The radius of gyration of a uniform rod of length l, about an axis passing through a point l/4 away from the centre of the rod, and perpendicular to it, is: (a) 1/8l (b) 1/4l (c) √(7/48) l (d) √(3/8) l

Answer: (c). I = (ml²)/(12) + m(l/4)² = ml²(4+3/48) = 7/48ml² I = mk² ⇒ k = √(7/48) l So the answer is (c) √(7/48) l.

The radius of gyration of a uniform rod of length l, about an axis… | JEE Main 2020 (7 Jan, Shift 1)

Q143JEE Main 2020 · 7 JAN · SHIFT 1Easy

Moment of Inertia & Radius of Gyration

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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Answer: (c)

$I = \frac{ml^2}{12} + m\left(\frac{l}{4}\right)^2 = ml^2\left(\frac{4+3}{48}\right) = \frac{7}{48}ml^2$
$I = mk^2 \Rightarrow k = \sqrt{\frac{7}{48}}\,l$
So the answer is (c) $\sqrt{\frac{7}{48}}\,l$ .

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