Fluid Mechanics

JEE Main level, single correct, medium

117 of 192

Question 117 of 192

Eight equal drops of water are falling through air with a steady speed of 10cms−110cm{s}^{-1}. If the drops coalesce, the new velocity is:-

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Answer: (B) 40cms−140cm{s}^{-1}

The terminal velocity can be given by
vT=2(ρ−σ)r2g9η{v}_{T}=\dfrac{2(\rho -\sigma ){r}^{2}g}{9\eta }.
Clearly, vT∝r2{v}_{T}\propto {r}^{2}
Let the radius of the small drops be r1{r}_{1} and the radius of the big drop be r2{r}_{2}.
So,
8(4π3)r13=(4π3)r23⇒r1=r228\left(\dfrac{4\pi }{3}\right){r}_{1}^{3}=\left(\dfrac{4\pi }{3}\right){r}_{2}^{3} \Rightarrow {r}_{1}=\dfrac{{r}_{2}}{2}
Hence, taking the ratio
(vT)1(vT)2=(r1r2)2\dfrac{{\left({v}_{T}\right)}_{1}}{{\left({v}_{T}\right)}_{2}}={\left(\dfrac{{r}_{1}}{{r}_{2}}\right)}^{2}
⇒10(vT)2=14\Rightarrow \dfrac{10}{{\left({v}_{T}\right)}_{2}}=\dfrac{1}{4}
⇒(vT)2=40cms−1\Rightarrow {\left({v}_{T}\right)}_{2}=40cm{s}^{-1}

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