Question

Energy needed in breaking a liquid drop of radius \(R\), into \(n\) smaller drops each of radius \(r\), is \([T=\text{surface tension of the liquid}]\):

• A. \((4\pi r^2n-4\pi R^2)T\)
• B. \(\left(\dfrac{4}{3}\pi r^3n-\dfrac{4}{3}\pi R^3\right)T\)
• C. \((4\pi R^2-4\pi r^2)nT\)
• D. \((4\pi R^2-14\pi r^2)T\)

Hint:

Whenever a liquid drop breaks into smaller drops, total surface area increases. The required energy equals: \[ \text{Surface tension}\times \text{increase in surface area}. \]

Answer 1

The Correct Option is A

Step 1: Recall the relation between energy and surface tension.
The energy required to create a new liquid surface is \[ E=T\times (\text{increase in surface area}) \] Thus, \[ E=T\Delta A \]

Step 2: Find the initial surface area.
The initial drop has radius \[ R \] Surface area of a sphere is \[ 4\pi R^2 \] Therefore, \[ A_i=4\pi R^2 \]

Step 3: Find the final surface area.
Each smaller drop has radius \[ r \] Surface area of one small drop: \[ 4\pi r^2 \] Since there are \[ n \] drops, \[ A_f=n(4\pi r^2) \] \[ A_f=4\pi r^2n \]

Step 4: Find increase in surface area.
\[ \Delta A=A_f-A_i \] \[ \Delta A=4\pi r^2n-4\pi R^2 \]

Step 5: Find the required energy.
\[ E=T\Delta A \] \[ E=(4\pi r^2n-4\pi R^2)T \]

Step 6: Final conclusion.
Hence, the required energy is \[ \boxed{(4\pi r^2n-4\pi R^2)T} \]