Question

A vertical capillary with inside diameter 0.50mm is submerged in to water so that the length of its part emerging outside the water surface is equal to 25mm.Find the radius of curvature of the meniscus. Surface tension of water is $73\times {10}^{-3}N/m(g=9.8m/s^2)$

• A. 0.4mm
• B. 0.6mm
• C. 0.2mm
• D. 0.8mm

Answer 1

The correct option is B

0.6mm

$h=\frac{2T}{r\rho g}=\frac{2\times 73\times {10}^{-3}}{0.25\times {10}^{-3}\times {10}^3\times 9.8}$
=59mm >25mm lenght that is outside the water surface
Therefore radius of menscues >radius of capillary r. If R is the radius of meniscus then we have
$\frac{2T}{R}=h^{\prime }\rho gh{h}^1=25mm$
$\text{(or)}R=\frac{2T}{h^1\rho g}$
$=\frac{2\times 73\times {10}^{-3}}{25\times {10}^{-3}\times {10}^3\times 9.8}=0.6\times {10}^{-3}m=0.6mm$