A metal ball of radius 9 × 1 0 − 4 m and density 1 0 4 kg / m 3 falls freely under gravity through a distance 'h' and erfters a tank of water. Considering that the metal ball has constant ve

Question

A metal ball of radius 9 × 1 0 − 4 m and density 1 0 4 kg / m 3 falls freely under gravity through a distance 'h' and erfters a tank of water. Considering that the metal ball has constant velocity, the value of h is [coefficient of viscosity of water = 8.1 × 1 0 − 4 pa − s , g = 10 m / s 2 density of water = 1 0 3 kg / m 3 ]

TestHub · Accepted Answer

The terminal velocity of the spherical ball in water is given by:

$v_{t} = \frac{2}{9} \frac{r ^{2} ( ρ - σ ) g}{η}$

Substituting the given values:

$v_{t} = \frac{2}{9} \frac{( 9 \times 1 0 ^{- 4} ) ^{2} ( 1 0 ^{4} - 1 0 ^{3} ) \times 10}{8.1 \times 1 0 ^{- 4}}$

$v_{t} = \frac{2}{9} \frac{81 \times 1 0 ^{- 8} \times 9000 \times 10}{8.1 \times 1 0 ^{- 4}}$

$v_{t} = \frac{2}{9} \times \frac{729 \times 1 0 ^{- 4}}{8.1 \times 1 0 ^{- 4}}$

$v_{t} = \frac{2}{9} \times 90 = 20 m/s$

Since the ball falls freely through a height $h$ and enters the water with constant velocity, its velocity just before entering the water must be equal to the terminal velocity.

$v = 2 g h = v_{t}$

$2 \times 10 \times h = 20$

$20 h = 400$

$h = 20 m$

Answer: 20 m

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