Question

Mercury has angle of contact equal to ${135}^{\circ }$ with glass. A narrow glass tube of radius 0.5 mm is dipped in mercury. By what amount does the mercury dip down? (Surface tension $=408\times {10}^{-3}N/m,\rho =13.6\times {10}^{3}kg/m^3)$.

• A. 4.2 mm
• B. 3.2 mm
• C. 8.4 mm
• D. 7.6 mm

Answer 1

The correct option is C 8.4 mm

$h=\frac{2Tcos\theta }{r\rho g}=\frac{2\times 408\times {10}^{-3}\times cos135}{0.5\times \text{/}{10}^{-3}\times 13.6\times \text{/}{10}^3\times 10}=\frac{2\times 408\times {10}^{-3}}{2\times 13.6\times 10}\times (-\frac{1}{\sqrt{2}})=-\left(\frac{12}{\sqrt{2}}\right)=h=-8.4$ mm
(-ve sign indicates dip down)