A uniform capillary tube of inner radius is dipped vertically into a beaker filled with water. The water rises to a height in the capillary tube above the water surface in the beaker. [2018]

Question

A uniform capillary tube of inner radius $r$ is dipped vertically into a beaker filled with water. The water rises to a height $h$ in the capillary tube above the water surface in the beaker. The surface tension of water is $σ$ . The angle of contact between water and the wall of the capillary tube is $θ$ . Ignore the mass of water in the meniscus. Which of the following statements is (are) true? [2018]

Smart Achievers · Accepted Answer

(1, 3)

$As we know h = \frac{2 σ \cos θ}{r ρ g_{eff}}$

$As' r' increases, h decreases h \propto \frac{1}{r}$

[all other parameters remaining constant]

$Also h \propto σ$

Further if lift is going up with an acceleration $' a'$

$then g_{eff} = g + a . As g_{eff} increases, h decreases.$

$Also h \propto \cos θ not h \propto θ$

Solution

1 For a given material of the capillary tube, $h$ decreases with increase in $r$

2 For a given material of the capillary tube, $h$ is independent of $σ$

3 If this experiment is performed in a lift going up with a constant acceleration, then $h$ decreases

4 $h$ is proportional to contact angle $θ$

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Solution

1 For a given material of the capillary tube, h decreases with increase in r

2 For a given material of the capillary tube, h is independent of σ

3 If this experiment is performed in a lift going up with a constant acceleration, then h decreases

4 h is proportional to contact angle θ

Ans. (1, 3) As we know h=2σcosθrρgeff As 'r' increases, h decreases h∝1r [all other parameters remaining constant] Also h∝σ Further if lift is going up with an acceleration 'a' then geff=g+a. As geff increases, h decreases. Also h∝cosθ not h∝θ

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1 For a given material of the capillary tube, $h$ decreases with increase in $r$

2 For a given material of the capillary tube, $h$ is independent of $σ$

3 If this experiment is performed in a lift going up with a constant acceleration, then $h$ decreases

4 $h$ is proportional to contact angle $θ$