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A glass capillary tube is dipped in a liquid and the liquid rises to a height of 6 mm. The tube is then pushed down until only 4 mm projects above the liquid surface. In the compressed position, the meniscus makes an angle of 60 degrees with the capillary wall. Find the original contact angle between the glass and the liquid.
- cos⁻¹(3/4)
- cos⁻¹(3/7)
- 37 deg
- cos⁻¹(1/2)
Correct answer: cos⁻¹(3/4)
Solution
Original rise: h1 = 6 mm, contact angle theta0. Shortened: h2 = 4 mm (limited by tube length), new contact angle = 60 deg. Using h*cos(theta) = constant: 6*cos(theta0) = 4*cos(60 deg) = 4*(1/2) = 2. So cos(theta0) = 2/6 = 1/3? Wait: ratio gives 6*cos(theta0) = 4*cos(60). h1/h2 = cos(theta1)/cos(theta2) => h is proportional to cos(theta). So 6/4 = cos(theta2)/cos(theta0)... Let me redo: h = 2T*cos(theta)/(rho*g*r). h*1/cos(theta) = constant. So h1/cos(theta0) = h2/cos(60). 6/cos(theta0) = 4/(1/2) = 8. cos(theta0) = 6/8 = 3/4. theta0 = cos⁻¹(3/4).
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