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Question: If one end of a capillary tube is dipped into water, then the water rises up to 3 cm. If the surface...

If one end of a capillary tube is dipped into water, then the water rises up to 3 cm. If the surface tension of water is 75×10375 \times {10^{ - 3}} N/m, then the diameter of the capillary will be?
A. 0.10.1 mm
B. 0.50.5 mm
C. 1 mm
D. 2 mm

Explanation

Solution

Hint
Capillary action is caused by the surface tension of liquid, which makes it move upwards in a tube without the assistance of external forces like gravity. This question is a formula based exercise to find the unknown diameter.
h=2TcosθρRgh = \dfrac{{2T\cos \theta }}{{\rho Rg}}
where ρ{{\rho }} is the density of water, θ{{\theta }} is the angle of contact made by water with the capillary’s surface which is 0{\text{0}}^\circ and g{\text{g}} is the gravitational acceleration given as 9.81m/s29.81m/{s^2} .

Complete step by step answer
In the given question we are told that the capillary tube is dipped in water which rises to a certain height. We are asked to find the diameter of the said tube. The following data is provided to us:
Surface tension of water T=75×103T = 75 \times {10^{ - 3}} N/m
Height to which the water rises h=3cm=0.03mh = 3cm = 0.03m [1 m = 100 cm]
We also know the following properties of water:
Angle of contact for water θ=0\theta = {0^\circ }
Density of water ρ=1000kg/m3\rho = 1000kg/{m^3}
Acceleration due to gravity g=9.81m/s2g = 9.81m/{s^2}
We know that the rise of liquid in a capillary is given as:
h=2TcosθρRg\Rightarrow h = \dfrac{{2T\cos \theta }}{{\rho Rg}}
We are aware of all the values except RR . So, we put values in this equation to find the unknown as:
0.03=2×75×103×cos01000×R×9.81\Rightarrow 0.03 = \dfrac{{2 \times 75 \times {{10}^{ - 3}} \times \cos 0}}{{1000 \times R \times 9.81}}
We take RR on the LHS:
R=2×75×103×cos01000×0.03×9.81\Rightarrow R = \dfrac{{2 \times 75 \times {{10}^{ - 3}} \times \cos 0}}{{1000 \times 0.03 \times 9.81}}
R=150×106×10.03×9.81\Rightarrow \Rightarrow R = \dfrac{{150 \times {{10}^{ - 6}} \times 1}}{{0.03 \times 9.81}} [As cos0=1\cos 0 = 1 ]
Solving further gives us
R=5.09×104\Rightarrow R = 5.09 \times {10^{ - 4}} m
Converting this into mm:
R=0.5\Rightarrow R = 0.5 mm [ 1m=103mm1m = {10^3}mm ]
We calculate the diameter as:
D=2R=2×0.5=1\Rightarrow D = 2R = 2 \times 0.5 = 1 mm
Hence, the correct answer is option (C) i.e. 1 mm.

Note
As you can see, the rise of water in a capillary tube is inversely proportional to its radius. Hence, the smaller the radius of the tube, the more is the rise.