Question

The terminal velocity \((v_t)\) of the spherical rain drop depends on the radius (r) of the spherical rain drop as

• A. \(r^{\frac 12}\)
• B. \(r\)
• C. \(r^2\)
• D. \(r^3\)

Answer 1

The Correct Option is C

 To determine how the terminal velocity \((v_t)\) of a spherical raindrop depends on its radius \(r\), we need to understand the concept of terminal velocity and the forces acting on the raindrop.

**Explanation:**

Terminal velocity is the constant speed that a freely falling object eventually reaches when the resistance of the medium through which it is falling prevents further acceleration. For a raindrop, this involves the balance between gravitational force and drag force.

The gravitational force on a raindrop is given by:

\(F_g = mg = \rho V g = \rho \left(\frac{4}{3}\pi r^3\right)g\)

where:

• \(\rho\) is the density of water,
• \(V\) is the volume of the spherical drop,
• \(g\) is the acceleration due to gravity.

The drag force \(F_d\) acting on the raindrop in a fluid is given by Stoke's Law (for small, spherical objects and low Reynolds numbers):

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

where:

• \(\eta\) is the dynamic viscosity of air,
• \(v_t\) is the terminal velocity.

At terminal velocity, these two forces are equal:

\(\rho \left(\frac{4}{3}\pi r^3\right)g = 6 \pi \eta r v_t\)

Solving for \(v_t\), we get:

\(v_t = \frac{\rho g}{6 \eta} \cdot \frac{4}{3} \pi r^2\)

Thus, the terminal velocity \(v_t\) is directly proportional to \(r^2\).

Therefore, the correct answer is:

\(r^2\)

This reasoning shows why the terminal velocity of the raindrop depends on the square of its radius.