Two rain drops of same radius 'r' falling wit | vedclass

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(A) The terminal velocity '$V_t$' of a spherical drop falling through a viscous medium is given by the formula: $V_t = \frac{2r^2(\rho - \sigma)g}{9\e

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$V(R/r)^2$

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(A) The terminal velocity '$V_t$' of a spherical drop falling through a viscous medium is given by the formula: $V_t = \frac{2r^2(ho - \sigma)g}{9\eta}$.From this, we see that $V_t \propto r^2$.Let the terminal velocity of the small drops be $V$ and their radius be $r$. Thus, $V = k r^2$ (where $k$ is a constant).When two drops of radius '$r$' merge to form a bigger drop of radius '$R$',the volume remains conserved:$2 	imes (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$$2r^3 = R^3 \implies R = 2^{1/3}r$.The terminal velocity of the big drop '$V_{big}$' is proportional to the square of its radius '$R^2$':$V_{big} = k R^2$.Taking the ratio of the two velocities:$\frac{V_{big}}{V} = \frac{k R^2}{k r^2} = \frac{R^2}{r^2}$.Therefore, $V_{big} = V \frac{R^2}{r^2}$.

Two rain drops of same radius '$r$' falling with same terminal velocity '$V$' merge and form a bigger drop of radius '$R$'. The terminal velocity of the big drop is

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