Current Affairs JEE Main & Advanced

If n small drops of radius r coalesce to form a big drop of radius R then surface area of the liquid decreases. Amount of surface energy released = Initial surface energy ? final surface energy                                         \[E=n4\pi {{r}^{2}}T-4\pi {{R}^{2}}T\]    

Various formulae of released energy
\[4\pi T[n{{r}^{2}}-{{R}^{2}}]\]	\[4\pi {{R}^{2}}T({{n}^{1/3}}-1)\]	\[4\pi T{{r}^{2}}{{n}^{2/3}}({{n}^{1/3}}-1)\]	\[4\pi T{{R}^{3}}\left[ \frac{1}{r}-\frac{1}{R} \right]\]

(i) If this released energy is absorbed by a big drop, its temperature increases and rise in temperature can be given by \[\Delta \theta =\frac{3T}{JSd}\left[ \frac{1}{r}-\frac{1}{R} \right]\] (ii) If this released energy is converted into kinetic energy of a big drop without dissipation then by the law of conservation of energy. \[\frac{1}{2}m{{v}^{2}}=4\pi {{R}^{3}}T\left[ \frac{1}{r}-\frac{1}{R} \right]\]     \[\Rightarrow \] \[\frac{1}{2}\left[ \frac{4}{3}\pi {{R}^{3}}d \right]{{v}^{2}}=4\pi {{R}^{3}}T\left[ \frac{1}{r}-\frac{1}{R} \right]\]     \[\Rightarrow \] \[{{v}^{2}}=\frac{6T}{d}\left[ \frac{1}{r}-\frac{1}{R} \right]\]  \[\therefore \]  \[v=\sqrt{\frac{6T}{d}\left( \frac{1}{r}-\frac{1}{R} \right)}\]  

When a drop of radius R splits into n smaller drops, (each of radius r) then surface area of liquid increases. Hence the work is to be done against surface tension.   Since the volume of liquid remains constant therefore \[\frac{4}{3}\pi {{R}^{3}}=n\frac{4}{3}\pi {{r}^{3}}\]  \[\therefore \] \[{{R}^{3}}=n{{r}^{3}}\] Work done \[=T\times \Delta A=T\times \] [Total final surface area of \[n\] drops \[-\] surface area of big drop] \[=T[n4\pi {{r}^{2}}-4\pi {{R}^{2}}]\]  

Various formulae of work done
\[4\pi T[n{{r}^{2}}-{{R}^{2}}]\]	\[4\pi {{R}^{2}}T[{{n}^{1/3}}-1]\]	\[4\pi T{{r}^{2}}{{n}^{2/3}}[{{n}^{1/3}}-1]\]	\[4\pi T{{R}^{3}}\left[ \frac{1}{r}-\frac{1}{R} \right]\]

If the work is not done by an external source then internal energy of liquid decreases, subsequently temperature decreases. This is the reason why spraying causes cooling. By conservation of energy, Loss in thermal energy = work done against surface tension \[JQ=W\] \[\Rightarrow \] \[JmS\Delta \theta =4\pi T{{R}^{3}}\left[ \frac{1}{r}-\frac{1}{R} \right]\] \[\Rightarrow \] \[\frac{4}{3}\pi {{R}^{3}}d\,S\Delta \theta =4\pi {{R}^{3}}T\left[ \frac{1}{r}-\frac{1}{R} \right]\]                        \[[\text{As}\,\,\,m\,=V\times d=\frac{4}{3}\pi \,{{R}^{3}}\times d]\] \[\therefore \] Decrease in temperature  \[\Delta \theta =\frac{3T}{JSd}\left[ \frac{1}{r}-\frac{1}{R} \right]\] where \[J=\] mechanical equivalent of heat, \[S=\] specific heat of liquid, \[d=\] density of liquid.  

