The capillaries shown in figure have inner radii 0.5 mm, 1.0 mm and 1.5 mm respectively. The liquid in the beaker is water. Find the heights of water level in the capillaries. The surface tension of water is 7.5 × 10^{−2} N m^{−1}.

#### Solution

Given:

Surface tension of water T = 7.5 × 10^{−2} N/m

Taking cos θ = 1:

Radius of capillary A (r_{A}) = 0.5 mm = 0.5 × 10^{−3} m

Height of water level in capillary A:

\[\text{h}_\text{A } = \frac{2\text{T} \cos \theta}{\text{r}_\text{A} \rho \text{ g}}\]

\[ = \frac{2 \times 7 . 5 \times {10}^{- 2}}{0 . 5 \times {10}^{- 3} \times 1000 \times 10}\]

\[ = 3 \times {10}^{- 2}\text{ m = 3 cm}\]

Radius of capillary B (r_{B}) = 1 mm = 1 × 10^{−3} m

Height of water level in capillary B:

\[\text{h}_\text{B} = \frac{2\text{T}\cos \theta}{\text{r}_\text{B} \rho \text{ g}}\]

\[ = \frac{2 \times 7 . 5 \times {10}^{- 2}}{1 \times {10}^{- 3} \times {10}^3 \times 10}\]

\[ = 15 \times {10}^{- 3} \text{ m = 1 . 5 cm }\]

Radius of capillary C (r_{C}) = 1.5 mm = 1.5 × 10^{−3} m

Height of water level in capillary C:

\[\text{h}_\text{C} = \frac{2\text{T} \cos \theta}{\text{r}_\text{ C} \rho \text{ g}}\]

\[ = \frac{2 \times 7 . 5 \times {10}^{- 2}}{1 . 5 \times {10}^{- 3} \times {10}^3 \times 10}\]

\[ = \frac{15}{1 . 5} \times {10}^{- 3} \text{ m = 1 cm} \]