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Question: A drop of liquid having radius 2mm has a terminal velocity \[20\,{\text{cm}}\,{{\text{s}}^{ - 1}}\],...

A drop of liquid having radius 2mm has a terminal velocity 20 cm s−120\,{\text{cm}}\,{{\text{s}}^{ - 1}}, the terminal velocity of the drop 1 mm radius will be:
A. 40 cm s−140\,{\text{cm}}\,{{\text{s}}^{ - 1}}
B. 20 cm s−120\,{\text{cm}}\,{{\text{s}}^{ - 1}}
C. 10 cm s−110\,{\text{cm}}\,{{\text{s}}^{ - 1}}
D. 5 cm s−15\,{\text{cm}}\,{{\text{s}}^{ - 1}}

Explanation

Solution

Recall the expression for the terminal velocity. The terminal velocity is proportional to the square of the radius of the drop. Solve the two equations by keeping the other quantities the same.

Formula used:
Terminal velocity, vt=2r2(ρ−σ)g9η{v_t} = \dfrac{{2{r^2}\left( {\rho - \sigma } \right)g}}{{9\eta }}
Here, r is the radius of the drop, ρ\rho is the density of the drop, σ\sigma is the density of the medium through which the drop is falling, g is the acceleration due to gravity and η\eta is the coefficient of viscosity.

Complete step by step answer:
We have given that the terminal velocity of the first drop is v=20 cm s−1v = 20\,{\text{cm}}\,{{\text{s}}^{ - 1}}. The radius of the second drop is given r2=1 mm{r_2} = 1\,{\text{mm}}.We have the expression for the terminal velocity,
vt=2r2(ρ−σ)g9η{v_t} = \dfrac{{2{r^2}\left( {\rho - \sigma } \right)g}}{{9\eta }}
Here, r is the radius of the drop, ρ\rho is the density of the drop, σ\sigma is the density of the medium through which the drop is falling, g is the acceleration due to gravity and η\eta is the coefficient of viscosity.

Let’ express the terminal velocity for the drop of radius r1=2 mm{r_1} = 2\,{\text{mm}} as follows,
vt=2r12(ρ−σ)g9η{v_t} = \dfrac{{2r_1^2\left( {\rho - \sigma } \right)g}}{{9\eta }}
Substituting r1=2 mm{r_1} = 2\,{\text{mm}} in the above equation, we get,
vt=2(2)2(ρ−σ)g9η{v_t} = \dfrac{{2{{\left( 2 \right)}^2}\left( {\rho - \sigma } \right)g}}{{9\eta }}
⇒vt=4(2(ρ−σ)g9η)\Rightarrow {v_t} = 4\left( {\dfrac{{2\left( {\rho - \sigma } \right)g}}{{9\eta }}} \right) …… (1)
Let’ express the terminal velocity for the drop of radius r1=2 mm{r_1} = 2\,{\text{mm}} as follows,
vt′=2r12(ρ−σ)g9η{v'_t} = \dfrac{{2r_1^2\left( {\rho - \sigma } \right)g}}{{9\eta }}
Substituting r1=2 mm{r_1} = 2\,{\text{mm}} in the above equation, we get,
vt′=2(1)2(ρ−σ)g9η{v'_t} = \dfrac{{2{{\left( 1 \right)}^2}\left( {\rho - \sigma } \right)g}}{{9\eta }}
⇒vt′=2(ρ−σ)g9η\Rightarrow {v'_t} = \dfrac{{2\left( {\rho - \sigma } \right)g}}{{9\eta }} …… (2)

Using equation (1) in the above equation, we get,
vt′=vt4{v'_t} = \dfrac{{{v_t}}}{4}
Substituting 20 cm/s for vt{v_t} in the above equation, we get,
vt′=204{v'_t} = \dfrac{{20}}{4}
∴vt′=5 cm s−1\therefore {v'_t} = 5\,{\text{cm}}\,{{\text{s}}^{ - 1}}
Therefore, the terminal velocity of the drop of radius 1 mm will be 5 cm s−15\,{\text{cm}}\,{{\text{s}}^{ - 1}}.

So, the correct answer is option D.

Note: The terminal velocity of the liquid drop is the velocity at which the viscous force and buoyant force is balanced by the gravitational force. When the object sinks with terminal velocity, the net force acting on the object becomes zero. In the solution, we have taken the density of the drop and density of the medium the same since the same drop of smaller size descends in the air medium in both situations.