Question

The radius of gyration of a uniform rod of length $l$, about an axis passing through a point $\frac{1}{4}$ away from the centre of the rod, and perpendicular to it, is

• A. $\frac{1}{8}l$
• B. $\sqrt{\frac{3}{8}l}$
• C. $\sqrt{\frac{7}{48}l}$
• D. $\frac{1}{4}l$

Answer 1

The correct option is C $\sqrt{\frac{7}{48}l}$

As we know that moment of inertia of the rod about its centre of mas,

$I_{cm}=\frac{M{l}^2}{12}$

Using parallel axis theorem, Moment of inertia of the rod about an axis passing through a point $\frac{1}{4}$ away from centre of the rod,

$I=\frac{M{l}^2}{12}+M\times {\left(\frac{1}{4}\right)}^2$

$I=\frac{7M{l}^2}{48}$

If $K$ is radius of gyration, then

$M{K}^2=\frac{7M{l}^2}{48}$

$\Rightarrow K=\sqrt{\frac{7}{48}}l$
Final answer is (b)