When a liquid rises in a capillary tube, the height of the rise is inversely proportional to the density of the liquid and directly proportional to the surface tension.
The angle of contact (\( \theta \)) of the liquid with the capillary wall is also related to the density of the liquid. For a liquid with higher density, the angle of contact tends to be greater. In this case, we have three liquids with densities \( P_1 \), \( P_2 \), and \( P_3 \) such that \( P_1<P_2<P_3 \). According to the relationship between the angle of contact and density: - The liquid with the lowest density (\( P_1 \)) will have the highest angle of contact. - The liquid with the highest density (\( P_3 \)) will have the smallest angle of contact.
Therefore, the relation between the angles of contact is: \[ 0<\theta_3<\theta_2<\theta_1<\frac{\pi}{2} \]
Thus, the correct answer is option A.
Two soap bubbles of radius 2 cm and 4 cm, respectively, are in contact with each other. The radius of curvature of the common surface, in cm, is _______________.
200 ml of an aqueous solution contains 3.6 g of Glucose and 1.2 g of Urea maintained at a temperature equal to 27$^{\circ}$C. What is the Osmotic pressure of the solution in atmosphere units?
Given Data R = 0.082 L atm K$^{-1}$ mol$^{-1}$
Molecular Formula: Glucose = C$_6$H$_{12}$O$_6$, Urea = NH$_2$CONH$_2$