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

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

& A\propto {{r}^{2}} \\\ & or \\\ & r\propto {{A}^{\dfrac{1}{2}}} \\\ \end{aligned}$$ Substitute the value of r in the equation (2) $$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{aligned} & \dfrac{{{h}_{1}}}{{{h}_{2}}}=2 \\\ & {{h}_{1}}=2{{h}_{2}} \\\ & {{h}_{2}}=\dfrac{{{h}_{1}}}{2} \\\ \end{aligned}$$ **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.