Question:
Calculate the amount of work done in spraying a drop of a liquid of radius 1 mm into million identical droplets under isothermal conditions. [surface tension of liquid $=550\times10^{-3}\text{Nm}^{-1}$]
Calculate the amount of work done in spraying a drop of a liquid of radius 1 mm into million identical droplets under isothermal conditions. [surface tension of liquid $=550\times10^{-3}\text{Nm}^{-1}$]
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Shortcut formula for splitting drops: $W = 4\pi R^2 T (n^{1/3} - 1)$. For $10^6$ drops, $n^{1/3} = 100$, making the term $(100 - 1) = 99$.
Updated On: Jun 3, 2026
- $6.84\times10^{-4} \text{ J}$
- $5.50\times10^{-4} \text{ J}$
- $7.25\times10^{-4} \text{ J}$
- $3.42\times10^{-4} \text{ J}$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
The work done ($W$) in spraying a larger fluid drop into multiple smaller droplets is equal to the increase in surface energy: $W = T \cdot \Delta A = T(n \cdot 4\pi r^2 - 4\pi R^2)$, where $T$ is the surface tension, $R$ is the initial radius, $n$ is the number of small droplets, and $r$ is the radius of each droplet.
Step 2: Meaning
Since total volume remains constant during the conversion process, we have $\frac{4}{3}\pi R^3 = n \cdot \frac{4}{3}\pi r^3$, which simplifies to $R = n^{1/3}r$, or $r = R \cdot n^{-1/3}$.
Step 3: Analysis
Given $R = 1 \text{ mm} = 10^{-3} \text{ m}$, $n = 10^6$ (one million), and $T = 550 \times 10^{-3} \text{ Nm}^{-1}$. Thus, $r = 10^{-3} \times (10^6)^{-1/3} = 10^{-5} \text{ m}$. The total change in surface area is $\Delta A = 4\pi (n r^2 - R^2) = 4\pi (10^6 \times 10^{-10} - 10^{-6}) = 4\pi (10^{-4} - 10^{-6}) = 4\pi \times 10^{-6}(100 - 1) = 4\pi \times 99 \times 10^{-6} \text{ m}^2$. Now, calculating work done: $W = 550 \times 10^{-3} \times 4 \times 3.1416 \times 99 \times 10^{-6} \approx 6.84 \times 10^{-4} \text{ J}$.
Step 4: Conclusion
The total amount of work done under these isothermal scaling parameters equals $6.84 \times 10^{-4} \text{ J}$.
Final Answer: (A)
The work done ($W$) in spraying a larger fluid drop into multiple smaller droplets is equal to the increase in surface energy: $W = T \cdot \Delta A = T(n \cdot 4\pi r^2 - 4\pi R^2)$, where $T$ is the surface tension, $R$ is the initial radius, $n$ is the number of small droplets, and $r$ is the radius of each droplet.
Step 2: Meaning
Since total volume remains constant during the conversion process, we have $\frac{4}{3}\pi R^3 = n \cdot \frac{4}{3}\pi r^3$, which simplifies to $R = n^{1/3}r$, or $r = R \cdot n^{-1/3}$.
Step 3: Analysis
Given $R = 1 \text{ mm} = 10^{-3} \text{ m}$, $n = 10^6$ (one million), and $T = 550 \times 10^{-3} \text{ Nm}^{-1}$. Thus, $r = 10^{-3} \times (10^6)^{-1/3} = 10^{-5} \text{ m}$. The total change in surface area is $\Delta A = 4\pi (n r^2 - R^2) = 4\pi (10^6 \times 10^{-10} - 10^{-6}) = 4\pi (10^{-4} - 10^{-6}) = 4\pi \times 10^{-6}(100 - 1) = 4\pi \times 99 \times 10^{-6} \text{ m}^2$. Now, calculating work done: $W = 550 \times 10^{-3} \times 4 \times 3.1416 \times 99 \times 10^{-6} \approx 6.84 \times 10^{-4} \text{ J}$.
Step 4: Conclusion
The total amount of work done under these isothermal scaling parameters equals $6.84 \times 10^{-4} \text{ J}$.
Final Answer: (A)
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