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Two capillary tubes A and B of diameter 1 mm and 2 mm respectively are dipped vertically in a liquid. If the capillary rise in A is 6 cm, then the capillary rise in B is
A) 2 cm
B) 3 cm
C) 4 cm
D) 6 cm
E) 9 cm
Answer
461.1k+ views
Hint: When capillaries are dipped in liquid, the liquid either rises or drops in the tube due to capillary action. The rise in height is proportional to the diameter of the tube
Formula used: In this solution, we will use the following formula:
$h = \dfrac{{2T\cos \theta }}{{r\rho g}}$ where \[h\] is the rise in height of the liquid inside the capillary, $T$ is the tension in the liquid and $\rho $ is the density of the liquid, $\theta $ is the angle of contact, and $g$ is the gravitational acceleration acting on the object.
Complete step by step answer
We’ve been given that two capillary tubes A and B of diameter 1 mm and 2 mm respectively are dipped vertically in a liquid and the liquid rises by 6 cm in capillary tube A and we want to find the rise in height in liquid when capillary tube B is dipped in the liquid.
The rise in height of the liquid in a capillary tube can be calculated using the formula
$h = \dfrac{{2T\cos \theta }}{{r\rho g}}$
In our case, the only difference between the two tubes is their diameter. So we can say that $h \propto \dfrac{1}{r}$. Since the radius is equal to half of the diameter of the capillary tube, we can write
$\dfrac{{{h_1}}}{{{h_2}}} = \dfrac{{{r_2}}}{{{r_1}}}$
Substituting the values of ${h_1} = 6\,cm$, ${r_1} = \dfrac{1}{2} = 0.5mm$ and ${r_2} = \dfrac{2}{2} = 1\,mm$, we get
$\dfrac{6}{{{h_2}}} = \dfrac{1}{{0.5}}$
Solving for ${h_2}$, we get
${h_2} = 3\,cm$
Hence the rise of liquid in the second capillary tube will be 3 cm which corresponds to option (B).
Note
The angle of contact of the liquid, the tension of the surface of the liquid, its density will all remain constant in both cases. So the only change in liquid height inside the capillary will be caused by the change in the diameter of the capillary.
Formula used: In this solution, we will use the following formula:
$h = \dfrac{{2T\cos \theta }}{{r\rho g}}$ where \[h\] is the rise in height of the liquid inside the capillary, $T$ is the tension in the liquid and $\rho $ is the density of the liquid, $\theta $ is the angle of contact, and $g$ is the gravitational acceleration acting on the object.
Complete step by step answer
We’ve been given that two capillary tubes A and B of diameter 1 mm and 2 mm respectively are dipped vertically in a liquid and the liquid rises by 6 cm in capillary tube A and we want to find the rise in height in liquid when capillary tube B is dipped in the liquid.
The rise in height of the liquid in a capillary tube can be calculated using the formula
$h = \dfrac{{2T\cos \theta }}{{r\rho g}}$
In our case, the only difference between the two tubes is their diameter. So we can say that $h \propto \dfrac{1}{r}$. Since the radius is equal to half of the diameter of the capillary tube, we can write
$\dfrac{{{h_1}}}{{{h_2}}} = \dfrac{{{r_2}}}{{{r_1}}}$
Substituting the values of ${h_1} = 6\,cm$, ${r_1} = \dfrac{1}{2} = 0.5mm$ and ${r_2} = \dfrac{2}{2} = 1\,mm$, we get
$\dfrac{6}{{{h_2}}} = \dfrac{1}{{0.5}}$
Solving for ${h_2}$, we get
${h_2} = 3\,cm$
Hence the rise of liquid in the second capillary tube will be 3 cm which corresponds to option (B).
Note
The angle of contact of the liquid, the tension of the surface of the liquid, its density will all remain constant in both cases. So the only change in liquid height inside the capillary will be caused by the change in the diameter of the capillary.
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