Question

A spherical ball of radius $3\times {10}^{-4}m$ and density ${10}^{4}kg/m^3$ falls freely under the gravity through a distance h before entering a tank of water. If after entering the water the velocity of the ball doesn’t change, find h.

$[Viscosityofwater=9.8\times {10}^{-6}N-s/m^{2}andg=9.8m/s^2]$

• A. $\frac{81}{49}\times {10}^{3}m$
• B. $\frac{49}{36}\times {10}^{3}m$
• C. $\frac{36}{25}\times {10}^{3}m$
• D. $\frac{81}{64}\times {10}^{3}m$

Answer 1

Answer: (A)

$\frac{81}{49}\times {10}^{3}m$

The velocity acquired by the ball after falling through a height $h$ through the air is given by,

$u=\sqrt{2gh}$

The terminal velocity is given by,

$v_t=\frac{2}{9}\frac{r^2(\rho -\sigma )g}{\eta }$

According to the question,

$u=v_t$

$\Rightarrow \sqrt{2gh}=\frac{2}{9}\frac{r^2(\rho -\sigma )g}{\eta }$

$\Rightarrow h=\frac{2}{81}\frac{r^4(\rho -\sigma {)}^{2}g}{{\eta }^2}$

$\Rightarrow h=\frac{2}{81}\frac{(3\times {10}^{-4}{)}^4({10}^4-{10}^3{)}^2\times 9.8}{(9.8\times {10}^{-6}{)}^2}=\frac{81}{49}\times {10}^{3}m$.