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Two drops of equal radius are falling through air with a steady velocity of $5\,cm/s$. If the two drops coalesce, then its terminal velocity will be
${4^{\frac{1}{3}}} \times 5\,cm/s$
${4^{\frac{1}{3}}}\,cm/s$
${5^{\frac{1}{3}}} \times 4\,cm/s$
${4^{\frac{2}{3}}} \times 5\,cm/s$
Solution
When $n$ drops of radius $r$ combine to form single drop of radius $R,$ then mass remains same hence
$\left( {\frac{4}{3}} \right)\pi {R^3}\rho = \left( {\frac{4}{3}} \right)\pi n{r^3}\rho $
$\mathrm{R}^{3}=\mathrm{nr}^{3}$
Hence $n=2$ hence $R^{3}=2 r^{3}$ i.e. $R=2^{(1 / 3)} \cdot r$
When drop of radius r falls through a medium, then velocity $\vee \propto r^{2}$
$\therefore \left( {\frac{{{v_2}}}{{{v_1}}}} \right) = {\left( {\frac{{{r_2}}}{{{r_1}}}} \right)^2}$
Here $\mathrm{V}_{1}=5 \mathrm{cm} / \mathrm{s}$
$r_{2}=R$
$r_{1}=r$
$\therefore {V_2} = 5{\left( {\frac{R}{r}} \right)^2}$
$ = 5 \times \left[ {\frac{{\left( {{2^{\left( {1/3} \right)}}.r} \right)}}{r}} \right]$
$\mathrm{V}^{2}=5 \times 4^{(1 / 3)} \mathrm{cm} / \mathrm{s}$
