Question

In an experiment to verify Stokes law, a small spherical ball of radius r and density falls under gravity through a distance h in the air before entering a tank of water. If the terminal velocity of the ball inside water is the same as its velocity just before entering the water surface, then the value of h is proportional to (ignore viscosity of air)
1. \[{r^4}\]
2. \[r\]
3. \[{r^3}\]
4. \[{r^2}\]

Answer 1

Hint: When the net force acting on the body is zero then the body moves with constant velocity and this velocity is called the terminal velocity. When a body is moving inside the fluid then the fluid offers viscous resistive force which tries to resist the motion of the body. At terminal velocity the resultant of the resistive force and the weight of the body is zero.



Formula used:
\[v = \sqrt {{u^2} + 2gh} \], here v is the final velocity of the body moving under the influence of gravity with initial velocity u and after travelling the height h.



Complete answer:
As the viscosity of the air is ignored, so when the ball is in air then it moves with constant acceleration under the influence of gravitational force due to earth. So, we can use the equation of motion to find the velocity of the ball just before entering the water and after travelling the height h in the air.
When a body falls, then the initial velocity is assumed to be zero,
\[u = 0m/s\]
So, after travelling the height h in air, the velocity of the ball will be obtained using the equation of motion,
\[v = \sqrt {{0^2} + 2gh} \]
\[v = \sqrt {2gh} \]
Hence, the velocity of the ball at the surface of water is \[\sqrt {2gh} \].
It is given that the terminal velocity of the spherical ball inside the water is same as the velocity of the ball at the surface of the water.
\[{v_T} = v\]
\[{v_T} = \sqrt {2gh} \]
From Stokes law, we know that the terminal velocity is given as,
\[{v_T} = \dfrac{{2{r^2}\left( {\rho - \sigma } \right)}}{{9\eta }}\], here \[{v_T}\]is the terminal velocity of the spherical ball of radius r and of density \[\rho \]in a fluid of density \[\sigma \]and viscosity \[\eta \]
On equating the expression of the terminal velocity,
\[\sqrt {2gh} = \dfrac{{2{r^2}\left( {\rho - \sigma } \right)}}{{9\eta }}\]
On squaring both the sides, we get
\[2gh = {\left( {\dfrac{{2{r^2}\left( {\rho - \sigma } \right)}}{{9\eta }}} \right)^2}\]
\[h = \dfrac{{4{r^4}}}{{162g}}{\left( {\dfrac{{\rho - \sigma }}{{9\eta }}} \right)^2}\]
\[h = {r^4}\left[ {\dfrac{2}{{81g}}{{\left( {\dfrac{{\rho - \sigma }}{{9\eta }}} \right)}^2}} \right]\]
\[h \propto {r^4}\]
Hence, the height of fall of the spherical ball is proportional to \[{r^4}\]
Therefore, the correct option is (1).



Note: We should assume the ball to be perfectly spherical to use the Stokes law, The viscous force by the fluid acts in the direction opposite to the motion of the body inside the fluid and equal in magnitude with the weight of the body so that the net force acting on the body becomes zero and the terminal velocity is attained.