Question

Eight drops of a liquid of density $\rho$ and each radius a are falling through air with a constant velocity $3.75cm{s}^{-1}$. when the eight drops coalesce to from a single drop the terminal velocity of the new drop will be

• A. $2.4\times {10}^{-2}m{s}^{-1}$
• B. $15\times {10}^{-2}m{s}^{-1}$
• C. $0.75\times {10}^{-2}m{s}^{-1}$
• D. $25\times {10}^{-2}m{s}^{-1}$

Answer 1

The correct option is B $15\times {10}^{-2}m{s}^{-1}$

We know that the terminal speed $v$ is related to radius $r$ as follows $v\propto r^2$ ................(1)

mass of $8$ small drops is $8\times \rho \frac{4}{3}\pi a^3$

while mass of $1$ big drops of radius b(say ) will be $\rho \frac{4}{3}\pi b^3$

mass should be $conserved$ so equating both masses will lead to $8a^3=b^3$ or $b=2a$

Now from the equation-1 we can write $\frac{v_b}{v_a}=\frac{b^2}{a^2}=\frac{4a^2}{a^2}=4$

so the new $terminal$ velocity will be $v_b=4v_a=4\times 3.75=15cm/s=0.15m/s$