Question

Water rises to a height \[h\] on a capillary tube of cross-section area \[A\]. The height to which water rises in capillary tube of cross-section area \[4A\] will be:
A. \[h\]
B. \[\dfrac{h}{2}\]
C. \[\dfrac{h}{4}\]
D. \[4h\]

Answer 1

Hint: This problem is solved with the help of capillary rise equation:
\[h=\dfrac{2T\operatorname{Cos}\theta }{r\partial Xg}\]
\[r=Radius\,of\,capillary\]
\[h=Height\,of\,capillary\]
\[T=Surface\,Tension\]
\[g=gravity\]
\[\theta =Angle\,of\,contact\]

Complete step by step solution:
Use the capillary rise equation
\[h=\dfrac{2T\operatorname{Cos}\theta }{r\partial Xg}\] …..(1)
Therefore from the equation (1)
Height of capillary is inversely proportional to the radius of the capillary
\[h\propto \dfrac{1}{r}\]
Area of the capillary
\[A=\pi {{r}^{2}}\] …..(2)
Where \[r=Radius\,of\,capillary\] and \[A=Area\,of\,capillary\]
Area of capillary is directly proportional to the radius of the capillary
\[\begin{align}
  & A\propto {{r}^{2}} \\
 & or \\
 & r\propto {{A}^{\dfrac{1}{2}}} \\
\end{align}\]
Substitute the value of r in the equation (2)
\[h\propto \dfrac{1}{{{A}^{\dfrac{1}{2}}}}\]
\[{{h}_{1}}=Initial\,height\,of\,capillary\]
\[{{h}_{2}}=Final\,height\,of\,capillary\]
Then, \[{{A}_{2}}=4{{A}_{1}}\] (given in question)
\[\dfrac{{{h}_{1}}}{{{h}_{2}}}={{(\dfrac{{{A}_{1}}}{{{A}_{2}}})}^{\dfrac{1}{2}}}\]
\[\begin{align}
  & \dfrac{{{h}_{1}}}{{{h}_{2}}}=2 \\
 & {{h}_{1}}=2{{h}_{2}} \\
 & {{h}_{2}}=\dfrac{{{h}_{1}}}{2} \\
\end{align}\]

Thus, option B is correct.

Note: Students should focus to use the equation of the area of the capillary.
Measurements of height should be taken very carefully.
Students should have knowledge about the capillary.