MHT CET · Physics · Mechanical Properties of Fluids
Two capillary tubes of the same diameter are kept vertically in two different liquids whose densities are in the ratio \(4: 3\). The rise of liquid in two capillaries is ' \(h_1\) ' and ' \(h h_2\) ' respectively. If the surface tensions of liquids are in the ratio \(6: 5\), the ratio of heights \(\left(\frac{h_1}{h_2}\right)\) is
(Assume that their angles of contact are same)
- A \(0.4\)
- B \(0.5\)
- C \(0.8\)
- D \(0.9\)
Answer & Solution
Correct Answer
(D) \(0.9\)
Step-by-step Solution
Detailed explanation
Given: Density ratio: \(\frac{\rho_1}{\rho_2}=\frac{4}{3}\) and surface tension ratio: \(\frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{6}{5}\)
\(\therefore \quad\) Rise of liquid in a capillary:
\(
\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho \mathrm{g}}
\)
For constant \(\theta, \mathrm{r}\) and \(\mathrm{g}\),
\(
\begin{aligned}
\mathrm{h} & \propto \frac{\mathrm{T}}{\rho} \\
\therefore & \frac{\mathrm{h}_1}{\mathrm{~h}_2}=\frac{\mathrm{T}_1 \rho_2}{\mathrm{~T}_2 \rho_1}=\frac{6 \times 3}{5 \times 4}=0.9
\end{aligned}
\)
\(\therefore \quad\) Rise of liquid in a capillary:
\(
\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho \mathrm{g}}
\)
For constant \(\theta, \mathrm{r}\) and \(\mathrm{g}\),
\(
\begin{aligned}
\mathrm{h} & \propto \frac{\mathrm{T}}{\rho} \\
\therefore & \frac{\mathrm{h}_1}{\mathrm{~h}_2}=\frac{\mathrm{T}_1 \rho_2}{\mathrm{~T}_2 \rho_1}=\frac{6 \times 3}{5 \times 4}=0.9
\end{aligned}
\)
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