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What is the surface tension of a liquid, given the depression of the meniscus in a capillary tube dipped in the liquid is $0.03 \mathrm{~cm}$ ? The diameter of the capillary tube is $0.5 \mathrm{~mm}$ , the density of the liquid is $7.5 \times \frac{10^3 \mathrm{~kg}}{\mathrm{~m}^3}$ , the angle of contact of the liquid with the tube is $60^{\circ}$ , and the gravitational acceleration is $10 \frac{\mathrm{m}}{\mathrm{s}^2}$ .
Option: 1
$0.38 \frac{\mathrm{N}}{\mathrm{m}}$

Option: 2
$5.625 \times 10^{-3} \frac{\mathrm{N}}{\mathrm{m}}$

Option: 3
$0.48 \frac{\mathrm{N}}{\mathrm{m}}$

Option: 4
$0.54 \frac{\mathrm{N}}{\mathrm{m}}$

Answers (1)

best_answer

The surface tension of the liquid can be calculated using the expression:

$T=\frac{\rho g d r}{2 \cos \theta}$

Where $T$ is the surface tension of the liquid, $\rho$ is the density of the liquid, $g$ is the acceleration due to gravity, $\mathrm{d}$ is the depression of the meniscus in the capillary tube, $\mathrm{r}$ is the radius of the capillary tube, and $\theta$ is the angle of contact of the liquid with the tube.

Substituting the given values, we get:

$T=\frac{7.5 \times 10^3 \times 10 \times 0.0003 \times 0.00025}{2 \times \cos 60^{\circ}}$

$T=5.625 \times 10^{-3} \frac{N}{m}$

Therefore, the surface tension of the liquid is $0.42 \frac{\mathrm{N}}{\mathrm{m}}$ , and the correct answer is option B).

Posted by

Deependra Verma

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