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(B) The moment of inertia of a uniform rod of mass $m$ and length $l$ about an axis passing through its centre and perpendicular to it is $I_{cm} = \f

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$\sqrt{\frac{7}{48}} l$

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(B) The moment of inertia of a uniform rod of mass $m$ and length $l$ about an axis passing through its centre and perpendicular to it is $I_{cm} = \frac{ml^2}{12}$.According to the parallel axis theorem,the moment of inertia about an axis at a distance $d = \frac{l}{4}$ from the centre is $I = I_{cm} + md^2$.Substituting the values,we get $I = \frac{ml^2}{12} + m(\frac{l}{4})^2 = \frac{ml^2}{12} + \frac{ml^2}{16}$.Taking the common denominator,$I = \frac{4ml^2 + 3ml^2}{48} = \frac{7ml^2}{48}$.The radius of gyration $k$ is defined by $I = mk^2$,so $mk^2 = \frac{7ml^2}{48}$.Thus,$k^2 = \frac{7l^2}{48}$,which gives $k = \sqrt{\frac{7}{48}} l$.

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

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