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

Answer: Option B

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 $\propto \mu dv$ and one gets $v_t\propto d^2$. But this question specifies constant $C_D$ (Newton’s regime), so $v_t\propto \sqrt{d}$.