When a capillary tube is dipped into two liqu | vedclass

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(D) The height of a liquid in a capillary tube is given by the formula $h = \frac{2T \cos 	heta}{rdg}$,where $T$ is surface tension,$r$ is the radius

Vedclass · Accepted Answer

$\frac{9}{10}$

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(D) The height of a liquid in a capillary tube is given by the formula $h = \frac{2T \cos 	heta}{rdg}$,where $T$ is surface tension,$r$ is the radius of the tube,$d$ is the density,and $g$ is the acceleration due to gravity.Assuming the contact angle $	heta$ and the radius $r$ are the same for both liquids,the ratio of heights is given by:$\frac{h_1}{h_2} = \left( \frac{T_1}{T_2} ight) 	imes \left( \frac{d_2}{d_1} ight)$Given $T_1 = 60 \, dyne/cm$,$T_2 = 50 \, dyne/cm$,$d_1 = 0.8 \, g/cm^3$,and $d_2 = 0.6 \, g/cm^3$:$\frac{h_1}{h_2} = \left( \frac{60}{50} ight) 	imes \left( \frac{0.6}{0.8} ight)$$\frac{h_1}{h_2} = \left( \frac{6}{5} ight) 	imes \left( \frac{6}{8} ight) = \frac{36}{40} = \frac{9}{10}$.

When a capillary tube is dipped into two liquids having relative densities $0.8$ and $0.6$ and surface tensions $60 \, dyne/cm$ and $50 \, dyne/cm$ respectively,the ratio of the heights of the liquids in the capillary tube $\frac{h_1}{h_2}$ is:

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