JEE Advance - Physics (2014 - Paper 2 Offline - No. 10)
A glass capillary tube is of the shape of truncated cone with an apex angle $$\alpha$$ so that its two ends have cross sections of different radii. When dipped in water vertically, water rises in it to a height h, where the radius of its cross section is b. If the surface tension of water is S, its density is $$\rho$$, and its contact angle with glass is $$\theta$$, the value of h will be (g is the acceleration due to gravity)
Explanation
Let R be the radius of the meniscus formed with a contact angle $$\theta$$.
By geometry, this radius makes an angle $$\theta + {\alpha \over 2}$$ with the horizontal where
$$\cos \left( {\theta + {\alpha \over 2}} \right) = b/R$$. ...... (1)
Let P0 be the atmospheric pressure and P1 be the pressure just below the meniscus. Excess pressure on the concave side of meniscus of radius R is given by
$${P_0} - {P_1} = 2S/R$$ ..... (2)
The hydrostatic pressure gives
$${P_0} - {P_1} = h\rho g$$ ..... (3)
From (2) and (3) we get,
$${{2S} \over R} = h\rho g$$
or, $$h\rho g = {{2s} \over b}\cos \left( {\theta + {\alpha \over 2}} \right)$$ (By putting value of R)
$$h = {{2S} \over {b\rho g}}\cos \left( {\theta + {\alpha \over 2}} \right)$$
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