Question

A capillary tube of radius 0.5 mm is dipped vertically in water. Calculate the

surface tension of water if the water rises2.94 cm above the outer level.​

Answer 1

Solution  :-

As per the given data ,

• Radius (r) = 0.5 mm = 0.05 cm
• Height (h) = 2.94 cm
• Acceleration due to gravity (g) = 980 cm /s²
• Density of water (ρ) = 1 g / cm³
• Angle of contact (θ) = 0 ( for water )

Rise or fall of liquid in capillary tube ,

$\leadsto \boxed{\mathtt{ h = \dfrac{2T cos \theta}{r \rho g }}}$

On rearranging ,

$\leadsto \mathtt{ T= \dfrac{r \rho h g }{ 2 cos \theta}}$

Now let's substitute the given values in the above equation ,

$\leadsto \mathtt{ T= \dfrac{0.5 \times 1 \times 2.94 \times 980 }{ 2 cos 0}}$

$\leadsto \mathtt{ T= \dfrac{1440.6}{ 2}}$

$\leadsto\boxed{\mathtt{ T= 720.3\: dyne }}$

The surface tension of water is 720.3 dynes