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9 votes
9 votes
A steel ball at 27°C of radius 5cm and density 8000kgm falls

through air of density 1.29kgm until it attains terminal velocity of
25ms. Calculate the coefficient of viscosity of air in this case at
27°C.

User Casimiro
by
2.6k points

1 Answer

27 votes
27 votes

Answer:

eta: 1.76

Step-by-step explanation:

solution:

radius of steel ball(r)=5cm=0.05m

density of ball =8000kgm

terminal velocity(v)=25m/s^2

density of air( d) =1.29 kgm

now

volume of ball(V)=4/3pir^3=1.33×3.14×0.05^3=0.00052 m^3

density of ball= mass of ball/Volume of ball

or, 8000=m/0.00052

or, m=4.16 kg

weight of the ball (W)= mg=4.16×10=41.6 N

viscous force(F)=6 × pi × eta × r × v

=6×3.14×eta×0.05×25

=23.55×eta

To attain the terminal velocity,

Fiscous force=Weight

or, 23.55× eta = 41.6

or, eta = 1.76

whete eta is the coefficient of viscosity.

User Kasiem
by
2.5k points