Question

A copper ball of radius r is moving with a uniform velocity v in the mustard oil and the dragging force acting on the ball is F. The dragging force on the copper ball of radius 2r with uniform velocity 2v in the mustard oil is

• A. F
• B. 2F
• C. 4F
• D. 8F

Hint:

Stokes' law: drag force $F \propto r v$ for a sphere in viscous fluid.

Answer 1

The Correct Option is D

The problem involves understanding the relationship between the radius of a sphere, its velocity, and the drag force experienced when moving through a fluid. The general formula for the drag force \(F_d\) acting on a sphere moving through a fluid is given by Stoke's Law, when the flow is laminar:

\(F_d = 6 \pi \eta r v\)

where:

• \(\eta\) is the dynamic viscosity of the fluid
• \(r\) is the radius of the sphere
• \(v\) is the velocity of the sphere

According to the question, for the initial copper ball:

• The radius is \(r\)
• The velocity is \(v\)
• The drag force is \(F\)

Now consider a new situation:

• The radius of the ball becomes \(2r\)
• The velocity becomes \(2v\)

Substituting these into the formula for drag force:

\(F_d' = 6 \pi \eta (2r) (2v) = 24 \pi \eta r v\)

Notice that \(6 \pi \eta r v = F\) from the initial case. Hence, the new drag force \(F_d'\) will be:

\(F_d' = 4 \times (6 \pi \eta r v) = 4 \times F = 8F\)

Therefore, the dragging force on the copper ball of radius \(2r\) with uniform velocity \(2v\) in the mustard oil is 8F.