Three liquids have the same surface tension and densities $\rho_1, \rho_2$ and $\rho_3$ $(\rho_1 < \rho_2 < \rho_3)$. In three identical capillaries, the rise of liquid is the same. The corresponding angles of contact $\theta_1, \theta_2$ and $\theta_3$ are related as:
- A$\theta_1 < \theta_2 < \theta_3$
- B$\theta_1 > \theta_2 > \theta_3$
- C$\theta_1 = \theta_2 = \theta_3$
- D$\theta_1 > \theta_2 < \theta_3$
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Similar Questions
$A$ long capillary tube of mass $\pi \,g$,radius $2\,mm$ and negligible thickness,is partially immersed in a liquid of surface tension $0.1\,N/m$. Take the angle of contact as zero and neglect the buoyant force of the liquid. Find the force required to hold the tube vertically. $(g = 10\,m/s^2)$
Medium
View SolutionThree liquids of densities $\rho_1, \rho_2$ and $\rho_3$ (with $\rho_1 > \rho_2 > \rho_3$),having the same value of surface tension $T$,rise to the same height in three identical capillaries. The angles of contact $\theta_1, \theta_2$ and $\theta_3$ obey:
MediumNEET 2016
View SolutionLiquid $A$ rises to a height of $10 \ cm$ in a capillary tube and liquid $B$ falls to a depth of $2 \ cm$ in the same tube. The densities of $A$ and $B$ are $1 \ g/cm^3$ and $10 \ g/cm^3$ respectively. The contact angles of $A$ and $B$ with the tube are $0^{\circ}$ and $135^{\circ}$ respectively. If the surface tensions of $A$ and $B$ are $S_A$ and $S_B$, then the ratio $\frac{S_B}{S_A}$ is:
Two capillary tubes $A$ and $B$ of the same internal diameter are kept vertically in two different liquids whose densities are in the ratio $4:3$. If the surface tensions of these two liquids are in the ratio $6:5$,then the ratio of rise of liquid in capillary $A$ to that in $B$ is (assume their angles of contact are nearly equal).
$A$ capillary tube of radius $R$ is dipped in water,and water rises to a height $H$. The mass of water in the capillary tube is $M$. If a capillary tube of half the radius is dipped in water,what will be the mass of the water in the capillary tube?
Medium
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