Question

Two spheres each of mass $M$ and radius $R$ are connected with a massless rod of length $4 R$ . The moment of inertia of the system about an axis passing through the centre of one of the spheres and perpendicular to the rod will be

• A. $\frac{21}{5}\text{MR}^2$
• B. $\frac{84}{5}\text{MR}^2$
• C. $\frac{42}{5}\text{MR}^2$
• D. $\frac{5}{21}\text{MR}^2$

Hint:

Don't forget the Parallel Axis Theorem for the second sphere! It's not just a point mass; it's a rotating sphere.

Answer 1

The Correct Option is B

Step 1: Concept
Total Moment of Inertia $I = I_1 + I_2$. Use Parallel Axis Theorem: $I = I_c + Mh^2$.

Step 2: Meaning
Axis passes through center of sphere 1, so $I_1 = \frac{2}{5}MR^2$. For sphere 2, the center is at distance $d = R + 4R + R = 6R$ from the axis? Wait, length of rod is $4R$, so distance between centers is $h = 4R$? If centers are $4R$ apart: $I_2 = \frac{2}{5}MR^2 + M(4R)^2$.

Step 3: Analysis
$I = \frac{2}{5}MR^2 + \frac{2}{5}MR^2 + 16MR^2 = \frac{4}{5}MR^2 + \frac{80}{5}MR^2 = \frac{84}{5}MR^2$.

Step 4: Conclusion
The moment of inertia is $\frac{84}{5}MR^2$. Final Answer: (B)