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

Hint: 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{align}
 &\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{align} $
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{align}
 &\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{align} $
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{align}
 &\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{align} $
 $ \begin{align}
 &\Rightarrow {{\text{V}}_{1}}=0\cdot 625\times {{10}^{-3}} \\
 &\Rightarrow \text{ }=6\cdot 25\times {{10}^{-4}}\text{cm }{{\text{s}}^{-1}} \\
\end{align} $
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.