question_answer

A certain number of spherical drops of a liquid of radius r coalesce to form a single drop of radius R and volume V. If T is the surface tension of the liquid, then                                                                          [NEET 2014]

A)  Energy \[=4VT\left( \frac{1}{r}-\frac{1}{R} \right)\] is released

B)  Energy \[=3VT\left( \frac{1}{r}+\frac{1}{R} \right)\] absorbed

C)

Energy \[=3VT\left( \frac{1}{r}-\frac{1}{R} \right)\] is released D)  Energy is neither released nor absorbed Correct Answer: C Solution :

Here, if the surface are a changes, it will change the surface energy as well. If the surface area decreases, it means that energy is released and vice-versa.

Change in surface energy \[\Delta A\times T\]                      (i)

Let we have 'n' number of drops initially.

So,       \[\Delta A=4\pi {{R}^{2}}-n(4\pi {{r}^{2}})\]               (ii)

Volume is constant

So,       \[n=\frac{4}{3}\pi {{r}^{3}}=\frac{4}{3}\pi {{R}^{3}}=V\]               (iii)

From Eqs. (ii) and (iii)

\[\Delta A=\frac{3}{R}\frac{4\pi }{3}\times {{R}^{3}}-\frac{3}{r}\left( n\frac{4\pi }{3{{r}^{3}}} \right)\]

\[=\frac{3}{R}\times V-\frac{3}{r}V\]

\[\Delta A=3V\left( \frac{1}{R}-\frac{1}{r} \right)=-ve\] value.

As \[R>r,\] so \[\Delta \,A\] is negative. It means surface area is decreased, so energy must be released.

Energy released \[=\Delta A\times T=-3VT\left( \frac{1}{r}-\frac{1}{R} \right)\]

Above expression shows the magnitude of energy released.