Question

The moment of inertia of a thin uniform rod of mass ' (M) ' and length ' (L) ', about an axis perpendicular to length and at a distance ' (L/4) ' from one end is

• A. (\frac{ML^2}{6})
• B. (\frac{ML^2}{12})
• C. (\frac{7ML^2}{24})
• D. [suspicious link removed]

Hint:

$I = \frac{ML^2}{3}$ about one end and $\frac{ML^2}{12}$ about the center.

Answer 1

The Correct Option is D

Step 1: Parallel Axis Theorem
$I = I_{cm} + Mh^2$.

Step 2: Geometry
$I_{cm} = \frac{ML^2}{12}$.
Distance from end is $L/4$, so distance from center ($h$) is $|L/2 - L/4| = L/4$.

Step 3: Calculation
$I = \frac{ML^2}{12} + M(L/4)^2 = \frac{ML^2}{12} + \frac{ML^2}{16}$.
$I = \frac{4ML^2 + 3ML^2}{48} = \frac{7ML^2}{48}$.
Final Answer: (D)