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CGPET 2015 Solved Paper

Section: Physics

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Question : 8 of 150

Marks: +1, -0

Water rises to a height of

10\ \mathrm{cm}

in a certain capillary tube. In the same tube, the level of mercury is depressed by

3.42\ \mathrm{cm}

. The ratio of surface tensions of water and mercury is (specific gravity of mercury is

13.6

, the angle of contact for water is zero and that for mercury is

135^{\circ}

$1:1$
1 : $13.6$
$1:6.57$
$1:0.342$

Solution:

Height or depth of liquid column in the capillary tube

y = \frac{2S \cos \theta}{r \delta g}

where, S = surface tension of the liquid

θ = angle of contact

r = radius of the tube

ρ = density of the liquid

For water,

h = \frac{2S_w \cos \theta}{r \rho_w g}

......(i)

For mercury,

d = \frac{2S_m \cos \theta}{r \rho_m g}

..... (ii)

From Eqs. (i) and (ii),

\frac{h}{d} = \frac{S_w}{S_m} \times \frac{\rho_m}{\rho_w} \times \frac{\cos 0^{\circ}}{\cos 135^{\circ}}

⇒

\frac{10}{-3.42} = \frac{S_w}{S_m} \times 13.6 \times \frac{1}{\frac{-1}{\sqrt{2}}}

⇒

\frac{10}{-3.42} = \frac{S_w}{S_m} \times 13.6 \times \sqrt{2}

or,

\frac{S_w}{S_m} = \frac{10}{3.42} \times \frac{1}{13.6} \times \frac{1}{\sqrt{2}}

=1:6.57

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