For a sphere falling in the constant drag co-efficient regime, its terminal velocity depends on its diameter (d) as

Solution (By Examveda Team)

Given: A sphere falling through a fluid in the constant drag-coefficient (Newton’s law) regime, i.e., high Reynolds number where C_D ≈ constant.

Forces acting at terminal velocity v_t:

Gravitational (effective weight) – Buoyancy = Drag

Write each term:

Effective weight = \( \left(\rho_p - \rho_f\right) g \times \text{(volume)} = \left(\rho_p - \rho_f\right) g \left(\frac{\pi}{6} d^3\right) \)

Drag (Newton’s regime) = \( \frac{1}{2} C_D \rho_f A v_t^2 \), where frontal area \( A = \frac{\pi}{4} d^2 \)

Force balance at terminal velocity:

\( \left(\rho_p - \rho_f\right) g \left(\frac{\pi}{6} d^3\right) \;=\; \frac{1}{2} C_D \rho_f \left(\frac{\pi}{4} d^2\right) v_t^2 \)

Simplify (cancel \( \pi \) and powers of \( d \)):

\( \left(\rho_p - \rho_f\right) g \left(\frac{1}{6} d^3\right) \;=\; \frac{1}{8} C_D \rho_f d^2 \, v_t^2 \)

Multiply both sides by \( \frac{8}{C_D \rho_f d^2} \):

\( v_t^2 \;=\; \frac{8}{C_D \rho_f d^2} \cdot \left(\rho_p - \rho_f\right) g \left(\frac{1}{6} d^3\right) \;=\; \frac{4}{3} \cdot \frac{\left(\rho_p - \rho_f\right) g}{C_D \rho_f} \, d \)

Take square root:

\( v_t \;=\; \sqrt{ \frac{4}{3} \cdot \frac{\left(\rho_p - \rho_f\right) g}{C_D \rho_f} } \;\sqrt{d} \)

Conclusion: Since the bracketed factor is constant in the constant-\( C_D \) regime, \( v_t \propto \sqrt{d} \).

Contrast with Stokes (creeping flow) regime: There, drag \( \propto \mu d \, v \) and one gets \( v_t \propto d^2 \). But this question specifies constant \( C_D \) (Newton’s regime), so \( v_t \propto \sqrt{d} \).