Question

If ball of steel (density $\text{ }\rho\text{ }=7\cdot 8\text{ c}{{\text{m}}^{-3}}$ ) attains a terminal velocity of $10\text{ cm}{{\text{s}}^{-1}}$ when falling in a tank of water (coefficient of viscosity q water $=8\cdot 5\times {{10}^{-6}}\text{ P}$ a.s) then its terminal velocity in glycerin $\left( \text{ }\rho\text{ }=1\cdot 2\text{ gc}{{\text{m}}^{-3}}\text{q}=13\cdot 2\text{ P a}\text{.s} \right)$ would be nearly:
(A) $1\cdot 5\times {{10}^{-3}}\text{ cm}{{\text{s}}^{-1}}$
(B) $1\cdot 6\times {{10}^{-5}}\text{cm}{{\text{s}}^{-1}}$
(C) $6\cdot 25\times {{10}^{-4}}\text{ cm}{{\text{s}}^{-1}}$
(D) $6\cdot 45\times {{10}^{-4}}\text{ cm}{{\text{s}}^{-1}}$

Answer 1

Terminal velocity is a steady speed achieved by an object freely falling through a gas or liquid. An object dropped from rest will increase its speed until it reaches terminal velocity; an object forced to move faster than its terminal velocity will, upon release, slow down to this constant velocity. The terminal velocity of a sphere of radius ‘r’ and density ‘ $\text{ }\rho\text{ }$ ’ immersed in a liquid of density $\text{ }\rho\text{ }$ and viscosity $\text{ }\eta\text{ }$ is given by
$\Rightarrow \text{V}=\left( {{\text{ }\rho\text{ }}_{\text{b}}}-{{\text{ }\rho\text{ }}_{1}} \right)\text{ g}=\text{6 }\pi\text{ }\eta\text{ rv}$

Complete step by step solution
Terminal velocity of ball when falling in liquid is given by
$\Rightarrow \left( {{\text{ }\rho\text{ }}_{\text{b}}}-{{\text{ }\rho\text{ }}_{1}} \right)\text{ Vg}=\text{6 }\pi\text{ }\eta\text{ rv}$
Case 1:
When a ball falls in a tank of water of density ${{\text{ }\rho\text{ }}_{\text{water}}}=1\text{ g c}{{\text{m}}^{-3}}$
$\begin{aligned} &\Rightarrow {{\text{ }\rho\text{ }}_{\text{ball}}}=7\cdot 8\text{ g c}{{\text{m}}^{-3}} \\\ &\Rightarrow {{\text{ }\eta\text{ }}_{\text{water}}}=8\cdot 5\times {{10}^{-4}}\text{ }\rho\text{ } \\\ \end{aligned}$
And terminal velocity $\text{V}=10\text{cm }{{\text{s}}^{-1}}$
Put all the above value in equation (1)
$\Rightarrow\left( 7\cdot 8-1 \right)\text{Vg}=6\text{ }\pi\text{ }\times \text{8}\cdot \text{5}\times \text{1}{{\text{0}}^{-4}}\text{r}\times \text{10}$ …. (2)
Case 2:
When ball falling in tank of glycerin
$\begin{aligned} &\Rightarrow {{\text{ }\rho\text{ }}_{\text{ball}}}=7\cdot 8\text{g c}{{\text{m}}^{-3}} \\\ &\Rightarrow {{\text{ }\eta\text{ }}_{\text{glycerin}}}=13\cdot 2\text{ }\rho\text{ } \\\ \end{aligned}$
And terminal velocity ${{\text{V}}_{1}}$
Put all the value in equation (1)
$\Rightarrow \left( 7\cdot 8-1\cdot 2 \right)\text{Vg}=6\text{ }\pi\text{ }\times \text{13}\cdot \text{2r}\times {{\text{v}}_{1}}$ … (3)
Now divide (3) by equation (2)
$\begin{aligned} &\Rightarrow \dfrac{\left( 7\cdot 8-1\cdot 2 \right)\text{Vg}}{\left( 7\cdot 8-1 \right)\text{Vg}}=\dfrac{13\cdot 2\text{r}\times 6\text{ }\pi\text{ }\times {{\text{V}}_{1}}}{8\cdot 5\times {{10}^{-4}}\text{r}\times 6\text{ }\pi\text{ }\times 10} \\\ &\Rightarrow \dfrac{7\cdot 8-1\cdot 2}{7\cdot 8-1}=\dfrac{13\cdot 2\text{r}\times {{\text{V}}_{1}}}{8\cdot 5\times {{10}^{-4}}\text{r}\times 10} \\\ &\Rightarrow \dfrac{6\cdot 6}{6\cdot 8}\times \dfrac{8\cdot 5\times {{10}^{-3}}}{13\cdot 2}={{\text{V}}_{1}} \\\ &\Rightarrow \dfrac{56\cdot \times {{10}^{-3}}}{89\cdot 76}={{\text{V}}_{1}} \\\ \end{aligned}$
$\begin{aligned} &\Rightarrow {{\text{V}}_{1}}=0\cdot 625\times {{10}^{-3}} \\\ &\Rightarrow \text{ }=6\cdot 25\times {{10}^{-4}}\text{cm }{{\text{s}}^{-1}} \\\ \end{aligned}$
Terminal velocity of ball in glycerin is $6\cdot 25\times {{10}^{-4}}\text{cm }{{\text{s}}^{-1}}$
Hence, option C is correct.

Note
The concept of terminal velocity is important in our daily life. This is the maximum velocity attainable by an object as it falls through a fluid. From terminal velocity we can find viscosity and density of liquid. We cannot fall faster than terminal velocity because the maximum speed we obtain when falling is called terminal velocity.