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A small iron ball falls through a viscous liquid at a constant speed of 20cm/sec. If the steel ball is pulled upwards with a force equal to twice its effective weight, how will it move upward?

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with constantly increasing speed

with constantly decreasing speed

with the same speed

will be at rest without any movement

answer is C.

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Detailed Solution

For the first case, $B + 6 πηrv = mg$
For the second case, $F + B = mg + 6 {πηrv}_{t}$

$\begin{array}{l} 2 mg - 2 B + B = mg + 6 {πηrv}_{t} \\ \Rightarrow mg - B = 6 {πηrv}_{t} \end{array}$

Solving we get, v= 20 cm/sec

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