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Eight equal drops of water are falling through air with a steady speed of 10 \mathrm{~cm} / \mathrm{s}. If the drops collapse, the new velocity is :-

Option: 1

10 \mathrm{~cm} / \mathrm{s}


Option: 2

40 \mathrm{~cm} / \mathrm{s}


Option: 3

16 \mathrm{~cm} / \mathrm{s}


Option: 4

5 \mathrm{~cm} / \mathrm{s}


Answers (1)

best_answer

8 \times \frac{4}{3} \pi r^{3}=\frac{4}{3} \pi R^{3}
\mathrm{R}=2 \mathrm{r}
\mathrm{V}=\frac{2 r^{2}}{9 \eta}\left(\rho_{b}-\rho_{a i}\right)
V \propto r^{2}
\frac{V_{1}}{V_{2}}=\left(\frac{r}{R}\right)^{2}


\mathrm{V}_{2}=\mathrm{V}_{1} \times 4=10 \times 4=40 \mathrm{~km} / \mathrm{s}^{-1}

Posted by

Pankaj Sanodiya

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