Question

The angle of contact between glass and water is $ {{0}^{0}} $ and water (surface tension \[70\ dyne/cm\]) rises in a glass capillary up to 6cm. Another liquid of surface tension\[140\ dyne/cm\], angle of contact $ {{60}^{0}} $ and relative density 2 will rise in the same capillary up to
A. 12cm
B. 24cm
C. 3cm
D. 6cm

Answer 1

Hint: To solve this problem, we must find the rise in the capillary tube with different angles of contact. The formula to find the height of rise in the capillary is: \[h=\dfrac{2T\cos \theta }{r\rho g}\]. Here, T refers to the surface tension of the liquid, r to the radius of the capillary tube, \[\rho \]to the density of the fluid and g to the acceleration due to gravity.

Complete step-by-step answer:
Let’s start by understanding how a liquid rises in a capillary tube. The rise of a liquid in a capillary tube is due to the surface tension of the liquid, and this action is known as the capillary action. An example of such an example in daily life is, a tissue paper. Just touching a tip of the tissue paper to water, we will see that the water will start to rise in the tissue paper, soaking the whole paper slowly due to capillary action as just a small tip of tissue paper touches water.
Let’s use the formula to find the height of rise in the capillary is: \[h=\dfrac{2T\cos \theta }{r\rho g}\]. Here, T refers to the surface tension of the liquid, r to the radius of the capillary tube, \[\rho \]to the density of the fluid and g to the acceleration due to gravity.
For the first case when the rise in the capillary tube is 6cm. \[{{h}_{1}}=\dfrac{2{{T}_{1}}\cos {{\theta }_{1}}}{r{{\rho }_{1}}g}\Rightarrow 6=\dfrac{2(70)\cos 0}{r{{\rho }_{1}}g}\Rightarrow {{\rho }_{1}}=\dfrac{70}{3rg}\].
Similarly, for the second case, the rise in the capillary tube will be $ \left( {{h}_{2}} \right) $ given by: \[{{h}_{2}}=\dfrac{2{{T}_{2}}\cos {{\theta }_{2}}}{r{{\rho }_{2}}g}\Rightarrow {{h}_{2}}=\dfrac{2(140)\cos 60}{r{{\rho }_{2}}g}\Rightarrow {{\rho }_{2}}=\dfrac{280}{2rg{{h}_{2}}}\Rightarrow {{\rho }_{2}}=\dfrac{140}{rg{{h}_{2}}}\].
The values of r and g will be constant, since the same capillary tube is used and g value remains constant too. Now, taking the ratio of the densities, and using the information regarding the density given in the problem, we get: \[\dfrac{{{\rho }_{1}}}{{{\rho }_{2}}}=\dfrac{\left( \dfrac{70}{3rg} \right)}{\left( \dfrac{140}{rg{{h}_{2}}} \right)}\Rightarrow \dfrac{1}{2}=\dfrac{70}{3rg}\times \dfrac{rg{{h}_{2}}}{140}\Rightarrow 1=\dfrac{70}{3}\times \dfrac{{{h}_{2}}}{70}\Rightarrow {{h}_{2}}=3cm\].
Therefore, the other liquid will rise up to 3cm in the capillary tube, given by Option C.

Note: Another example of capillary action is the transport phenomenon of water and nutrition transport systems in plants. The water received by the roots in the soil, are transported to the whole plant or the tree, is through capillary action through the veins in the plants. These capillaries also transport nutrients by mixing the nutrients in the water being transported.
We must remember that the capillary action is only possible when the radius of the capillary tube is small. Since, for very wide capillary tubes, having a large radius the surface tension of the liquid will not be able to cause the liquid to rise significantly at all.