Question

Moment of inertia of a thin rod of mass $M$ and length $L$ about an axis passing through its center is $\frac{M{L}^2}{12}$. Its moment of inertia about a parallel axis at a distance of $\frac{L}{4}$ from this axis is given by

• A. $\frac{M{L}^2}{48}$
• B. $\frac{M{L}^3}{48}$
• C. $\frac{M{L}^2}{12}$
• D. $\frac{7M{L}^2}{48}$

Answer 1

The correct option is D $\frac{7M{L}^2}{48}$

By parallel axis theorem

$I_1=I_0+m{d}^2$

$I_1=\frac{M{L}^2}{12}+M(\frac{L}{H}{)}^2=\frac{4M{L}^2+3M{L}^2}{48}=\frac{7M{L}^2}{48}$