First slide

Viscosity

Question

A small steel ball falls through a syrup at a constant speed of 10 $cm s^{- 1}$ . If the steel ball is pulled upwards with a force equal to twice its effective weight, how fast will it move upward?

Moderate

A

10 $cm s^{- 1}$
B

20 $cm s^{- 1}$
C

5 $cm s^{- 1}$
D

zero

Solution

First case:
$B + 6 πηrv = mg$ --------(i)
For second case:

$F + B = m g + 6 {πηrv}_{t}$
$\Rightarrow 2 m g - 2 B + B = m g + 6 {πηrv}_{t}$
$mg - B = 6 {πηrv}_{t}$ -----(ii)
From (i) and (ii), $v_{t} = v = 10 cm / s$

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