Physics - Fluid Question with Solution | TestHub

Given,

The radius of capillary tube$=0.5\mathrm{ }\mathrm{mm}$

$=0.5\times {10}^{-3} \mathrm{m}$

Level inside tube$=0.8\mathrm{ }\mathrm{cm}$

$=0.8\times {10}^{-2} \mathrm{m}$

The angle of contract,$\theta ={120}^o$

The mass density of mercury,

$\rho =13.6\times {10}^3 \mathrm{kg} {\mathrm{m}}^{-2}$

Acceleration due to gravity,$g=10 \mathrm{m} {\mathrm{s}}^{-2}$

$h=\frac{2T\cos \theta }{r\rho g}$

$T=\frac{hr \rho g}{2\cos \theta }$

$=\frac{0.8\times {10}^{-2}\times 0.5\times {10}^{-3}\times 13.6\times {10}^3\times 10}{2\times \cos 120}$

$=\frac{0.8\times {10}^{-2}\times 0.5\times {10}^{-3}\times 13.6\times {10}^3\times 10}{2\times \frac{1}{2}}$

$=0.8\times 0.5\times 13.6\times {10}^{-1}$

$=0.544\mathrm{N} {\mathrm{m}}^{-1}$