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A small steel ball falls through a syrup at a constant speed of $10 {cms}^{- 1}$ . If the steel ball is pulled upwards with a force equal to twice its effective weight, find the terminal velocity of the ball (in cm/s) in this case.

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Detailed Solution

case(i):

$B + 6 π η rv = mg - - - - - (i)$

Case (ii) :

$F + B = mg + 6 π η r v_{t}$

$2 mg - 2 B + B = mg + 6 π η r v_{t}$

$mg - B = 6 π η {rv}_{t} . . . . . . . . . (i i)$

From eqn (i) & (ii)

$v_{t} = v = 10 cm / s$

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