Q.Water is observed to rise to a height of $6.0\ \text{cm}$ in a clean capillary tube dipped vertically in water, the angle of contact being taken as $0°$. Taking the surface tension of water as $0.072\ \text{N/m}$ and its density as $1000\ \text{kg/m}^3$, calculate the internal radius of the capillary tube. (Take $g = 9.8\ \text{m/s}^2$.)

Given: $h = 6.0\ \text{cm} = 0.060\ \text{m}$, $T = 0.072\ \text{N/m}$, $\rho = 1000\ \text{kg/m}^3$, $\theta = 0°$ so $\cos\theta = 1$, $g = 9.8\ \text{m/s}^2$.

Rearranging $h = \dfrac{2T\cos\theta}{\rho g r}$ to make $r$ the subject: …

Rearrange the capillary-rise formula to make the radius $r$ the subject, then substitute the given surface tensio …

• Forgetting to convert the given height from centimetres to metres before substituting. …