Question

If ball of steel (density $\rho =7.8{cm}^{-3}$) attains a terminal velocity of $10cm{s}^{-1}$ when falling in a tank of water (coefficient of viscosity ${\eta }_{water}=8.5\times {10}^{-4}Pa.s$) then its terminal velocity in glycerin ($\rho =1.2g{cm}^{-3}\eta =13.2Pa.s$) would be nearly:

• A. $1.5\times {10}^{-5}cm{s}^{-1}$
• B. $1.6\times {10}^{-5}cm{s}^{-1}$
• C. $6.25\times {10}^{-4}cm{s}^{-1}$
• D. $6.45\times {10}^{-4}cm{s}^{-1}$

Answer 1

The correct option is C $6.25\times {10}^{-4}cm{s}^{-1}$

Terminal velocity of ball when falling in liquid is given by

$({\rho }_b-{\rho }_l)Vg=6\pi \eta rv$...eq(1)

1. case

when ball falling in tank of water of density ${\rho }_{water}=1gc{m}^{-3}$, ${\rho }_{ball}=7.8gc{m}^{-3}$${\eta }_{water}=8.5\times {10}^{-4}Pa.s$ and terminal velocity $v=10cm{s}^{-1}$

Put all the above value in eq(1)

$\Rightarrow (7.8-1)Vg=6\pi \times 8.5\times {10}^{-4}r\times 10$...eq(2)

2. case

when ball falling in tank of glycerin

${\rho }_{ball}=7.8gc{m}^{-3}$${\eta }_{glycerin}=13.2Pa.s$ and terminal velocity $v_1$

put all the value in eq(1)

$\Rightarrow (7.8-1.2)Vg=6\pi \times 13.2r\times v_1$...eq(3)

divide eq(3) b eq(1)

$\frac{7.8-1.2}{7.8-1}=\frac{13.2\times v_1}{8.5\times {10}^{-4}\times 10}$

$\Rightarrow v_1=\frac{6.6\times {8.510}^{-4}\times 10}{6.8\times 13.2}=6.25\times {10}^{-4}cm{s}^{-1}$

terminal velocity of ball in glycerin is $6.25\times {10}^{-4}cm{s}^{-1}$

Hence C option is correct.