1. Given data

• Number of spheres $=4$
• Mass of each sphere $m=0.5\,\text{kg}$
• Diameter of each sphere $=\sqrt{5}\,\text{cm}$
• Hence radius of each sphere, $r=\frac{\sqrt{5}}{2}\,\text{cm}$
• Side of square $a=4\,\text{cm}$

We need the moment of inertia of the system about a diagonal of the square.

2. Identify distances of sphere centers from the axis

A diagonal of the square passes through the centers of two opposite spheres.

So:

• For 2 spheres lying on the diagonal, distance from axis $=0$
• For the other 2 spheres, perpendicular distance from the diagonal is the distance from a corner to the other diagonal.

Let us compute that distance.

Take the square corners as $(0,0)$, $(4,0)$, $(4,4)$, $(0,4)$ cm. One diagonal is $y=x$.

Distance of point $(4,0)$ from line $y=x$ is $d=\frac{|4-0|}{\sqrt{2}}=\frac{4}{\sqrt{2}}=2\sqrt{2}\,\text{cm}$

Thus for the other two spheres, $d=2\sqrt{2}\,\text{cm}$

3. Moment of inertia of one solid sphere about the given axis

For a solid sphere, moment of inertia about any diameter through its center is $I_{\text{cm}}=\frac{2}{5}mr^2$

If the axis does not pass through the center, use parallel axis theorem: $I=I_{\text{cm}}+md^2$

4. Contribution of spheres on the diagonal

For each of these 2 spheres, $d=0$, so $I_1=\frac{2}{5}mr^2$

Total for 2 such spheres: $I_{\text{diag spheres}}=2\cdot \frac{2}{5}mr^2$

Now, $r^2=\left(\frac{\sqrt{5}}{2}\right)^2=\frac{5}{4}\,\text{cm}^2$

So for one sphere, $\frac{2}{5}mr^2=\frac{2}{5}\cdot 0.5\cdot \frac{5}{4}=0.25\,\text{kg-cm}^2$

Hence for two spheres, $I_{\text{diag spheres}}=2(0.25)=0.5\,\text{kg-cm}^2$

5. Contribution of the other two spheres

For each of these spheres, $I_2=\frac{2}{5}mr^2+md^2$

We already have $\frac{2}{5}mr^2=0.25\,\text{kg-cm}^2$

Also, $d^2=(2\sqrt{2})^2=8\,\text{cm}^2$ $md^2=0.5\times 8=4\,\text{kg-cm}^2$

Therefore for one such sphere, $I_2=0.25+4=4.25\,\text{kg-cm}^2$

For two spheres, $I_{\text{off-diag spheres}}=2(4.25)=8.5\,\text{kg-cm}^2$

6. Total moment of inertia

$I_{\text{total}}=0.5+8.5=9\,\text{kg-cm}^2$

Convert to SI units: $1\,\text{kg-cm}^2 = 10^{-4}\,\text{kg-m}^2$

Hence, $I_{\text{total}}=9\times 10^{-4}\,\text{kg-m}^2$

So in the form $N\times 10^{-4}\,\text{kg-m}^2$, $N=9$

7. Comparison with stored answer

Stored correct answer: $9$

Our derived answer also gives $9$, so they agree.