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