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A spherical steel ball released at the top of a long column of glycerine of length L, falls through a distance \mathrm{L / 2} with accelerated motion and the remaining distance \mathrm{ L / 2} with a uniform velocity. If \mathrm{ t 1} and \mathrm{ t2} denote the times taken to cover the first and second half and W1 and W2 the work done against gravity in the two halves, then

Option: 1

\mathrm{t_1<t_2 ; W_1>W_2}


Option: 2

\mathrm{t_1>t_2 ; W_1<W_2}


Option: 3

\mathrm{t_1=t_2 ; W_1=W_2}


Option: 4

\mathrm{t_1>t_2 ; W_1=W_2}


Answers (1)

best_answer

The average velocity in the first half of the distance \mathrm{=\frac{0+v}{2}=\frac{v}{2}}; while in the second half, the average velocity is v. Therefore, t_1>t_2. The work done against gravity in both halves \mathrm{=m g h=m g L / 2}.

Hence the correct choice is (d).

Posted by

manish painkra

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