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Q.

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

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a

t1W2

b

t1 > t2, W1 

c

t1 = t2, W1 = W2

d

t1 >t2,W1=W2

answer is B.

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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.
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