Q. Typical silt(hard mud) particle of radius $20\mu m$ is on the top of lake water, its density is $2000\ kg\ m^{- 3}$ and the viscosity of lake water is 1.0 mPa, density is $1000\ kg\ m^{- 3}$. If the lake is still(has no internal fluid motion). The terminal speed with which the particle hits the bottom of the lake is ………. ${mms}^{- 1}$
A) 0.67
B) 0.77
C) 0..87
D) 0.97
Explanation
(c) $v = \frac{2}{9}\frac{r^{2}(\rho - \sigma)g}{ɳ}$
$\ \ \ = \frac{2}{9} \times \frac{\left( 20 \times 10^{- 6} \right)^{2}(2000 - 1000) \times 9.8}{1.0 \times 10^{- 3}}$
$= 8.7 \times 10^{- 4}{ms}^{- 1} = 0.87{mm\ s}^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.