1. A liquid of density 830kg/m3 rises through 0.0893 m in a capillary tube of diameter 1.68 \times 10^{-4} m. Determine the surface tension of the liquid. Take angle of contact as 00.
h=\frac{2TCos\theta }{r \rho g}
T=\frac{rh\rho g}{2Cos\theta}
\rho= 830kg/m3
h=0.0893 m
radius, r = 0.84 \times 10^{-4}
\theta = 00
T=\frac{0.84 \times 10^{-4} \times 0.0893 \times 830 \times 9.8}{2Cos0^{0}}
T = 0.0305 N m-1
2. Two capillary tubes of radii 0.2cm and 0.4 cm are dipped inside the same liquid. Determine the ratio of the heights of the liquid in the two tubes.
From Jurin’s law
rh=Constant
r_{1}h_{1}=r_{2}h_{2}
\frac{h_{1}}{h_{2}}=\frac{r_{2}}{r_{1}}
r_{1} = 0.2 cm, r_{2}= 0.4 cm
\frac{h_{1}}{h_{2}}=\frac{0.4}{0.2}=\frac{2}{1}
h_{1}:h_{2}=2:1
3. If a capillary tube of radius 0.1 mm is dipped in water of surface tension 7.2 \times 10^{-2}N/m then find the rise of water through the capillary tube. Assume angle of contact as 00.
h=\frac{2TCos\theta }{r \rho g}
T = 7.2 \times 10^{-2}N/m
\theta = 00.
r= 0.1 mm = 0.1 \times 10^{-3} m
\rho = 10^{3} kg/m3
h=\frac{2\times 7.2 \times 10^{-2} \times Cos0^{0}}{0.1 \times 10^{-3}\times 10^{3}\times 9.8}
h=0.146 m = 146 mm
4. 1000 water droplets having radius 0.01 cm each coalesce to form a single big drop. What will be the decrease in energy? (Surface tension of water = 72 dyne/cm)
Radius of the droplets, r = 0.01 cm
Let R be the radius of the big droplet formed by 1000 small droplets.
The volume of the liquid will be same. So,
\frac{4}{3} \pi R^{3}=1000\times \frac{4}{3} \pi r^{3}
R^{3}=1000\times r^{3}
R=10r = 10\times 0.01 = 0.1 cm
The decrease in surface area,
\Delta A= (1000\times 4\pi r^{2}) - 4\pi R^{2}
\Delta A= 4\pi (1000r^{2} - R^{2})
\Delta A= 4\pi (1000r^{2} - (10r)^{2})
\Delta A= 4\pi (1000r^{2} - 100r^{2})
\Delta A= 4\pi (900r^{2})
\Delta A= 4\pi (900 \times (0.01)^{2})=1.130 m^{2}
So, decrease in energy = decrease in area \times surface tension
= 1.130 m^{2}\times 72 = 81.43 erg