Correct Answer - Option 3 : 15 mm
Concept:
Capillary rise of water due to surface tension: If a glass tube of small diameter is partially immersed in water, the water will wet the surface of the tube and it will rise in the tube to some height, above the normal water surface. So, there will be a height difference in water surfaces within and outside the capillary tube. This phenomena occurs due to surface tension of water and known as capillary rise of water.
The capillary rise of water h is given by,
\({\rm{h}} = {\rm{}}\frac{{4{\rm{\sigma }}\cos {\rm{\theta }}}}{{{{\rm{\gamma }}_{\rm{w}}}{\rm{d}}}}\)
Where, σ is surface tension of water, θ is the contact angle between water and glass surface, is unit weight of water and d is the diameter of the capillary tube.
Calculation:
Given, surface tension of water = \(75 \times {10^{ - 3}}\) N/m.
If nothing is mentioned about contact angle, then contribution from contact angle may be ignored.
Unit weight of water, \({{\rm{\gamma }}_{\rm{w}}} = 9.81{\rm{\;kN}}/{\rm{\;}}{{\rm{m}}^3}{\rm{\;}} = 9810{\rm{\;N}}/{{\rm{m}}^3}\)
Note: Ideally in the question statement it should be clearly mentioned whether it is internal bore diameter or internal bore radius.
Let’s assume given 1 mm is the internal bore radius. \(\therefore {\rm{\;Diameter}},{\rm{\;d\;}} = {\rm{\;}}2{\rm{\;mm}} = 2 \times {10^{ - 3}}{\rm{\;m}}\) .
\(\therefore {\rm{h}} = {\rm{\;}}\frac{{4{\rm{\sigma }}\cos {\rm{\theta }}}}{{{{\rm{\gamma }}_{\rm{w}}}{\rm{d}}}} \approx {\rm{\;}}\frac{{4{\rm{\sigma }}}}{{{{\rm{\gamma }}_{\rm{w}}}{\rm{d}}}} = \frac{{4 \times 75 \times {{10}^{ - 3}}}}{{9810 \times 2 \times {{10}^{ - 3}}}} = 0.015{\rm{\;m}} = 15{\rm{\;mm}}\)
∴ So, the difference in water surface will be 15 mm.
Note: If given 1 mm is taken as internal bore diameter then the answer would have been 30 mm and does not match with the options provided.
