Q145 / 150 · MHT CET 2024 (16 May Shift 1) · Physics
When capillary is dipped vertically in water, rise of water in capillary is ' h '. The angle of contact is zero. Now the tube is depressed so that its length above the water surface is $\frac{\mathrm{h}}{3}$. The new apparent angle of contact is $\left(\cos 0^{\circ}=1\right)$
- A$\cos ^{-1}\left(\frac{1}{2}\right)$
- Correct answerB$\quad \cos ^{-1}\left(\frac{1}{3}\right)$
- C$\quad \cos ^{-1}\left(\frac{1}{4}\right)$
- D$\quad \cos ^{-1}\left(\frac{1}{6}\right)$
Solution
Surface tension in terms of capillary rise $h$ is
$\begin{array}{ll}
& \mathrm{T}=\frac{\text { rh } \rho g}{2 \cos \theta} \Rightarrow \frac{\mathrm{rh}^{\prime} \rho g}{2 \cos \theta^{\prime}} \\
\therefore \quad & \frac{\cos \theta^{\prime}}{\cos \theta}=\frac{\mathrm{h}^{\prime}}{\mathrm{h}}=\frac{1}{3} \\
\therefore \quad & \cos \theta^{\prime}=\frac{1}{3} \cos 0^{\circ}=\frac{1}{3} \\
\therefore \quad & \theta^{\prime}=\cos ^{-1}\left(\frac{1}{3}\right)
\end{array}$
Hence, the correct answer is option: 'B'.