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

Hint: 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.