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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 $L / 2$ with accelerated motion and the remaining distance $L / 2$ with a uniform velocity. If $t_{1}$ and $t_{2}$ denote the times taken to cover the first and second half and $W_{1}$ and $W_{2}$ the work done against gravity in the two halves, then

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$t_{1} < t_{2}; W_{1} > W_{2}$

$t_{1} > t_{2}; W_{1} < W_{2}$

$t_{1} = t_{2}; W_{1} = W_{2}$

$t_{1} > t_{2}; W_{1} = W_{2}$

answer is D.

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Detailed Solution

The average velocity in the first half of the distance $= \frac{0 + v}{2} = \frac{v}{2}$ ; while in the second half, the average velocity is $v$ . Therefore, $t_{1} > t_{2}$ . The work done against gravity in both halves $= m g h = m g L / 2$ .

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