A spherical steel ball released at the top of a long column of glycerine of length l, falls through a distance l/2 with accelerated motion and the remaining distance l/2 with uniform velocity. Lett1 and t2 denote the times taken to cover the first and second half and W1 andW2 the work done against viscous force in the two halves, then
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
a
t1W2
b
t1 > t2, W1
c
t1 = t2, W1 = W2
d
t1 >t2,W1=W2
answer is B.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
initially as it starts from rest and moves with decreasing acceleration it takes more time for the first half but work done against viscous force will be maximum in the second half as it cross the distance with maximum velocity and viscous force is proportional to velocity.