Question

In a falling-ball viscometer, the ball attains terminal velocities of 0.01 m/s for oil A and 0.002 m/s for oil B. Assuming the oils have the same density and oil A has kinematic viscosity of $5\times {10}^{-3}{m}^2/s$, the kinematic viscosity of oil B in $m^2/s$ is

• A. $15\times {10}^{-3}$
• B. $20\times {10}^{-3}$
• C. $25\times {10}^{-3}$
• D. $30\times {10}^{-3}$

Answer 1

The correct option is C $25\times {10}^{-3}$

In case of falling sphere viscometer,

$\text{kinematic viscosity (V)}\propto \frac{1}{\text{terminal velocity}\left(V_T\right)}$

$\therefore \frac{V_A}{V_B}=\frac{(V_T{)}_B}{(V_T{)}_A}$

$V_B=\frac{V_A\times (V_T{)}_A}{(V_T{)}_B}=\frac{5\times {10}^3\times 0.01}{0.002}$

$=25\times {10}^{-3}{m}^2/sec$