(C) The height of the liquid rise in a capillary tube is given by the formula: $h = \frac{2T \cos \theta}{r \rho g}$.Given that $h, T, r$,and $g$ are

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$\theta_1 < \theta_2 < \theta_3$

(C) The height of the liquid rise in a capillary tube is given by the formula: $h = \frac{2T \cos \theta}{r \rho g}$.Given that $h, T, r$,and $g$ are constant for all three liquids.Therefore,$\frac{\cos \theta}{\rho} = \text{constant}$.This implies: $\frac{\cos \theta_1}{\rho_1} = \frac{\cos \theta_2}{\rho_2} = \frac{\cos \theta_3}{\rho_3}$.Since $\rho_1 > \rho_2 > \rho_3$,it follows that $\cos \theta_1 > \cos \theta_2 > \cos \theta_3$.Because the cosine function is a decreasing function in the range $[0, \pi/2]$,a larger cosine value corresponds to a smaller angle.Therefore,$\theta_1 < \theta_2 < \theta_3$.

Class 11 Physics Fluid Mechanics and Surface Tension Capillary Tube and Capillarity

Easy

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 the liquid is the same. The corresponding angles of contact $\theta_1, \theta_2$,and $\theta_3$ are related as:

MHT CET 2023 All MHT CET

A
$\theta_1 > \theta_2 > \theta_3$
B
$\theta_1 > \theta_3 > \theta_2$
C
$\theta_1 < \theta_2 < \theta_3$
D
$\theta_1 = \theta_2 = \theta_3$

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Fluid Mechanics and Surface Tension

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Capillary Tube and Capillarity

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Water rises in a capillary tube to a certain height such that the upward force due to surface tension is balanced by $75 \times 10^{-4} \, N$ force due to the weight of the liquid. If the surface tension of water is $6 \times 10^{-2} \, N m^{-1}$,the inner circumference of the capillary must be

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The heat evolved for the rise of water when one end of the capillary tube of radius $r$ is immersed vertically into water is (Assume surface tension $= T$ and density of water to be $\rho$)

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Two capillary tubes of same diameter are put vertically one each in two liquids whose relative densities are $0.8$ and $0.6$ and surface tensions are $60$ and $50 \text{ dyne/cm}$ respectively. The ratio of the heights of the liquids in the two tubes $\frac{h_1}{h_2}$ is:

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$A$ 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?

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$A$ glass capillary tube of inner diameter $0.28 \,mm$ is lowered vertically into water in a vessel. The pressure to be applied on the water in the capillary tube so that the water level in the tube is the same as that in the vessel is ............ $\times 10^3 \,N/m^2$ (surface tension of water $= 0.07 \,N/m$ and atmospheric pressure $= 10^5 \,N/m^2$).

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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 the liquid is the same. The corresponding angles of contact $\theta_1, \theta_2$,and $\theta_3$ are related as:

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Water rises in a capillary tube to a certain height such that the upward force due to surface tension is balanced by $75 \times 10^{-4} \, N$ force due to the weight of the liquid. If the surface tension of water is $6 \times 10^{-2} \, N m^{-1}$,the inner circumference of the capillary must be

The heat evolved for the rise of water when one end of the capillary tube of radius $r$ is immersed vertically into water is (Assume surface tension $= T$ and density of water to be $\rho$)

Two capillary tubes of same diameter are put vertically one each in two liquids whose relative densities are $0.8$ and $0.6$ and surface tensions are $60$ and $50 \text{ dyne/cm}$ respectively. The ratio of the heights of the liquids in the two tubes $\frac{h_1}{h_2}$ is:

$A$ 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?

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