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(A) The terminal velocity $V_T$ of a spherical drop of radius $r$ falling through a viscous medium is given by $V_T = \frac{2r^2(\rho - \sigma)g}{9\et

Vedclass · Accepted Answer

$5 	imes (4)^{1/3}$

Vedclass · Answer

(A) The terminal velocity $V_T$ of a spherical drop of radius $r$ falling through a viscous medium is given by $V_T = \frac{2r^2(ho - \sigma)g}{9\eta}$.Thus,$V_T \propto r^2$.Let the radius of each small drop be $r$ and the radius of the new coalesced drop be $R$.Since the volume is conserved,the volume of the new drop is equal to the sum of the volumes of the two small drops:$\frac{4}{3}\pi R^3 = 2 	imes \frac{4}{3}\pi r^3$$R^3 = 2r^3 \Rightarrow R = 2^{1/3}r$.Now,the ratio of the new terminal velocity $V'$ to the initial terminal velocity $V$ is:$\frac{V'}{V} = \frac{R^2}{r^2} = \frac{(2^{1/3}r)^2}{r^2} = (2^{1/3})^2 = 2^{2/3} = (2^2)^{1/3} = 4^{1/3}$.Given $V = 5 \, cm/s$,the new terminal velocity $V'$ is:$V' = 5 	imes 4^{1/3} \, cm/s$.

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

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