Question

The lower end of capillary tube is immersed in mercury. The level of mercury in the tube is found to be 2 cm below the outer level. If the same tube is immersed in water, upto what height(in cm) will the water rise in the capillary?

(Density of mercury $=13.6g/c{m}^3$ ,Density of wate$r=1g/c{m}^3$,Surface tension of Mercury $=484.5dyne/cm$, Surface tension of water=$72.7dyne/cm)$

• A. 5.9
• B. 4.9
• C. 2.9
• D. 1.9

Answer 1

The correct option is A 5.9

Given data:

Level of mercury in the tube is, $h^{\prime }=2cm$.

Level of mercury in the tube is given as:

$h^{\prime }=\frac{2T^{\prime }cos{\theta }^{\prime }}{r{\rho }^{\prime }g}$ ................(1)

where ${\rho }^{\prime }$ is the density of mercury.

Level of water in the tube is given as:

$h=\frac{2Tcos\theta }{r\rho g}$ ................(2)

where $\rho$ is the density of water.

Here, the same capillary is immersed in both the liquids hence, the radius of the capillary is same i.e. $r$

Using the standard values for variables and taking ${\theta }^{\prime }={135}^o$, in equation (1) gives:

$2=\frac{2\left(484.5\right)cos{135}^{\circ }}{r\left(13.6\right)\left(980\right)}$

$r=\frac{685.083}{\left(26656\right)}$

$r=0.025cm$

Using value of $r$ and $\theta =0^o$ in equation (2), gives:

$h=\frac{2\left(72.7\right)cos{0}^{\circ }}{\left(0.025\right)\left(1\right)\left(980\right)}$

$h=\frac{145.4}{\left(24.5\right)}$

$h=5.9cm$

Hence, level of water in the tube is $5.9$ cm.