Question

A steel shot of diameter $ 2mm $ is dropped in a viscous liquid filled in a drum. Find the terminal speed of the shot. The density of the material of the shot $ = 8.0 \times {10^3}kg{m^{ - 3}} $ , density of the liquid $ = 1.0 \times {10^3}kg{m^{ - 3}} $ . Coefficient of viscosity of the liquid $ = 1.0kg{m^{ - 1}}{s^{ - 1}} $ , $ g = 10m{s^{ - 2}} $ .

Answer 1

Hint: To solve this question let us first define terminal velocity, which is the maximum velocity attained by an object when it is dropped into a column of fluid, be it liquid or gas. It depends on the density of the object dropped and the fluid on which it is dropped, dimensions of the object, the viscosity of the fluid, and the acceleration due to gravity acting on the body.

Formula used:
 $ {v_t} = \dfrac{2}{9}\dfrac{{({\rho _o} - {\rho _f})}}{\eta }g{R^2} $
where, $ {v_t} $ is the terminal velocity of the shot,
 $ {\rho _o} $ is the density of the object dropped,
 $ {\rho _f} $ is the density of the fluid,
 $ g $ is the acceleration due to gravity, and
 $ R $ is the radius of the shot.

Complete step by step answer:
From the question, we have the values given as,
The diameter of the shot $ = 2mm = 2 \times {10^{ - 3}}m $
Therefore, the radius of the shot, $ R = {10^{ - 3}}m $
The density of the material of the shot, $ {\rho _o} = 8.0 \times {10^3}kg{m^{ - 3}} $
The density of fluid, $ {\rho _f} = 1.0 \times {10^3}kg{m^{ - 3}} $
Coefficient of viscosity of fluid, $ \eta = 1.0kg{m^{ - 1}}{s^{ - 1}} $
Acceleration due to gravity, $ g = 10m{s^{ - 2}} $ .
We know that the formula to calculate the terminal velocity is,
 $ {v_t} = \dfrac{2}{9}\dfrac{{({\rho _o} - {\rho _f})}}{\eta }g{R^2} $
Substituting the given values in the above equation we get,
 $ {v_t} = \dfrac{2}{9}\dfrac{{\{ (8.0 \times {{10}^3}) - (1.0 \times {{10}^3})\} }}{{1.0}} \times 10 \times {({10^{ - 3}})^2} $
 $ \Rightarrow {v_t} = \dfrac{2}{9} \times {10^3}(8 - 1) \times 10 \times {10^{ - 6}} $
Upon solving this equation further we get,
 $ {v_t} = \dfrac{2}{9} \times 7 \times {10^{ - 2}} $
 $ \Rightarrow {v_t} = 1.55 \times {10^{ - 2}}m/s $
We know that $ 1cm/s = {10^{ - 2}}m/s $ .
Therefore can say that the terminal speed of the shot, $ {v_t} = 1.55cm/s $ .
Hence, the correct answer is $ 1.55cm/s $ .

Note: Terminal velocity is a property of fluid dynamics. It is applicable both in the case of liquids and gases. Diving into the atmosphere using a parachute, raindrops falling from the clouds, falling of tiny droplets of mist, etc. are some of the everyday life examples of the application of terminal velocity in gases.