Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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:

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

Energy $= 3VT\left(\frac{1}{r}+\frac{1}{R}\right)$ is released

Energy $= 3VT\left(\frac{1}{r}-\frac{1}{R}\right)$ is released

Energy is neither released nor absorbed

Step-by-Step Solution

Principle: When smaller drops coalesce to form a larger drop, the total volume remains conserved, but the total surface area decreases. The decrease in surface energy appears as released energy .
Formulas:

Volume of a sphere: $V = \frac{4}{3}\pi R^3 = n \cdot \frac{4}{3}\pi r^3$ (where $n$ is the number of small drops).
Surface Area: $A = 4\pi R^2$ .
Surface Energy: $E = T \times A$ .

Derivation:

Initial Surface Area ( $A_i$ ): $n(4\pi r^2)$ . Using $V = n\frac{4}{3}\pi r^3$ , we can substitute $n = \frac{3V}{4\pi r^3}$ . Thus, $A_i = \left(\frac{3V}{4\pi r^3}\right)4\pi r^2 = \frac{3V}{r}$ .
Final Surface Area ( $A_f$ ): $4\pi R^2$ . Using $V = \frac{4}{3}\pi R^3$ , we get $A_f = \frac{3V}{R}$ .
Change in Energy ( $\Delta E$ ): $T(A_i - A_f) = T\left(\frac{3V}{r} - \frac{3V}{R}\right) = 3VT\left(\frac{1}{r} - \frac{1}{R}\right)$ .

Conclusion: Since $r < R$ , the term $(\frac{1}{r} - \frac{1}{R})$ is positive, indicating energy is released due to the decrease in surface area.

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