(1) If the initial radius of liquid drop is \[{{r}_{1}}\] and final radius of liquid drop is \[{{r}_{2}}\] then \[W=T\times \]  Increment in surface area \[W=T\times 4\pi [r_{2}^{2}-r_{1}^{2}]\]                             [drop has only one free surface] (2) In case of soap bubble \[W=T\times 8\pi [r_{2}^{2}-r_{1}^{2}]\]                             [Bubble has two free surfaces]  

The molecules on the liquid surface experience net downward force. So to bring a molecule from the interior of the liquid to the free surface, some work is required to be done against the intermolecular force of attraction, which will be stored as potential energy of the molecule on the surface. The potential energy of surface molecules per unit area of the surface is called surface energy. Unit : \[Joule/{{m}^{2}}\] (S.I.) \[erg/c{{m}^{2}}\] (C.G.S.) Dimension : \[[M{{T}^{2}}]\] If a rectangular wire frame ABCD, equipped with a sliding wire LM dipped in soap solution, a film is formed over the frame. Due to the surface tension, the film will have a tendency to shrink and thereby, the sliding wire LM will be pulled in inward direction. However, the sliding wire can be held in this position under a force F, which is equal and opposite to the force acting on the sliding wire LM all along its length due to surface tension in the soap film. If T is the force due to surface tension per unit length, then \[F=T\times 2l\] Here l is length of the sliding wire \[LM\]. The length of the sliding wire has been taken as \[2l\] for the reason that the film has got two free surfaces. Suppose that the sliding wire LM is moved through a small distance x, so as to take the position \[L'M'\]. In this process, area of the film increases by \[2l\times x\] (on the two sides) and to do so, the work done is given by \[W=F\times x=(T\times 2l)\times x=T\times (2lx)=T\times \Delta A\] \[\therefore \] \[W=T\times \Delta A\] [\[\Delta A=\] Total increase in area of the film] If temperature of the film remains constant in this process, this work done is stored in the film as its surface energy. From the above expression \[T=\frac{W}{\Delta A}\] or \[T=W\] [If  \[\Delta A=1\]] i.e. surface tension may be defined as the amount of work done in increasing the area of the liquid surface by unity against the force of surface tension at constant temperature.

The maximum distance upto which the force of attraction between two molecules is appreciable is called molecular range \[(\approx {{10}^{-9}}m)\]. A sphere with a molecule as centre and radius equal to molecular range is called the sphere of influence. The liquid enclosed between free surface (PQ) of the liquid and an imaginary plane (RS) at a distance r (equal to molecular range) from the free surface of the liquid form a liquid film. To understand the concept of tension acting on the free surface of a liquid, let us consider four liquid molecules like A, B, C and D. Their sphere of influence are shown in the figure. (1) Molecule A is well within the liquid, so it is attracted equally in all directions. Hence the net force on this molecule is zero and it moves freely inside the liquid. (2) Molecule B is little below the free surface of the liquid and it is also attracted equally in all directions. Hence the resultant force acts on it is also zero. (3) Molecule C is just below the upper surface of the liquid film and the part of its sphere of influence is outside the free liquid surface. So the number of molecules in the upper half (attracting the molecules upward) is less than the number of molecule in the lower half (attracting the molecule downward). Thus the molecule C experiences a net downward force. (4) Molecule D is just on the free surface of the liquid. The upper half of the sphere of influence has no liquid molecule. Hence the molecule D experiences a maximum downward force. Thus all molecules lying on surface film experiences a net downward force. Therefore, free surface of the liquid behaves like a stretched membrane.  

(1) The oil and grease spots on clothes cannot be removed by pure water. On the other hand, when detergents (like soap) are added in water, the surface tension of water decreases. As a result of this, wetting power of soap solution increases. Also the force of adhesion between soap solution and oil or grease on the clothes increases. Thus, oil, grease and dirt particles get mixed with soap solution easily. Hence clothes are washed easily. (2) The antiseptics have very low value of surface tension. The low value of surface tension prevents the formation of drops that may otherwise block the entrance to skin or a wound. Due to low surface tension, the antiseptics spreads properly over wound. (3) Surface tension of all lubricating oils and paints is kept low so that they spread over a large area. (4) Oil spreads over the surface of water because the surface tension of oil is less than the surface tension of cold water. (5) A rough sea can be calmed by pouring oil on its surface. (6) In soldering, addition of ?flux? reduces the surface tension of molten tin, hence, it spreads.