Two drops of equal radius are falling through | vedclass

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Question

(A) When $n$ drops of radius $r$ combine to form a single drop of radius $R$,the volume remains constant.$\frac{4}{3}\pi R^3 = n \left( \frac{4}{3}\pi

Vedclass · Accepted Answer

${4^{\frac{1}{3}}} 	imes 5\,cm/s$

Vedclass · Answer

(A) When $n$ drops of radius $r$ combine to form a single drop of radius $R$,the volume remains constant.$\frac{4}{3}\pi R^3 = n \left( \frac{4}{3}\pi r^3 ight)$$R^3 = n r^3$Given $n = 2$,we have $R^3 = 2r^3$,which implies $R = 2^{1/3} r$.The terminal velocity $v$ of a drop falling through a viscous medium is given by Stokes' Law,$v = \frac{2r^2(ho - \sigma)g}{9\eta}$,which implies $v \propto r^2$.Therefore,$\frac{v_2}{v_1} = \left( \frac{R}{r} ight)^2$.Given $v_1 = 5\,cm/s$,we substitute $R = 2^{1/3} r$:$v_2 = v_1 \left( \frac{2^{1/3} r}{r} ight)^2 = 5 	imes (2^{1/3})^2 = 5 	imes 2^{2/3} = 5 	imes 4^{1/3}\,cm/s$.

Two drops of equal radius are falling through air with a steady velocity of $5\,cm/s$. If the two drops coalesce,then its terminal velocity will be

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