Solveeit Logo
Login

Question

Physics Question on mechanical properties of fluid

Eight drops of water, each of radius 2 mm2 \,mm are falling through air at a terminal velocity of 8 cm s−18\, cm \,s^{-1} . If they coalesce to form a single drop, then the terminal velocity of combined drop will be

A

32 cm s−132 \, cm \, s^{-1}

B

30 cm s−130\, cm \, s^{-1}

C

28 cm s−128\, cm \, s^{-1}

D

24 cm s−124\, cm \, s^{-1}

Answer

32 cm s−132 \, cm \, s^{-1}

Explanation

Solution

Let the radius of bigger drop is RR and smaller drop is rr then 43πR3=8×43×πr3\frac{4}{3} \pi R^{3}=8\times\frac{4}{3}\times\pi r^{3} or R=2r…(i)R=2r \ldots\left(i\right) Terminal velocity, v∝r2v \propto r^{2} ∴v′v=R2r2=(2rr)2=4\therefore \frac{v'}{v}=\frac{R^{2}}{r^{2}}=\left(\frac{2r}{r}\right)^{2}=4 (Using (i)) or v′=4v=4×8=32 cm s−1v' =4v=4\times8=32\, cm \, s^{-1}