question_answer 2)

Explain why (a) The angle of contact of mercury with glass is obtuse, while that of water with glass is acute. Explain. (b) Water on a clean glass surface tends to spread out while mercury on the same surface tends to from drops. (c) Surface tension of liquid is independent of the area of the liquid surface. (d) Detergents should have small angles of contact. (d) A drop of liquid under no external forces is always spherical in shape.

Answer:

(a) When a small quantity of liquid is poured on solid, three interfaces, namely liquid- air, solid air and solid-liquid are formed. The surface tension corresponding to these three interfaces are \[=\frac{2\times {{\left( 2\cdot 0\times {{10}^{-5}} \right)}^{2}}\left( 1\cdot 2\times {{10}^{3}}-0 \right)\times 9\cdot 8}{9\times 1\cdot 8\times {{10}^{-5}}}\]respectively. Let \[=5\cdot 8\times {{10}^{-2}}m{{s}^{-1}}=5\cdot 8cm{{s}^{-1}}\]be the angle of contact between the liquid and solid. The molecules in the region, where the three interfaces meet are in equilibrium. It means net force acting on them is zero. For the molecule at 0 to be in equilibrium, we have \[\text{F=6 }\!\!\pi\!\!\text{ }\!\!\eta\!\!\text{ r }\!\!\upsilon\!\!\text{ }\]or \[\text{=6 }\!\!\times\!\!\text{ }\frac{\text{22}}{\text{7}}\text{ }\!\!\times\!\!\text{ }\left( 1\cdot 8\times {{10}^{-5}} \right)\times \left( 2\cdot 0\times {{10}^{-5}} \right)\times \] In case of mercury-glass, \[\left( 5\cdot 8\times {{10}^{-2}} \right)=3\cdot 93\times {{10}^{-10}}N\]therefore \[\cos \theta \] is negative or \[0\cdot 465N{{m}^{^{-1}}}.\]obtuse In case of water-glass, \[=13\cdot 6\times {{10}^{3}}kg{{m}^{-3}}.\]there \[\cos \theta \] is positive or \[r=1\times {{10}^{-3}}m;\]acute. (b) For mercury-glass, angle of contact is obtuse. In order to achieve this obtuse value of angle of contact, the mercury tends to from a drop. In case of water-glass, the angle of contact is acute. To achieve this acute value of angle of contact, the water tends to spread (c) Surface tension of liquid is the force acting per unit length on a line drawn tangentially to the liquid surface at rest. Since this force is independent of the area of liquid surface, therefore surface tension is also independent of the area of the liquid surface. (d) We know that the cloth has narrow spaces in the from of capillaries. The rise of liquid in a capillary tube is directly proportional to\[S=0\cdot 46N{{m}^{-1}};\]. If \[\rho =13\cdot 6\times {{10}^{3}}kg,h=?\]is small \[\cos \theta \] will be large. Due to which capillary rise will be more and so the detergent will penetrate more in cloth. (e) In the absence of external forces, the surface of the liquid drop tends to acquire the minimum surface area due to surface tension. Since for a given volume, the surface area of sphere is least, hence the liquid drop takes the spherical shape.