Question:
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:
Correct Option: , 3
Solution:
(3) Moment inertia of the rod passing through a point away from the centre of the rod
$I=I g+m \ell^{2}$
$\Rightarrow I=\frac{M I^{2}}{12}+M \times\left(\frac{I^{2}}{16}\right)=\frac{7 M I^{2}}{48}$
Using $I=M K^{2}=\frac{7 M I^{2}}{48} \quad(K=$ radius of gyration $)$
$\Rightarrow K=\sqrt{\frac{7}{48}} I$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.