Question

A spherical ball of radius 0.3 mm falls vertically through water. Find the coefficient of viscosity of water given that terminal velocity acquired by the ball = 9.8×10^-2 m/s,density of water =10^3 kg/m^3 , density of ball = 1.5×10^3 kg/m^3.

Answer 1

Explanation:

Here, r=0.3mm=0.03cm,

Here, r=0.3mm=0.03cm,v=1ms

Here, r=0.3mm=0.03cm,v=1ms −1

Here, r=0.3mm=0.03cm,v=1ms −1 =100cms

Here, r=0.3mm=0.03cm,v=1ms −1 =100cms −1

Here, r=0.3mm=0.03cm,v=1ms −1 =100cms −1 ,

Here, r=0.3mm=0.03cm,v=1ms −1 =100cms −1 ,η=18×10

Here, r=0.3mm=0.03cm,v=1ms −1 =100cms −1 ,η=18×10 −5

Here, r=0.3mm=0.03cm,v=1ms −1 =100cms −1 ,η=18×10 −5 poise.

Here, r=0.3mm=0.03cm,v=1ms −1 =100cms −1 ,η=18×10 −5 poise.According to stokes law, force of viscosity on rain drop is

Here, r=0.3mm=0.03cm,v=1ms −1 =100cms −1 ,η=18×10 −5 poise.According to stokes law, force of viscosity on rain drop isF=6πηrv

Here, r=0.3mm=0.03cm,v=1ms −1 =100cms −1 ,η=18×10 −5 poise.According to stokes law, force of viscosity on rain drop isF=6πηrv=6×3.142×18×10

Here, r=0.3mm=0.03cm,v=1ms −1 =100cms −1 ,η=18×10 −5 poise.According to stokes law, force of viscosity on rain drop isF=6πηrv=6×3.142×18×10 −5

Here, r=0.3mm=0.03cm,v=1ms −1 =100cms −1 ,η=18×10 −5 poise.According to stokes law, force of viscosity on rain drop isF=6πηrv=6×3.142×18×10 −5 ×0.03×100 dyne

Here, r=0.3mm=0.03cm,v=1ms −1 =100cms −1 ,η=18×10 −5 poise.According to stokes law, force of viscosity on rain drop isF=6πηrv=6×3.142×18×10 −5 ×0.03×100 dyne=1.018×10

Here, r=0.3mm=0.03cm,v=1ms −1 =100cms −1 ,η=18×10 −5 poise.According to stokes law, force of viscosity on rain drop isF=6πηrv=6×3.142×18×10 −5 ×0.03×100 dyne=1.018×10 −3

Here, r=0.3mm=0.03cm,v=1ms −1 =100cms −1 ,η=18×10 −5 poise.According to stokes law, force of viscosity on rain drop isF=6πηrv=6×3.142×18×10 −5 ×0.03×100 dyne=1.018×10 −3 dyne.