Q86 / 90 · JEE Main 2024 (08 Apr Shift 2) · Physics

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

Correct answer: 40

Solution

$\mathrm{m}=$ mass of small drop $\mathrm{M}=$ mass of bigger drop $\begin{aligned} & V_t=\frac{2}{9} \frac{R^2(\rho-\sigma) g}{\eta} \\ & 8 \propto m=M \\ & 8 r^3=R^3 \Rightarrow R=2 R\end{aligned}$ as $\mathrm{V}_{\mathrm{t}} \times \mathrm{R}^2 \quad \because$ Radius double so $\mathrm{V}_{\mathrm{t}}$ becomes 4 time $\therefore 4 \times 10=40 \mathrm{~cm} / \mathrm{s}$

Hence, the correct answer is 40.

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