Question

Small water droplets of radius 0.01 mm are formed in the upper atmosphere and falling with a terminal velocity of 10 cm/s. Due to condensation, if 8 such droplets are coalesced and formed a larger drop, the new terminal velocity will be ………..cm/s.  

Answer 1

Correct answer is : 40 

\(\mathrm{m}=\) mass of small drop

M= mass of bigger drop

\(\mathrm{V}_{\mathrm{t}}=\frac{2}{9} \frac{\mathrm{R}^{2}(\rho-\sigma) \mathrm{g}}{\eta}\)

\(8 \propto \mathrm{m}=\mathrm{M}\)

\(8 r^{3}=R^{3} \Rightarrow R=2 R\)

as \(\mathrm{V}_{\mathrm{t}} \times \mathrm{R}^{2} \ \because\)  Radius double so \(\mathrm{V}_{\mathrm{t}}\) becomes 4 time

\(\therefore \ 4 \times 10=40 \mathrm{~cm} / \mathrm{s}\)