Q.
A solid sphere rests on a horizontal surface. A horizontal impulse is applied at height h from centre. The sphere starts rolling just after the application of impulse. The ratio h/R will be
see full answer
Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!
Real Strategies. Real People. Real Success Stories - Just 1 call away
An Intiative by Sri Chaitanya
a
12
b
25
c
15
d
23
answer is B.
(Unlock A.I Detailed Solution for FREE)
Ready to Test Your Skills?
Check your Performance Today with our Free Mock Test used by Toppers!
Take Free Test
Detailed Solution
Rolling is rotation about point of contact. Applying impulse momentum equation about P.JR + h = IPω and J = mvAs sphere roll v = ωR, and I=25mR2+mR2=75mR2After solving, we get hR=25