A large number water drops radius |HEATTRANSFERJEEMain2021

JEE MAIN 2021

26-02-2021 S1

Question

A large number of water drops, each of radius r, combine to have a drop of radius R. If the surface tension is T and mechanical equivalent of heat is J, the rise in heat energy per unit volume will be:

Select the correct option:

$\frac{2 T}{J}\left(\frac{1}{r}-\frac{1}{R}\right)$

$\frac{2 T}{r J}$

$\frac{3 T}{r J}$

$\frac{3 T}{J}\left(\frac{1}{r}-\frac{1}{R}\right)$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$\begin{aligned} & n \times \frac{4}{3} \pi r^3=\frac{4}{3} \pi R^3 \\ & \quad \therefore n^{1 / 3} r=R \\ & \quad \begin{aligned} & \text { Total change in surface energy } \\ & \quad=\left(n\left(4 \pi r^2\right)-4 \pi R^2\right) T\end{aligned} \\ & \quad \Rightarrow \\ & \quad 4 \pi T\left(n r^2-R^2\right) \\ & \quad \therefore \text { Heat energy } \\ & \quad= \\ & \quad \frac{4 \pi T\left(n r^2-R^2\right)}{J \times \frac{4}{3} \pi R^3}=\frac{3 T}{J}\left(\frac{n r^2}{R^3}-\frac{1}{R}\right) \\ & \text { Put nr³ }=R^3 \\ & \quad \therefore \frac{3 T}{J}\left(\frac{1}{r}-\frac{1}{R}\right)\end{aligned}$

Question Tags

JEE Main

Physics

Hard

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