Question

Three solid spheres each of mass $m$ and diameter $d$ are stuck together such that the lines connecting the centres form an equilateral triangle of side of length $d$. The ratio $\frac{I_0}{I_A}$ of moment of inertia $I_0$ of the system about an axis passing the centroid and about centre of any of the spheres $I_A$ and perpendicular to the plane of the triangle, is:

• A. $\frac{13}{23}$
• B. $\frac{15}{13}$
• C. $\frac{23}{13}$
• D. $\frac{13}{15}$

Answer 1

The correct option is A $\frac{13}{23}$


Moment of inertia of a sphere, about an axis passing through $O$, is,

$I_1=\frac{2}{5}m{\left(\frac{d}{2}\right)}^2+m(AO{)}^2$

and $AO=\frac{d}{\sqrt{3}}$

Moment of inertia of the system about $O$, is

$I_0=3I_1=3[\frac{2}{5}m{\left(\frac{d}{2}\right)}^2+m{\left(\frac{d}{\sqrt{3}}\right)}^2]$

$\Rightarrow I_0=\frac{13}{10}m{d}^2$

Similarly, Moment of inertia of the system about $A$, is
$I_A=2[\frac{2}{5}m{\left(\frac{d}{2}\right)}^2+m{d}^2]+\frac{2}{5}m{\left(\frac{d}{2}\right)}^2$

$\Rightarrow I_A=\frac{23}{10}m{d}^2$

$\therefore \frac{I_0}{I_A}=\frac{\frac{13}{10}m{d}^2}{\frac{23}{10}m{d}^2}=\frac{13}{23}$

Hence, $\left(A\right)$ is the correct answer.