Fluid Mechanics

JEE Main level, numerical answer, medium

171 of 192

Question 171 of 192

Small water droplets of radius 0.01 mm0.01 \mathrm{~mm} are formed in the upper atmosphere and falling with a terminal velocity of 10 cm/s10 \mathrm{~cm} / \mathrm{s}. Due to condensation, if 8 such droplets are coalesced and formed a larger drop, the new terminal velocity will be _____cm/s\mathrm{cm} / \mathrm{s}.

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Answer: 40

m=\mathrm{m}= mass of small drop
M=\mathrm{M}= mass of bigger drop
Vt=29R2(ρ−σ)gη8∝m=M8r3=R3⇒R=2R\begin{aligned} & V_t=\frac{2}{9} \frac{R^2(\rho-\sigma) g}{\eta} \\ & 8 \propto m=M \\ & 8 r^3=R^3 \Rightarrow R=2 R\end{aligned}
as Vt×R2∵\mathrm{V}_{\mathrm{t}} \times \mathrm{R}^2 \quad \because Radius double so Vt\mathrm{V}_{\mathrm{t}} becomes 4 time
∴4×10=40 cm/s\therefore 4 \times 10=40 \mathrm{~cm} / \mathrm{s}

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