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Question
The lower end of a 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, up to what height will the water rise in the capillary?
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Solution
Let T be the surface tension, r be the inner radius of the capillary tube and ρ bethe density of the liquid.
For cos θ = 1, height (h) of the liquid level is given as:
\[\text{h} = \frac{2\text{T}\cos\theta}{\text{r}\rho \text{ g}}\]
Now, for mercury:
\[\text{h}_{\text{Hg}} = \frac{2\text{T}_{\text{Hg}}cosθ}{\text{r}\rho_{\text{Hg}} \text{ g}}\] .........(i)
For water:
\[\text{h}_\text{w} = \frac{2 \text{T}_\text{w}cosθ}{\text{r} \rho_\text{w g}}\] ...(ii)
Dividing (ii) by (i), we get:
\[\frac{h_w}{h_{Hg}} = \frac{T_w}{T_{Hg}} \times \frac{\rho_{Hg}}{\rho_w}\] ` xx (cos0°)/(cos 140°)`
`h_w/(- 0.02) = (7.5 xx 10^-2 xx 1 xx 13.6 xx 10^3)/(10^3 xx cos 140 xx 0.465)`
hw = 0.0576 m
hw = 5.73 cm
Hence, the required rise in the water level in the capillary tube is 5.73 cm.
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