A steel ball of diameter 2mm falls at constant rate 0f 10cm in 3.4secs through oil in a relatively large jar. calculate the coefficient of viscosity assuming the density of the oil and ball to be 920 and 7700kgm^3 respectively?

1 Answer
Mar 1, 2018

The coefficient of viscosity is #=0.065kgm^-1s^-1#

Explanation:

The forces acting on the ball are

#"Frictional force acting upwards" = "weight " - "upthrust"#

#6xxpixxetaxxrxxv_0=4/3xxr^3xxgxx(rho-sigma)#

The viscisity is #eta=?#

The radius of the ball is #r=0.002m#

The terminal velocity is

#v_0=d/t=0.1/3.4=0.029ms^-1#

The acceleration due to gravity is

#g=9.8ms^-2#

The density of the ball bearing material is #rho=7700kgm^-3#

The density of the oil is #sigma=920kgm^-3#

Therefore,

#6*pi*eta*0.029=4/3*0.002^2*9.8*(7700-920)#

#5.466eta=0.354#

#eta=0.354/5.466=0.065kgm^-1s^-1#