Question

A metal ball of radius 1mm and density ${10^4}kg.{m^{ - 3}}$ falls freely in air through a height $h$ before falling in a tank full of water. If on falling in the water tank, its velocity remains unchanged, then the value of $h$ will be (coefficient of viscosity of water= ${10^{ - 3}}Pa.s$ and $g = 10m.{s^{ - 2}}$ )
A) 10m
B) 15m
C) 25m
D) 20m.

Answer 1

The above question gives us two conditions. First, when the ball is falling in the air, and second when the ball is falling into the water tank. But the important point to note here is that the velocity of the ball is unchanged. The ball is falling freely in the air from the height $h$ we can calculate the speed of a free-falling object in the air and since the speed is not changed after falling in the water tank then the terminal velocity also can be calculated and will equal to the velocity in the air. From there we can calculate the height from which the ball starts to fall.

Complete step by step solution:
Step 1: First, let us understand what terminal velocity is. Terminal velocity is the maximum velocity attainable by an object as it falls freely in the air or fluid.
Express the formula for the velocity of the free-falling object in the air
$\Rightarrow {v_{air}} = \sqrt {2gh}$ , where $g$ is the acceleration due to gravity.…… equation (1).
Now express the formula for the terminal velocity of a sphere in a fluid according to stock’s law.
$\Rightarrow {v_t} = \dfrac{2}{9}{r^2}\dfrac{{(\rho - \sigma )g}}{\eta }$………equation (2).
Where, $r$ is the radius of the ball, $\rho$ is the density of the material of the ball, $\sigma$ is the density of water and $\eta$ is the coefficient of viscosity of water.
Step 2: Since the velocity of the ball is unchanged therefore we can say that the velocity of the ball in the air is equal to the terminal velocity of the ball. Therefore,
$\Rightarrow {v_{air}} = {v_t}$………. Equation (3).
Step 3: first calculate the terminal velocity of the ball. Put the given values in equation 2. ${10^{ - 3}}m$ for $r$ , ${10^4}kg.{m^{ - 3}}$ for $\rho$ , ${10^3}$ for $\sigma$ , and $g = 10m.{s^{ - 2}}$ .
$\Rightarrow {v_t} = \dfrac{2}{9} \times {({10^{ - 3}})^2}\dfrac{{({{10}^4} - {{10}^3})10}}{{{{10}^{ - 3}}}}$
$\Rightarrow \Rightarrow {v_t} = \dfrac{2}{9} \times 9 \times 10$
$\Rightarrow {v_t} = 20m/s$
Step 4: now put this value in equation 1
$\Rightarrow \sqrt {2gh} = 20$
Squaring on both sides
$\Rightarrow 2gh = 400$
$\Rightarrow gh = 200$
Now put $g = 10m/{s^2}$
$\Rightarrow 10h = 200$
$\Rightarrow h = \dfrac{{200}}{{10}} = 20m$

Hence the correct option is option D.

Note: Here the viscosity of the air is neglected. When the ball enters the water tank, because of the viscosity of the water a force acts upward on the ball called upthrust. Drag is a force acting opposite to the relative motion of any object moving with respect to a surrounding fluid. Terminal velocity occurs when the sum of the upthrust and drag is equal to the downward force of gravity.