Four solid spheres each of diameter $\sqrt{5} \mathrm{~cm}$ and mass $0.5 \mathrm{~kg}$ ar placed with their…
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
