Question 5.C-A.67: A capillary tube when immersed in a liquid of density d, the...
A capillary tube when immersed in a liquid of density d, the capillary rise observed is h. When another capillary tube of 4 times cross sectional area as the first tube is immersed in another liquid of density 2d, what is the observed capillary rise. Assume angle of contact of capillary tube with both liquids is 0° the ratio of surface tensions of the two liquids is 1 : 2.
Capillary rise in second capillary tube is
h=\frac{2 T_{1} \cos 0^{\circ}}{r_{1} d_{1} g} (1)
(given angle of contact θ = 0°)
here r_{1} is radius of the bole of second capillary tube d_{1} = d is density of first liquid.
T_{1} = is surface tension of first liquid.
Capillary rise in second capillary tube is
\frac{2 T_{2} \cos 0^{\circ}}{r_{2} d_{2} g} (2) [given θ = 0° ]
here r_{2} is radius of the bole of second capillary tube d_{1} = d is density of second liquid.
T_{2} is surface tension of second liquid.
Dividing equation (1) and (2)
\frac{h}{h_{2}}=\frac{\frac{2 T_{1}}{r_{1} d_{1} g}}{\frac{2 T_{2}}{r_{2} d_{2} g}} [since \cos 0^{\circ}=1 ]
\frac{h_{1}}{h_{2}}=\frac{T_{1}}{T_{2}} \times \frac{r_{2} d_{2}}{r_{1} d_{1}} \Rightarrow \frac{h}{h_{2}}=\frac{1}{2} \times \frac{r_{2}}{r_{1}} \times \frac{2 d}{d}
\frac{h}{h_{2}}=\frac{r_{2}}{r_{1}} (3)
Given: Area of cross section for second capillary tube A_{2}=4 A_{1}
[ A_{1} is area of cross section of first capillary tube]
r_{2}^{2}=4 \quad r_{1}^{2} \therefore r_{2}=2 r_{1} \therefore \frac{h}{h_{2}}=\frac{2 r_{1}}{r_{2}} \Rightarrow h_{2}=\frac{h}{2}