MHT CET · Physics · Mechanical Properties of Fluids
In capillary tube having area of cross-section ' \(\mathrm{A}\) ', water rises to a height ' \(h\) '. If cross sectional area is reduced to \(\frac{A}{9}\), the rise of water in the capillary tube is
- A \(\mathrm{h}\)
- B \(4 h\)
- C \(3 \mathrm{~h}\)
- D \(2 \mathrm{~h}\)
Answer & Solution
Correct Answer
(C) \(3 \mathrm{~h}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r \rho g}} \\ & \Rightarrow \mathrm{h} \propto \frac{1}{\mathrm{r}}\end{aligned}\)
And \(\mathrm{A}=\pi \mathrm{r}^2 \Rightarrow \mathrm{r} \propto \sqrt{\mathrm{A}}\)
\(\Rightarrow \mathrm{h} \propto \frac{1}{\sqrt{\mathrm{A}}}\)
Now, \(\frac{h_2}{h_1}=\sqrt{\frac{A_1}{A_2}}\)
\(\begin{aligned} & \Rightarrow \frac{h_2}{h}=\sqrt{\frac{A}{A / 9}}=3 \\ & \Rightarrow h_2=3 h\end{aligned}\)
And \(\mathrm{A}=\pi \mathrm{r}^2 \Rightarrow \mathrm{r} \propto \sqrt{\mathrm{A}}\)
\(\Rightarrow \mathrm{h} \propto \frac{1}{\sqrt{\mathrm{A}}}\)
Now, \(\frac{h_2}{h_1}=\sqrt{\frac{A_1}{A_2}}\)
\(\begin{aligned} & \Rightarrow \frac{h_2}{h}=\sqrt{\frac{A}{A / 9}}=3 \\ & \Rightarrow h_2=3 h\end{aligned}\)
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