JEE Mains · Physics · STD 11 - Fluid mechanics
- JEE Mains 2024
- Hard
- Numerical
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 \(8 \mathrm{such}\) droplets are coalesced and formed a larger drop, the new terminal velocity will be _______ \(\mathrm{cm} / \mathrm{s}\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(40\)
Step-by-step Solution
Detailed explanation
\(\mathrm{m}=\text { mass of small drop }\)
\(\mathrm{M}=\text { mass of bigger drop }\)
\(\mathrm{V}_{\mathrm{t}}=\frac{2}{9} \frac{\mathrm{R}^2(\rho-\sigma) \mathrm{g}}{\eta}\)
\(8 \propto \mathrm{m}=\mathrm{M}\)
\(8 \mathrm{r}^3=\mathrm{R}^3 \Rightarrow \mathrm{R}=2 \mathrm{R}\)
\(\text { as } \mathrm{V}_{\mathrm{t}} \times \mathrm{R}^2 \because \text { Radius double so } \mathrm{V}_{\mathrm{t}} \text { becomes } 4 \text { time }\)
\(\therefore 4 \times 10=40 \mathrm{~cm} / \mathrm{s}\)
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