A spherical ball of radius m and density falls freely under gravity through a distance before entering a tank of water. If after entering in water the velocity of the ball does not change, then the value of is approximately

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Solution

Q.
A spherical ball of radius $1 \times 10^{- 4}$ m and density $10^{5}$ $k g / m^{3}$ falls freely under gravity through a distance $h$ before entering a tank of water. If after entering in water the velocity of the ball does not change, then the value of $h$ is approximately

(The coefficient of viscosity of water is $9.8 \times 10^{- 6}$ $N s / m^{2}$ ) [2024]

1 2296 m

2 2249 m

3 2518 m

4 2396 m

Ans.
(3)

Terminal velocity, $v_{T} = \frac{2 g}{9} \frac{R^{2} [ρ_{B} - ρ_{L}]}{η}$

$\Rightarrow v_{T} = \frac{2}{9} \times \frac{9.8 \times {(10^{- 4})}^{2}}{9.8 \times 10^{- 6}} [10^{5} - 10^{3}] = 220 m / s$

When ball fall from height $(h), v = \sqrt{2 g h}$

Now $h = \frac{v^{2}}{2 g} = \frac{{(220)}^{2}}{2 \times 9.8} \approx 2518 m$

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