Question

A spherical ball of radius $1 \times {10^{ - 4}}\;{\rm{m}}$ and density ${10^4}\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}$ falls freely under gravity through a distance h before entering a tank of water. If after entering the water the velocity of the ball does not change, find h in m (Write answer to nearest integer). (The viscosity of water is $9.8 \times {10^{ - 6}}\;{\rm{N - s/}}{{\rm{m}}^{\rm{2}}}$ )

Answer 1

The above problem can be resolved using the mathematical formula for the terminal velocity, along with the basic concepts of free-falling motion. When any object is dropped from a certain height, it is made to achieve motion under free fall. And if the object is landed to the medium containing some amount of liquid, then the object's velocity with which further projections happens will be the terminal velocity.

Complete step by step solution
Given:
The radius of the ball is, $r = 1 \times {10^{ - 4}}\;{\rm{m}}$.
The density of the ball is, $\rho = {10^4}\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}$.
The viscosity of water is, $\eta = 9.8 \times {10^{ - 6}}\;{\rm{N - s/}}{{\rm{m}}^{\rm{2}}}$.
The ball dropped with some magnitude of linear velocity and undergoing free fall, such that after entering into water, there will be no variation in the magnitude of velocity of the ball. Thus, one can say that the ball has achieved the magnitude of terminal velocity.
The expression for the terminal velocity of the ball is,
${v_c} = \dfrac{{2{r^2}\left( {\rho - {\rho _0}} \right)g}}{{9\eta }}$
Here, g is the gravitational acceleration and its value is $9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$ and ${\rho _0}$ is the density of water and its value is $1000\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}$.
Solve by substituting the values in above equation as,

{v_c} = \dfrac{{2{r^2}\left( {\rho - {\rho _0}} \right)g}}{{9\eta }}\\\ {v_c} = \dfrac{{2 \times {{\left( {1 \times {{10}^{ - 4}}\;{\rm{m}}} \right)}^2}\left( {{{10}^4}\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}} - {{10}^3}\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}} \right) \times 9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}}}{{9 \times 9.8 \times {{10}^{ - 6}}\;{\rm{N - s/}}{{\rm{m}}^{\rm{2}}}}}\\\ {v_c} = 20\;{\rm{m/s}} \end{array}$$ The velocity achieved by the ball after reaching the height h is given as, $$h = \dfrac{{v_c^2}}{{2g}}$$ Substitute the numerical values in above equation as, $$\begin{array}{l} h = \dfrac{{v_c^2}}{{2g}}\\\ \Rightarrow h = \dfrac{{{{\left( {20\;{\rm{m/s}}} \right)}^2}}}{{2 \times 9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}}}\\\ \Rightarrow h = 20.4\;{\rm{m}} \end{array}$$ **Therefore, the value of h in metres is 20.4 metres.** **Note:** To solve the given problem, one must remember the basic concept behind the terminal velocity and the fundamentals using the free-fall conditions. The terminal velocity of the ball is related to the radius of the object and the difference in the medium's density and the density of the surrounding. Moreover, the terminal velocity is also related to the viscosity, as on increasing the viscosity, the terminal velocity reduces.