MHT CET · Physics · Motion In One Dimension
Two spherical rain drops reach the surface of the earth with terminal velocities having ratio \(16: 9\). The ratio of their surface area is
- A \(4: 3\)
- B \(64: 27\)
- C \(16: 9\)
- D \(9: 16\)
Answer & Solution
Correct Answer
(C) \(16: 9\)
Step-by-step Solution
Detailed explanation
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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