Answer
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Hint: Capillary tubes are called tubes that have very small diameters. If these narrow tubes are immersed in a liquid, it is observed that the liquid in the capillary either increases (or falls) relative to the level of the liquid surrounding it. This phenomenon is called the action of capillaries, and these tubes are called capillaries.
Formula used:
The following relation gives the height to which the water rises in the capillary tube:
$ H = \dfrac{{2S\cos \theta }}{R} $
Where, $ H $ is the height to which water will rise in the capillary
$ S $ is the surface tension
$ R $ is the radius of the capillary tube
$ \theta $ is the angle of contact.
Complete step by step answer:
Now, according to the question, the mass of the water is $ m $
The area of the capillary tube is given by $ \pi {R^2} $
And the volume of the water in the capillary can be given by the area multiplied by height, that is
$ \pi {R^2} \times H $
Let the density of the liquid be $ \rho $
We know that $ mass = density \times volume $
Now the mass of the water in the capillary tube is given by
$ M = \rho \times (\pi {R^2} \times H) $
$ M = \rho \pi {R^2} \times \left( {\dfrac{{2S\cos \theta }}{R}} \right) $
We can rewrite the above expression as
$ M = \left( {\pi \rho S\cos \theta } \right)R $
In the above equation, $ \rho $ , $ S $ , and $ \theta $ are constant.
So, we can say that mass is directly proportional to the radius of the capillary tube, that is
$ M \propto R $
Now according to the question, if we double the diameter of the capillary, then the radius of the capillary will also get doubled.
That means that if the radius is doubled (or the diameter is the doubled), the mass of the liquid will also be doubled.
Hence, the correct option is (B); $ 2m $.
Note:
We can define capillary action as a phenomenon in which liquid ascension takes place through a tube or cylinder. Due to adhesive and cohesive forces, capillary action mainly occurs.
Formula used:
The following relation gives the height to which the water rises in the capillary tube:
$ H = \dfrac{{2S\cos \theta }}{R} $
Where, $ H $ is the height to which water will rise in the capillary
$ S $ is the surface tension
$ R $ is the radius of the capillary tube
$ \theta $ is the angle of contact.
Complete step by step answer:
Now, according to the question, the mass of the water is $ m $
The area of the capillary tube is given by $ \pi {R^2} $
And the volume of the water in the capillary can be given by the area multiplied by height, that is
$ \pi {R^2} \times H $
Let the density of the liquid be $ \rho $
We know that $ mass = density \times volume $
Now the mass of the water in the capillary tube is given by
$ M = \rho \times (\pi {R^2} \times H) $
$ M = \rho \pi {R^2} \times \left( {\dfrac{{2S\cos \theta }}{R}} \right) $
We can rewrite the above expression as
$ M = \left( {\pi \rho S\cos \theta } \right)R $
In the above equation, $ \rho $ , $ S $ , and $ \theta $ are constant.
So, we can say that mass is directly proportional to the radius of the capillary tube, that is
$ M \propto R $
Now according to the question, if we double the diameter of the capillary, then the radius of the capillary will also get doubled.
That means that if the radius is doubled (or the diameter is the doubled), the mass of the liquid will also be doubled.
Hence, the correct option is (B); $ 2m $.
Note:
We can define capillary action as a phenomenon in which liquid ascension takes place through a tube or cylinder. Due to adhesive and cohesive forces, capillary action mainly occurs.
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