Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

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

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya

answer is 1.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

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
score_test_img

courses

No courses found

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon
personalised 1:1 online tutoring