A steel ball of radius 0.05cm and density $7.8 g c m^{- 3}$ is dropped into a tank of water. The terminal velocity of the steel ball is (Density of water = $1 g c m^{- 3}$ and viscosity of water=0.001 Pa-s)

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$3.42 m s^{- 1}$

$1.81 m s^{- 1}$

$5.11 m s^{- 1}$

$3.77 m s^{- 1}$

answer is D.

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Detailed Solution

Given r = 0.05cm $= 5 \times 10^{- 4} m$
Density of ball $ρ = 7.8 g c m^{- 3} = 7.8 \times 10^{3} k g / m^{3}$
Density of water $σ = 1 g c m^{- 3} = 10^{3} k g m^{- 3}$
Coefficient of viscosity, $η = 0.001 P a - s = 10^{- 3} P a - s$
$∴$ terminal velocity if ball
$v = \frac{2 g r^{2} (ρ - σ)}{9 η} = \frac{2 \times 10 \times {(5 \times 10^{- 4})}^{2} (7.8 \times 10^{3} - 10^{3})}{9 \times 10^{- 3}} = 3.77 m s^{- 1}$