Question

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.
Figure

Answer 1

Surface Tension is 1 cm.

Explanation:

Given,  

Surface Tension of Water, $T  = 7.5 \times 10^-^2 \frac {N}{m}$

taking $cos \theta = 1$

Radius of capillary$(r_A) = 0.5 mm = 0.5 \times 10^-^3 m$

Height of Water level in capillary A:

$h_A = \frac {2T cos \theta}{r_A\rho g}$

$\frac {2 \times 7.5 \times 10^-^2}{0.5 \times 10^-^3\times 1000 \times 10}$

= $3 \times 10^-^2 m$

= $3\ cm$

Radius of capillary $B(r_B)= 1\ mm= 1 \times 10^-^3 m$

Height of water in capillary $B h_B= \frac {2T cos \theta}{r_B \rho g}$

= $\frac {2 \times 7.5 \times 10^-^2}{1 \times 10^-^3 \times 10^3 \times 10}$

Radius of capillary = $C (r_C) = 1.5 mm =1.5 \times 10^-^3 m$

Height of Water level in capillary C:

$h_C= \frac{2T cos \theta}{r_C \rho g}$

= $\frac {2 \times 7.5 \times 10^-^3}{1.5 \times 10^-^3 \times 10^3 \times 10}$

= $\frac {15}{2.5} \times 10^-^3 m =1 m$