Question

A steel ball with radius 0.3 mm is falling with velocity of 2 m/s at a time t, through a tube filled with glycerin, having coefficient of viscosity 0.833 Ns/m^2. Viscous force acting on the steel ball at that time will be
1 point
a) 8.42 × 10^-3 N
b) 9 × 10^-3 N
c) 10.5 × 10^-3 N
d) 9.42 × 10^-3 N​

Answer 1

Answer :

Radius of ball = 0.3mm

Velocity = 2m/s

Coefficient of viscosity = 0.833

We have to find viscous force acting on the steel ball$.$

★ Formula of viscous force (F) acting on the body of radius r moving with velocity v in liquid of coefficient of viscosity η is given by

• F = 6πηrv

⧪ Conversion :

• 1mm = 0.001m
• 0.3mm = 0.0003m

⧪ Calculation :

➙ F = 6πη × rv

➙ F = 6(3.14)(0.833) × (0.0003)(2)

➙ F = 15.69 × 0.0006

➙ F = 0.0094

➙ F = 9.4 × 10‾³ N

Hence, (D) is the correct answer!

Cheers!

Answer 2

Answer:

Option D is the Answer

There for the force acting on a steel ball is 9.42×10^-3N