Question

A capillary tube of diameter 0.4 mm is dipped in a beaker containing mercury. If density of mercury is \[13.6\times {{10}^{3}}kg{{m}^{3}}\], surface tension 0.49 N/m, and angle of contact of mercury ${{135}^{o}}$, the depression of the meniscus in the capillary tube will be:
A. 0.13 m
B. 0.013 m
C. 0.026 m
D. 0.0026 m

Answer 1

Hint: We are given the diameter of a capillary tube dipped in mercury. We are also given the density, surface tension and angle of contact of mercury in the tube. To find the depression of the meniscus we have an equation to find the level of rise in the capillary tube. By substituting the known values we will get the solution.
Formula used:
$h=\dfrac{2T\cos \theta }{r\rho g}$

Complete answer:
In the question, it is said that a capillary tube is dipped in a beaker with mercury.
The diameter of the capillary tube is given as,
Diameter, $d=0.4mm$
Therefore the radius of the capillary tube will be,
$\begin{align}
  & r=\dfrac{d}{2} \\
 & \Rightarrow r=\dfrac{0.4mm}{2}=0.2mm \\
\end{align}$
By converting this into meters, we will get
$\Rightarrow r=0.2\times {{10}^{-3}}m$
We are also given the density of mercury, surface tension and the angle of contact of mercury.
Density of mercury, $\rho =13.6\times {{10}^{3}}kg{{m}^{3}}$
Surface tension, $T=0.49N/m$
Angle of contact of mercury, $\theta ={{135}^{o}}$
Now we need to find the depression in the meniscus of the capillary tube.
We have the equation for height of the level in capillary tube as,
$h=\dfrac{2T\cos \theta }{r\rho g}$, were ‘T’ is surface tension, ‘$\theta $ ‘ is the angle of contact, ‘r’ is the radius of the capillary tube, ‘$\rho $’ is density and ‘g’ is acceleration due to gravity.
By substituting the known values in the above equation, we get the depression in the meniscus as,
$h=\dfrac{2\times 0.49\times \cos 135}{0.2\times {{10}^{-3}}\times 13.6\times {{10}^{3}}\times 9.8}$
We know that, \[\cos \left( {{135}^{0}} \right)=-\dfrac{1}{\sqrt{2}}\]. Therefore, we get
\[\Rightarrow h=\left( \dfrac{\left( 2\times 0.49 \right)\times \left( -\dfrac{1}{\sqrt{2}} \right)}{\left( 0.2\times {{10}^{-3}}\times 13.6\times {{10}^{3}}\times 9.8 \right)} \right)\]

$\Rightarrow h=\left( -\dfrac{\left( 2\times 0.49 \right)\times \left( 0.707126.656 \right)}{26.656} \right)$
By solving this, we get
$\Rightarrow h=-0.0259m\approx -0.026m$
We got the value of ‘h’ as negative because it is the depression of the meniscus.
Thus the depression in the meniscus of the capillary tube is 0.026 m.

So, the correct answer is “Option C”.

Note:
Capillary rise is the rise of a liquid above zero pressure in a glass tube due to the total upward force formed by the attraction of liquid molecules to the surface of the glass tube. This phenomenon of attraction is known as adhesion.
Surface tension of a fluid is its tendency to shrink minimum possible surface area. The surface film of a fluid experiences a tension due to attraction of surface layer particles. And thus the surface layer shrinks.
Angle of contact is the angle between the tangents drawn to the surface of a liquid. For concave meniscus of a fluid, the angle of contact will be acute and for a convex meniscus the angle of contact will be obtuse.