Question

Mercury has an angle of contact equal to ${140}^o$ with soda lime glass. A narrow tube of radius 1 mm made of this glass is dipped in a trough containing mercury. The surface tension of mercury at the temperature of the experiment is $0.465N{m}^{-1}$. The distance by which the mercury dip down in the tube relative to the mercury surface outside is: (Density of mercury = $13.6\times {10}^{3}kg{m}^{-3}$)

• A. 5.34 mm
• B. 2.35 mm
• C. 6.25 mm
• D. 1.44 mm

Answer 1

The correct option is A 5.34 mm

Here, $\theta ={140}^o,r=1\times {10}^{-3}kgm$

$S=0.465N{m}^{-1},\rho =13.6\times {10}^{3}kg{m}^{-3}$

As $h=\frac{2Scos\theta }{r\rho g}=\frac{2\times 0.465\times (cos{180}^{\circ }-{140}^{\circ })}{{10}^{-3}\times 13.6\times {10}^3\times 9.8}$

$=-5.34\times {10}^{-3}=-5.34mm$