Question

What is the velocity v of a metallic ball of radius r falling in a tank of liquid at the instant when its acceleration is one-half that of a freely falling body? (The densities of metal and of liquid are \[\rho \] and \[\sigma \] respectively, and the viscosity of the liquid is \[\eta \])
A. \[\dfrac{{{r^2}g}}{{9\eta }}\left( {\rho - 2\sigma } \right) \\ \]
B. \[\dfrac{{{r^2}g}}{{9\eta }}\left( {2\rho - \sigma } \right) \\ \]
C. \[\dfrac{{{r^2}g}}{{9\eta }}\left( {\rho - \sigma } \right) \\ \]
D. \[\dfrac{{2{r^2}g}}{{9\eta }}\left( {\rho - \sigma } \right)\]

Answer 1

Hint:Here the force acting on the metallic ball are the viscous force and buoyant force and gravity. The gravitational force is the gravity acting on the ball and the viscous force is the frictional force between the ball and the fluid. The buoyant force is the force due to the density difference between the ball and the fluid.

Formula Used:
The force acting on the ball is,
\[mg - {F_v} - {F_b} = ma\]
Where, m is mass of the metallic ball, g is acceleration due to gravity, a is acceleration, \[{F_v}\] is viscous force and \[{F_b}\] is buoyant force.

Complete step by step solution:
Consider a metallic ball of radius r which is falling in a tank of liquid. At that instant, its acceleration is one-half that of a freely falling body. Then we need to find the velocity of the metallic ball. (Given the densities of metal and of liquid are \[\rho \] and \[\sigma \] respectively, and the viscosity of the liquid is \[\eta \]).

Then the forces which are acting on the ball are in the Upward direction, the viscous force, and Buoyant force, and in the downward direction, weight due to gravity is acting on the metallic ball (mg). We know that,
\[\rho = \dfrac{m}{V} \\ \]
Then,
\[m = \rho V\]
\[\Rightarrow m = \rho \dfrac{4}{3}\pi {r^3} \\ \]
The viscous force is,
\[{F_v} = 6\pi \eta rV\]
And Buoyant force is,
\[{F_b} = \sigma \dfrac{4}{3}\pi {r^3}\]

The force acting on the ball is,
\[mg - {F_v} - {F_b} = ma\]
Substitute the value of mass, viscous force and buoyant force in above equation, we obtain,
\[\rho \dfrac{4}{3}\pi {r^3}g - 6\pi \eta rV - \sigma \dfrac{4}{3}\pi {r^3}g = \rho \dfrac{4}{3}\pi {r^3}\dfrac{g}{2} \\ \]
\[\Rightarrow - 6\pi \eta rV + \left( {\rho - \sigma } \right)\dfrac{4}{3}\pi {r^3}g = \rho \dfrac{4}{3}\pi {r^3}\dfrac{g}{2} \\ \]
\[\Rightarrow 6\pi \eta rV = \left( {\rho - \sigma } \right)\dfrac{4}{3}\pi {r^3}g - \rho \dfrac{4}{3}\pi {r^3}\dfrac{g}{2} \\ \]
\[\Rightarrow 6\eta V = \left( {\dfrac{\rho }{2} - \sigma } \right)\dfrac{4}{3}{r^2}g \\ \]
\[\Rightarrow 3\eta V = \left( {\dfrac{{\rho - 2\sigma }}{2}} \right)\dfrac{2}{3}{r^2}g \\ \]
\[\Rightarrow V = \dfrac{1}{{9\eta }}\left( {\rho - 2\sigma } \right){r^2}g \\ \]
\[\therefore V = \dfrac{{{r^2}g}}{{9\eta }}\left( {\rho - 2\sigma } \right)\]
Therefore, the velocity of a metallic ball is \[\dfrac{{{r^2}g}}{{9\eta }}\left( {\rho - 2\sigma } \right)\].

Hence, option A is the correct answer.

Note:Here, one should notice that the Archimedes principle is only valid for fluids, where we observe the buoyant force. And, the displaced fluid is equal to the weight of the object immersed in the water. Here, gravity also plays an important role.