Q112 / 150 · MHT CET 2024 (15 May Shift 1) · Physics
The angle of contact between glass and water is $0^{\circ}$ and water rises in a glass capillary upto 6 cm (Surface tension of water is T). Another liquid of surface tension ' $2 \mathrm{~T}^{\prime}$, angle of contact $60^{\circ}$ and relative density 2 will rise in the same capillary up to $\left(\cos 0^{\circ}=1, \cos 60^{\circ}=0.5\right)$
- A1.5 cm
- B2 cm
- Correct answerC3 cm
- D4.0 cm
Solution
$\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{r \rho g}$
For water: $\begin{array}{r} 6=\frac{2 T \cos (0)}{r \times 1 \times g} \\ \therefore \quad r=\frac{2 T}{6 g}=\frac{T}{3 g} \end{array}$
For other liquid: $h=\frac{4 \mathrm{~T} \cos 60}{\mathrm{r} \times 2 \times \mathrm{g}}=\frac{4 \mathrm{~T} \times 0.5}{\frac{\mathrm{~T}}{3 \mathrm{~g}} \times 2 \times \mathrm{g}}=3 \mathrm{~cm}$
For water: $\begin{array}{r} 6=\frac{2 T \cos (0)}{r \times 1 \times g} \\ \therefore \quad r=\frac{2 T}{6 g}=\frac{T}{3 g} \end{array}$
For other liquid: $h=\frac{4 \mathrm{~T} \cos 60}{\mathrm{r} \times 2 \times \mathrm{g}}=\frac{4 \mathrm{~T} \times 0.5}{\frac{\mathrm{~T}}{3 \mathrm{~g}} \times 2 \times \mathrm{g}}=3 \mathrm{~cm}$
Hence, the correct answer is option: 'C'.