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Two spherical rain drops reach the surface of the earth with terminal velocities
having ratio $16: 9$. The ratio of their surface area is
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having ratio $16: 9$. The ratio of their surface area is
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Verified Answer
The correct answer is:
$16: 9$
The terminal velocity of rain drop is
$\begin{array}{l}
v_{T}=\frac{2(\sigma-\rho) r^{2} g}{9 \eta} \\
\Rightarrow v_{T} \propto r^{2} \ldots(i)
\end{array}$
Also, surface area of rain drop, $A=4 \pi r^{2}$ $\Rightarrow \mathrm{A} \propto \mathrm{r}^{2} \quad$...(ii)
From Eqs. (i) and (ii), we get
$\begin{array}{l}
\frac{A_{1}}{A_{2}}=\frac{V_{T_{1}}}{V_{T_{2}}} \\
=\frac{16}{9} \text { or } 16: 9
\end{array}$
$\begin{array}{l}
v_{T}=\frac{2(\sigma-\rho) r^{2} g}{9 \eta} \\
\Rightarrow v_{T} \propto r^{2} \ldots(i)
\end{array}$
Also, surface area of rain drop, $A=4 \pi r^{2}$ $\Rightarrow \mathrm{A} \propto \mathrm{r}^{2} \quad$...(ii)
From Eqs. (i) and (ii), we get
$\begin{array}{l}
\frac{A_{1}}{A_{2}}=\frac{V_{T_{1}}}{V_{T_{2}}} \\
=\frac{16}{9} \text { or } 16: 9
\end{array}$
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