Question

Four solid spheres each of diameter $\sqrt{5}$ cm and mass 0.5 kg
are placed with their centres at the comers of a square of side
4 cm. The moment of inertia of the system about the diagonal
of the square is $N\times {10}^{-4}kg-m_2$ , then N is.

Answer 1

$r=\frac{d}{2}=\frac{\sqrt{5}}{2}cm=\frac{\sqrt{5}}{2}\times {10}^{-2}m$
m= 0.5 kg
$a=4cm=4x10^{-2}m$
$I_{X}X=I_1+I_2+I_3+I_4$
$=[\frac{2}{5}m{r}^2+m(\frac{a}{\sqrt{2}}{)}^2]+\frac{2}{5}m{r}^2$
$+[\frac{2}{5}m{r}^2+m(\frac{a}{\sqrt{2}}{)}^2]+\frac{2}{5}m{r}^2$
Substituting the values, we get
$I_{X}X=9\times {10}^{-4}kg-m^2\therefore N=9$
Answer is 9.