Question

A liquid rises to a height of $50\,cm$ in a capillary tube of diameter $0.04\,mm$. If density of liquid is $0.08 \times {10^3}\,kg{m^{ - 3}}$, angle of contact is ${20^ \circ }$, the surface tension of the liquid is: ($\cos {20^ \circ } = 0.94$).
(A) $4.17\,N{m^{ - 1}}$
(B) $0.417\,N{m^{ - 1}}$
(C) $4.17\, \times {10^{ - 2}}N{m^{ - 1}}$
(D) $4.17 \times {10^{ - 3}}\,N{m^{ - 1}}$

Answer 1

Hint: The surface tension of the liquid in the capillary tube is determined by using the formula of the height of the liquid in the capillary tube. In this formula, by keeping the surface tension of the liquid on one side and taking the other terms to the other side, then the surface tension can be determined.

Useful formula
The height of the liquid in the capillary tube is given by,
$h = \dfrac{{2T\cos \theta }}{{\rho gr}}$
Where, $h$ is the height of the liquid in the capillary tube, $T$ is the surface tension of the liquid, $\theta $ is the angle of the contact of the liquid in the capillary tube, $\rho $ is the density of the liquid, $g$ is the acceleration due to gravity and $r$ is the radius of the capillary tube.

Complete step by step solution
Given that,
The height of the liquid in the capillary tube is, $h = 50\,cm = 50 \times {10^{ - 2}}\,m$
The diameter of the capillary tube is, $d = 0.04\,mm$
Then the radius of the capillary tube is, $r = \dfrac{d}{2} = \dfrac{{0.04}}{2} = 0.02\,mm = 0.02 \times {10^{ - 3}}\,m$
The density of the liquid is, $\rho = 0.08 \times {10^3}\,kg{m^{ - 3}}$
The angle of the contact of the liquid in the tube is, $\theta = {20^ \circ }$
The cosine value of the angle of contact is, $\cos {20^ \circ } = 0.94$
Now,
The height of the liquid in the capillary tube is given by,
$h = \dfrac{{2T\cos \theta }}{{\rho gr}}\,.......................\left( 1 \right)$
By keeping the surface tension in one side and the other terms in the other side, then the equation (1) is written as,
$T = \dfrac{{h\rho gr}}{{2\cos \theta }}$
By substituting the height of the liquid, density of the liquid, acceleration due to gravity, radius of the tube and angle of the contact in the above equation, then
$T = \dfrac{{50 \times {{10}^{ - 2}} \times 0.08 \times {{10}^3} \times 9.8 \times 0.02 \times {{10}^{ - 3}}}}{{2 \times \cos {{20}^ \circ }}}$
By multiplying the terms in the above equation, then
$T = \dfrac{{7.84 \times {{10}^{ - 3}}}}{{2\cos {{20}^ \circ }}}$
By substituting the cosine value in the above equation, then
$T = \dfrac{{7.84 \times {{10}^{ - 3}}}}{{2 \times 0.94}}$
By multiplying the terms in the above equation, then
$T = \dfrac{{7.84 \times {{10}^{ - 3}}}}{{1.88}}$
By dividing the terms, then the above equation is written as,
$T = 4.17 \times {10^{ - 3}}N{m^{ - 1}}$

Hence, the option (D) is the correct answer.

Note: The surface tension is directly proportional to the height of the liquid, density of the liquid, acceleration due to gravity and radius of the tube. And the surface tension is inversely proportional to the angel of the contact between the liquid and the surface of the tube.