2M}}{{\sqrt 3 \pi }}
\end{array}$

Moment 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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