Four solid spheres each of diameter $\sqrt{5} \mathrm{~cm}$ and mass $0.5 \mathrm{~kg}$ ar placed with their centres at the corners of a square of side $4 \mathrm{~cm}$. The moment of inertia of the system about the diagonal of the square is $N \times 10^{-4} \mathrm{~kg}-\mathrm{m}^2$, then $N$ is

Solution

$$ \begin{aligned} r & =\frac{d}{2}=\frac{\sqrt{5}}{2} \mathrm{~cm} \\ & =\frac{\sqrt{5}}{2} \times 10^{-2} \mathrm{~m} \\ m & =0.5 \mathrm{~kg} \\ a & =4 \mathrm{~cm} \\ & =4 \times 10^{-2} \mathrm{~m} \\ I_{X X} & =I_1+I_2+I_3+I_4 \end{aligned} $$
$$ \begin{aligned} = & {\left[\frac{2}{5} m r^2+m\left(\frac{a}{\sqrt{2}}\right)^2\right]+\frac{2}{5} m r^2 } \\ & +\left[\frac{2}{5} m r^2+m\left(\frac{a}{\sqrt{2}}\right)^2\right]+\frac{2}{5} m r^2 \end{aligned} $$ Substituting the values, we get $$ \begin{aligned} & I_{X X} & =9 \times 10^{-4} \mathrm{Kgm}^{-2} \\ \therefore \quad & N & =9 \end{aligned} $$ Answer is 9. Analysis of Question (i) Question is simple. (ii) Only theorem of parallel axes is to be used properly.

Hence, the correct answer is 9.