JEE Advance - Physics (2019 - Paper 1 Offline - No. 5)
A cylindrical capillary tube of 0.2 mm radius is made by joining two capillaries T1 and T2 of different materials having water contact angles of 0$$^\circ $$ and 60$$^\circ $$, respectively. The capillary tube is dipped vertically in water in two different configurations, case I and II as shown in figure. Which of the following option(s) is (are) correct? [Surface tension of water = 0.075 N/m, density of water = 1000 kg/m3, take g = 10 m/s2]
For case I, if the joint is kept at 8 cm above the water surface, the height of water colomn in the tube will be 7.5 cm. (Neglect the weight of the water in the meniscus).
For case I, if the capillary joint is 5 cm above the water surface, the height of water column raised in the tube will be more than 8.75 cm. (Neglect the weight of the water in the meniscus).
The correction in the height of water column raised in the tube, due to weight of water contained in the meniscus, will be different for both cases.
For case II, if the capillary joint is 5 cm above the water surface, the height of water column raised in the tube will be 3.75 cm. (Neglect the weight of the water in the meniscus).
Explanation
Heights if only single material tubes are used of sufficient length,
$${h_1} = {{2R\cos \theta } \over {\rho rg}}$$
$$ = {{2 \times 0.075 \times \cos 0^\circ } \over {1000 \times 2 \times {{10}^{ - 4}} \times 10}} = 7.5\,cm$$
$${h_2} = {{2T\cos \theta '} \over {\rho rg}}$$
$$ = {{2 \times 0.075 \times \cos 60^\circ } \over {1000 \times 2 \times {{10}^{ - 4}} \times 10}} = 3.75\,cm$$
Case - I
Case - II
2. Liquid will rise only upto height of 5 cm and meniscus will adjust by changing is radius of curvature. If the liquid goes up in tube 2 then it will not be able to support the weight of the liquid.
3. Weight of water in meniscus will be different in two cases because angle of contact is different.
$${h_1} = {{2R\cos \theta } \over {\rho rg}}$$
$$ = {{2 \times 0.075 \times \cos 0^\circ } \over {1000 \times 2 \times {{10}^{ - 4}} \times 10}} = 7.5\,cm$$
$${h_2} = {{2T\cos \theta '} \over {\rho rg}}$$
$$ = {{2 \times 0.075 \times \cos 60^\circ } \over {1000 \times 2 \times {{10}^{ - 4}} \times 10}} = 3.75\,cm$$
Case - I
Case - II
2. Liquid will rise only upto height of 5 cm and meniscus will adjust by changing is radius of curvature. If the liquid goes up in tube 2 then it will not be able to support the weight of the liquid.
3. Weight of water in meniscus will be different in two cases because angle of contact is different.
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