Fine particles of sand are shaken up in water contained in a tell cylinder. If the depth of water in the cylinder is 24 cm, calculate the size of the largest particle of sand that can remain suspended after the expiry of 40 minutes. Given density of sand =`2.6 g cm^(-3)` and viscosity of water = 0.01 poise. Assume that all the particles are spherical and are of different sizes.

Answer

`5.36`

Explanation

Largest particle which remains suspended is that which just covers 24 cm in 40 minutes . So terminal velocity of largest particle is
` v = ("Distance")/("Time") = (24 cm)/( 40 xx 60 s) = 0.01 cms^(-1)`
Also `eta 0 0.01 ` poise, ` rho = 2 . 6 g cm ^(-3) , rho . = 1 g cm^(-3), g = 980 cm s^(-2)`
As ` v = (2)/( 9) ( r^(2))/( eta) ( rho - rho.) g or r^(2) = (9)/( 2) ( eta v)/( ( rho - rho.) g)`
`:. r = sqrt(( 9 eta v)/(2 ( rho - rho.)g)) = sqrt(( 9 xx 0.01 xx 0.01)/( 2 xx ( 2 . 6 - 1) xx 980))`
`= (3 xx 0.01 )/( 2 xx 4 xx 7) = 5 . 357 xx 10 ^(-4) ` cm