Q.
The height of a solid cylinder is four times its radius, lt is kept vertically at time t = 0 on a belt which is moving in the horizontal direction with a velocity v = 2.45 t2 where v is in ms-1 and t is in second. If the cylinder does not slip, it will topple over at time t (in seconds) equal to…….[ Take g=9.8 ms-2 ]
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
answer is 1.
(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
The cylinder will topple when the torque mgr equals the torque mah2 (see figure)or a=2grh=g2 (∵h=4r) …….(i)Now v=2.45t2∴ a=dvdt=ddt2.45t2=4.9t ……..(ii)Equating (i) and (ii), we get t=g2×4.9=9.89.8=1s
Watch 3-min video & get full concept clarity