A spherical ball of radius 3.0 x 10^-4m and density 10⁴ kg/m³ falls freely under gravity through a distance h before entering a tank of water. After entering the water the velocity of the ball does not change. If h = $\frac{81}{P}$ x 10³m then find value of P. (Viscosity of water is 9.8 x 10^-6 N - s/m²)

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answer is 49.

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Detailed Solution

Before entering the water the velocity of ball is $\sqrt{2 gh} .$ If after entering the water this velocity does not change then this value should be equal to the terminal velocity.
Therefore, $\sqrt{2 gh} = \frac{2}{9} \frac{r^{2} (ρ - σ) g}{η}$
$\begin{array}{l} h = \frac{{\{\frac{2}{9} \frac{r^{2} (ρ - σ) g}{η}\}}^{2}}{2 g} \\ = \frac{2}{81} \times \frac{r^{4} (ρ - σ)^{2} g}{η^{2}} \\ = \frac{2}{81} \times \frac{{(3 \times 10^{- 4})}^{4} {(10^{4} - 10^{3})}^{2} \times 9.8}{{(9.8 \times 10^{- 6})}^{2}} \\ = \frac{81}{49} \times 10^{3} m \end{array}$