Two capillary tubes of same diameter are kept | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The height of a liquid column in a capillary tube is given by the formula $h = \frac{2T \cos 	heta}{r ho g}$,where $T$ is surface tension,$	he

Vedclass · Accepted Answer

$\frac{9}{10}$

Vedclass · Answer

(B) The height of a liquid column in a capillary tube is given by the formula $h = \frac{2T \cos 	heta}{r ho g}$,where $T$ is surface tension,$	heta$ is the angle of contact,$r$ is the radius of the tube,$ho$ is the density of the liquid,and $g$ is the acceleration due to gravity.Since the diameter (and thus radius $r$) is the same for both tubes,and the angle of contact $	heta$ is the same,the height $h$ is proportional to $\frac{T}{ho}$.Therefore,the ratio of heights is $\frac{h_1}{h_2} = \left(\frac{T_1}{T_2}ight) 	imes \left(\frac{ho_2}{ho_1}ight)$.Given $\frac{T_1}{T_2} = \frac{6}{5}$ and $\frac{ho_1}{ho_2} = \frac{4}{3}$,we have $\frac{ho_2}{ho_1} = \frac{3}{4}$.Substituting these values,we get $\frac{h_1}{h_2} = \left(\frac{6}{5}ight) 	imes \left(\frac{3}{4}ight) = \frac{18}{20} = \frac{9}{10}$.

Two capillary tubes of same diameter are kept vertically in two liquids whose densities are in the ratio $4:3$. If their surface tensions are in the ratio $6:5$,the ratio of heights $\left(\frac{h_1}{h_2}\right)$ of liquids in the two capillary tubes is (Their angle of contacts are same).

Similar Questions

If a capillary tube of radius $1 \,mm$ is immersed in water, the mass of water rising in the capillary tube is $m$. If the radius of the capillary tube is doubled, then the mass of water that rises in the same capillary tube will be

The height up to which water will rise in a capillary tube will be

In a capillary tube having area of cross-section $A$,the water rises to a height $h$. If the cross-sectional area is reduced to $\frac{A}{9}$,the rise of water in the capillary tube is

The radius of a capillary tube is $2 \times 10^{-3} \ m$. If a liquid of weight $6.28 \times 10^{-4} \ N$ can be supported in the capillary,what is the surface tension of the liquid?

$A$ capillary tube of radius $0.1 \,mm$ is dipped in water. The water rises to a height of $2 \,cm$ in the tube. If the surface tension of water is $0.072 \,N/m$, the contact angle between water and the wall of the tube is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module