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Question: A water drop of radius 1mm is sprayed into \({{10}^{6}}\)droplets of the same size at constant tempe...

A water drop of radius 1mm is sprayed into 106{{10}^{6}}droplets of the same size at constant temperature. If the surface tension of water is 72×10−3N/m72\times {{10}^{-3}}N/m, then the work done is
a)8.95×10−5ergs b)8.95×10−5J c)17.9×10−5J d)17.9×10−5ergs \begin{aligned} & a)8.95\times {{10}^{-5}}ergs \\\ & b)8.95\times {{10}^{-5}}J \\\ & c)17.9\times {{10}^{-5}}J \\\ & d)17.9\times {{10}^{-5}}ergs \\\ \end{aligned}

Explanation

Solution

It is given that a big drop let us say of radius R is sprayed into small drops of radius r. When the big drop is sprayed into small drops, the total surface area of the drops with respect to the initial big drop increases. Since the surface tension of the liquid remains the same, the total surface energy of the small drops with respect to the big drop increases. Hence we have to do the required work to increase the surface energy of the small drops. Therefore the difference in the surface energies of the big drop and the small drops will give us the required amount of work done.
Formula used:
Let us say a drop of liquid with tension(T) and radius r is in the form of a sphere. Therefore the surface energy of the drop is given by E=4πr2TE=4\pi {{r}^{2}}T. If there are n drops in the system than the total surface energy is equal to E=n4πr2TE=n4\pi {{r}^{2}}T

Complete answer:
Let the radius of the bigger drop be R and the radius of each small drops be r. When the liquid is sprayed the volume of the liquid remains constant. Therefore we can write volume of the bigger spherical drop is equal to the number of small drops i.e. N times the volume of each small drop. This can be mathematically represented as,
43πR3=N43πr3 ⇒R3=Nr3 ⇒R=(N)13r ⇒r=R(N)−13 \begin{aligned} & \dfrac{4}{3}\pi {{R}^{3}}=N\dfrac{4}{3}\pi {{r}^{3}} \\\ & \Rightarrow {{R}^{3}}=N{{r}^{3}} \\\ & \Rightarrow R={{\left( N \right)}^{\dfrac{1}{3}}}r \\\ & \Rightarrow r=R{{\left( N \right)}^{-\dfrac{1}{3}}} \\\ \end{aligned}
Now since we know that the work done the difference in the surface energies of the big drop and the small drops, this can be mathematically written as,
Work done=N4πr2T−4πR2T\text{Work done}=N4\pi {{r}^{2}}T-4\pi {{R}^{2}}T
Substituting for radius of the small drops in the above equation we get,
Work done=N4πr2T−4πR2T Work done=N4πR2((N)−13)2T−4πR2T ⇒Work done=4πR2T[N(N)−23−1] ⇒Work done=4πR2TN13 \begin{aligned} & \text{Work done}=N4\pi {{r}^{2}}T-4\pi {{R}^{2}}T \\\ & \text{Work done}=N4\pi {{R}^{2}}{{\left( {{\left( N \right)}^{-\dfrac{1}{3}}} \right)}^{2}}T-4\pi {{R}^{2}}T \\\ & \Rightarrow \text{Work done}=4\pi {{R}^{2}}T\left[ N{{\left( N \right)}^{-\dfrac{2}{3}}}-1 \right] \\\ & \Rightarrow \text{Work done}=4\pi {{R}^{2}}T{{N}^{\dfrac{1}{3}}} \\\ \end{aligned}
The radius of the bigger drop i.e. R= 1mm, T=72×10−3N/mT=72\times {{10}^{-3}}N/mand N=106N={{10}^{6}} is given to us in the question. Hence the value of work done is,
Work done=4πR2TN13 Work done=4π(1×10−3)2×72×10−3×(106)13 ⇒Work done=4π×72×10−6 ⇒Work done=4π×72×10−6 ⇒Work done=8.95×10−5J \begin{aligned} & \text{Work done}=4\pi {{R}^{2}}T{{N}^{\dfrac{1}{3}}} \\\ & \text{Work done}=4\pi {{\left( 1\times {{10}^{-3}} \right)}^{2}}\times 72\times {{10}^{-3}}\times {{\left( {{10}^{6}} \right)}^{\dfrac{1}{3}}} \\\ & \Rightarrow \text{Work done}=4\pi \times 72\times {{10}^{-6}} \\\ & \Rightarrow \text{Work done}=4\pi \times 72\times {{10}^{-6}} \\\ & \Rightarrow \text{Work done}=8.95\times {{10}^{-5}}J \\\ \end{aligned}

Hence the correct answer of the above question is option b.

Note:
It is to be noted that the Joules (J) is the SI unit of energy and the above quantities which are used to obtain the work done are in SI units. Hence the Energy obtained is in the form of SI units. It is to be noted that the energy having dimension ergs is in the CGS system.