Question

# A liquid rises to a height of $$50 \ cm$$ in a capillary tube of diameter $$0.04 \ mm$$. if density of a liquid is $$0.08\times10^3 kg/m^2$$, angle of contact is $$20 \ degrees$$, the surface tension of the liquid is :

A
4.17N/m
B
0.417N/m
C
4.17×102N/m
D
4.17×103N/m

Solution

## The correct option is D $$4.17\times10^{-3} N/m$$Given,$$h=50cm=0.5m$$$$d=0.04mm$$, $$r=0.02\times 10^{-2}m$$$$g=9.8m/s^2$$$$\rho=0.08\times 10^3kg/m^3$$$$\theta=20^0$$The height of the capillary tube,$$h=\dfrac{2Tcos\theta}{r\rho g}$$ $$T=\dfrac{hr\rho g}{2cos\theta}$$$$T=\dfrac{0.50\times 0.02\times 10^{-3}\times 0.08\times 10^3\times 9.8}{2\times cos20^0}$$$$T=4.17\times 10^{-3} N/m$$The correct option is D..Physics

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