Question

A spherical ball of diameter 1cm and density $5 \times {10^3}km.{m^{ - 3}}$ us dropped gently in a large tank containing viscous liquid of density $3 \times {10^3}kg.{m^{ - 3}}$ and coefficient of viscosity $0.1{\text{ }}Ns{\text{ }}{m^{ - 2}}$. The distance the ball moves in $1s$after attaining terminal velocity is ($g = 10m{s^{ - 2}}$)
A. $\dfrac{{10}}{9}m$
B. $\dfrac{2}{3}m$
C. $\dfrac{4}{9}m$
D. $\dfrac{4}{5}m$
E. $\dfrac{9}{{10}}m$

Answer 1

We should know the concept of viscosity, terminal velocity…
First we need to find terminal velocity since we know that for $1s$ the ball is going at terminal velocity and wee$=$ need to find the distance travelled when it moves in terminal velocity.

Complete step by step answer:
Terminal velocity is the maximum velocity attained by an object as it falls through a fluid.
The viscosity of a fluid is a measure of its resistance to deformation at a given rate. For liquids, it corresponds to the informal concept of thickness.

Given, diameter or spherical ball $\left( d \right)$
$d = 1cm{\text{ }} = 0.01m$.
Radius of spherical ball $\left( r \right) =$
Density of spherical ball($\rho$) $=$ $5 \times {10^3}km.{m^{ - 3}}$
Density of liquid ($\sigma$) = $$$$3 \times {10^3}kg.{m^{ - 3}}
Coefficient of viscosity ($\eta$) = $$$$0.1{\text{ }}Ns{\text{ }}{m^{ - 2}}

We know that ,

$$v = \dfrac{2}{9}\;\dfrac{{{\text{g}}{{\text{r}}^2}\left( {\rho - \sigma } \right)}}{\eta } \\\ v = \dfrac{{2 \times 10 \times {{\left( {0.01} \right)}^2}\left( {5 \times {{10}^3} - 3 \times {{10}^3}} \right)}}{{9 \times {{\left( 2 \right)}^2} \times 0.1}} \\\$$ $$v = \dfrac{{10}}{9}m{s^{ - 1}} \\\$$

Now, S = vt

$$S = \dfrac{{10}}{9} \times 1 \\\ S = \dfrac{{10}}{9}m \\\$$

So, the correct answer is “Option A”.

Note:
We should take care of units and numerical errors.
All $n = units$ should be in the SI system.
Also the terminal velocity is maximum attainable by the object.