Question

Four solid spheres, each of diameter \(\sqrt{5}\) cm and mass \(0.5\) kg are placed with their centres at the corners of a square of side \(4\) cm. The M.I. of the system about the diagonal of the square is \(N\times 10^{-4}\) kg-m\({}^{2}\), then \(N\) is

• A. \(9\)
• B. \(7\)
• C. \(5\)
• D. \(3\)

Hint:

Two spheres lie on the diagonal and two are at a perpendicular distance from it. Use the parallel axis theorem for the latter two.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Radius of each sphere: \(r=\dfrac{\sqrt{5}}{2}\) cm, so \(r^2=1.25\) cm\(^2\).
Half the diagonal of the square of side 4 cm is \(d=2\sqrt{2}\) cm, so \(d^2=8\) cm\(^2\).

Step 2: Spheres on the axis
Each of the two spheres on the diagonal has \(I=\dfrac{2}{5}mr^2\).

Step 3: Spheres off the axis
Each of the other two has \(I=\dfrac{2}{5}mr^2+md^2\) by the parallel axis theorem.

Step 4: Total
\(I=4\times\dfrac{2}{5}mr^2+2md^2\)
\(=4\times\dfrac{2}{5}(0.5)(1.25\times10^{-4})+2(0.5)(8\times10^{-4})\)
\(=1\times10^{-4}+8\times10^{-4}=9\times10^{-4}\) kg-m\(^2\)

Final Answer:
N equals 9. \[ \boxed{\text{(A) }9} \]