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 = 4 cm = 4 x 10^{-2} m $
$I_XX =I_1 +I_2 +I_3 +I_4 $
$ \, \, \, \, \, \, = \bigg[ \frac{2}{5}mr^2 +m \bigg( \frac{a}{\sqrt 2}\bigg)^2 \bigg] +\frac{2}{5}mr^2$
$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, +\bigg[ \frac{2}{5}mr^2 +m \bigg( \frac{a}{\sqrt 2}\bigg)^2 \bigg] +\frac{2}{5}mr^2 $
Substituting the values, we get
$ \, \, \, \, \, \, \, \, \, \, I_XX = 9 \times {10}^{ -4} kg -m^2 \, \, \therefore N =9 $
Answer is 9.