Two drops of equal radius are falling through | vedclass

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Question

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&nbs

Vedclass · Accepted Answer

${4^{\frac{1}{3}}} 	imes 5\,cm/s$

Vedclass · Answer

When $n$ drops of radius $r$ combine to form single drop of radius $R,$ then mass remains same hence

$\left( {\frac{4}{3}} ight)\pi {R^3}ho  = \left( {\frac{4}{3}} ight)\pi n{r^3}ho $

$\mathrm{R}^{3}=\mathrm{nr}^{3}$

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 $\vee \propto r^{2}$

$	herefore \left( {\frac{{{v_2}}}{{{v_1}}}} ight) = {\left( {\frac{{{r_2}}}{{{r_1}}}} ight)^2}$

Here $\mathrm{V}_{1}=5 \mathrm{cm} / \mathrm{s}$

$r_{2}=R$

$r_{1}=r$

$	herefore {V_2} = 5{\left( {\frac{R}{r}} ight)^2}$

$ = 5 	imes \left[ {\frac{{\left( {{2^{\left( {1/3} ight)}}.r} ight)}}{r}} ight]$

$\mathrm{V}^{2}=5 	imes 4^{(1 / 3)} \mathrm{cm} / \mathrm{s}$

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

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