Question

The radius of gyration of a uniform rod of length $l$ about an axis passing through a point $l/4$ away from the center of the rod, and perpendicular to it, is

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

Answer 1

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


Moment of inertia of rod about axis perpendicular to it passing through its centre is given by
$$\begin{array}{cc}& I=\frac{M{l}^2}{12}+M{\left(\frac{l}{4}\right)}^2\\ & =\frac{3M{l}^2+4M{l}^2}{48}\\ & =\frac{7M{l}^2}{48}\end{array}$$
Now, comparing with $I=M{k}^2$ where $k$ is the radius of gyration
$$\begin{array}{cc}& k=\sqrt{\frac{7l^2}{48}}\\ & k=l\sqrt{\frac{7}{48}}\end{array}$$