Question

If 27 small spheres each of Moment of Inertia $I$ about their diametrical axis are combined to form a big sphere, then the Moment of Inertia of the big sphere about its diameter is:
A) $243I$
B) $81I$
C) $27I$
D) $9I$

Answer 1

Hint: Moment of Inertia or rotational inertia is a physical quantity that determines the amount of torque required for a desired angular acceleration of a system. It is also defined as the property of a body due to which it resists angular acceleration. The formula for moment of inertia is;
$I = \sum {m_i}{r_i}^2$
Where ${m_i}$ is the mass of the $i$ particle and ${r_i}$ is its distance from the origin.

Formulae used:
The formula for moment of inertia is;
$I = \sum {m_i}{r_i}^2$
Where ${m_i}$ is the mass of the $i$ particle and ${r_i}$ is its distance from the origin.
The formula for moment of inertia of a sphere after integrating the formula is;
${I_s} = \dfrac{2}{5}M{R^2}$
Where ${I_s}$ is the moment of inertia of the sphere, $M$ is the mass of the sphere and $R$ is the radius of the sphere.

Complete step by step solution:
Let $M$ be the mass of the small spheres, $R$ be the radius of the small spheres, ${M_B}$ be the mass of the bigger sphere and ${R_B}$ be the radius of the bigger sphere. To find the moment of inertia of the combined sphere, we have to first calculate its mass and radius in terms of the mass and radius of the smaller spheres.
If $27$ small spheres each of mass $M$ and radius $R$ are combined to form a bigger sphere then:
The mass of the spheres would simply get added to give the bigger sphere’s mass, as in,
${M_B} = 27M$ $...\left( 1 \right)$
The volume of the spheres would also get combined to one bigger sphere with a volume ${V_B}$ which is $27$ times the volume of the smaller sphere, $V$ , that is;
${V_B} = 27V$
$ \Rightarrow \dfrac{{4\pi }}{3}{\left( {{R_B}} \right)^3} = 27\left[ {\dfrac{{4\pi }}{3}{R^3}} \right]$
$ \Rightarrow {R_B} = \sqrt[3]{{27{{(R)}^3}}}$
$ \Rightarrow {R_B} = 3R$ $...\left( 2 \right)$
Also, it is known that the Moment of Inertia of the small spheres is;
${I_{}} = \dfrac{2}{5}M{R^2}$ $...\left( 3 \right)$
Where $I$ is the moment of inertia of the small sphere, $M$ is the mass of the small sphere and $R$ is the radius of the small sphere.
In case of the bigger sphere, the moment of inertia would then be defined as;
${I_B} = \dfrac{2}{5}{M_B}{({R_B})^2}$
$ \Rightarrow {I_B} = \dfrac{2}{5}(27M){(3R)^2}$ From $...(1)$ and $...\left( 2 \right)$
$ \Rightarrow {I_B} = \left( {27} \right)\left( 3 \right)\dfrac{2}{5}M{(R)^2}$ From $...\left( 3 \right)$
$ \Rightarrow {I_B} = \left( {27} \right)\left( 3 \right)I$
$ \Rightarrow {I_B} = 243I$

Therefore the correct option is $\left( A \right), 243I.$

Note: A common mistake students make is equating the radius of the bigger sphere to $27$ time the radius of the smaller spheres instead of the volume. Also, practical loss in mass and volume during combination of spheres can be ignored in such questions.