The angle of contact between glass and water is $0^\circ $ and it rises in a capillary upto 6 cm when its surface tension is 70 dyne/cm. Another liquid of surface tension 140 dyne/cm, angle of contact $60^\circ $ and relative density 2 will rise in the same capillary by
A. 3 cm
B. 16 cm
C. 12 cm
D. 24 cm
Answer
267k+ views
Hint The increase of height in a capillary tube is given by $h = \dfrac{{2S\cos (\theta )}}{{r\rho g}}$ . In this question, all the variables expect the radius of the tube is different for both the cases. We need to find the ratio of the height of liquid rise in the 2nd tube to the height of the liquid rise in the 1st tube.
Complete Step by step solution
We know that rise in capillary tube is given by $h = \dfrac{{2S\cos (\theta )}}{{r\rho g}}$ , where S is the surface tension, $\theta $ is the angle of contact, $\rho $ is the density and r is the radius of capillary tube.
As the capillary tube is same, r and g will be same for both the cases,
Hence $\dfrac{{{h_1}}}{{{h_2}}} = \dfrac{{{S_1}\cos ({\theta _1})}}{{{S_2}\cos ({\theta _2})}} \times \dfrac{{{\rho _2}}}{{{\rho _1}}}$ ….. (i)
Given:
$
{S_1} = 70dyne/cm \\
{S_2} = 140dyne/cm \\
{\theta _1} = {0^o} \\
{\theta _2} = {60^o} \\
\dfrac{{{\rho _2}}}{{{\rho _1}}} = 2 \\
$
And
${h_1} = 6cm$
Now we put the all value in equation (1) and find the answer of question
${h_1} = 6cm$
after put all value in Equation (1) we get the value of capillary rise in second case
${h_2} = 3cm$
Hence the height of the capillary rise in second case will be 3 cm
Note We use the following formula $h = \dfrac{{2S\cos (\theta )}}{{r\rho g}}$ to get the rise in the capillary tube.
Applying this formula in both the cases, we get relation between heights of capillary rise in both the cases. And after putting all the values we get the rise in capillary for the second case that is \[{h_2} = 3cm\]
Complete Step by step solution
We know that rise in capillary tube is given by $h = \dfrac{{2S\cos (\theta )}}{{r\rho g}}$ , where S is the surface tension, $\theta $ is the angle of contact, $\rho $ is the density and r is the radius of capillary tube.
As the capillary tube is same, r and g will be same for both the cases,
Hence $\dfrac{{{h_1}}}{{{h_2}}} = \dfrac{{{S_1}\cos ({\theta _1})}}{{{S_2}\cos ({\theta _2})}} \times \dfrac{{{\rho _2}}}{{{\rho _1}}}$ ….. (i)
Given:
$
{S_1} = 70dyne/cm \\
{S_2} = 140dyne/cm \\
{\theta _1} = {0^o} \\
{\theta _2} = {60^o} \\
\dfrac{{{\rho _2}}}{{{\rho _1}}} = 2 \\
$
And
${h_1} = 6cm$
Now we put the all value in equation (1) and find the answer of question
${h_1} = 6cm$
after put all value in Equation (1) we get the value of capillary rise in second case
${h_2} = 3cm$
Hence the height of the capillary rise in second case will be 3 cm
Note We use the following formula $h = \dfrac{{2S\cos (\theta )}}{{r\rho g}}$ to get the rise in the capillary tube.
Applying this formula in both the cases, we get relation between heights of capillary rise in both the cases. And after putting all the values we get the rise in capillary for the second case that is \[{h_2} = 3cm\]
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