The radius of gyration of solid hemisphere of mass M and radius R about an axis parallel to the diameter at a distance 3R / 4 from center and lying in the same plane is given by (center of mass of the hemisphere of the hemisphere lies at a height 3R / 8 from the base)

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

$\frac{3 R}{\sqrt{10}}$

$\frac{5 R}{4}$

$\sqrt{\frac{2}{5}} R$

$\frac{5 R}{8}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$I_{XX'} = \frac{2}{5} {MR}^{2}$

The axes PQ and XX' are at an equal distance from center of mass of the hemisphere

$\Rightarrow I_{PQ} = I_{XX'}$

${MK}^{2} = \frac{2}{5} {MR}^{2} \Rightarrow K = \sqrt{\frac{2}{5}} R$

Watch 3-min video & get full concept clarity

No courses found

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test