The energy needed for breaking a drop of radius R into n drops each of radius r is

Options

A. `(4pir^(2)-4piR^(2))T`
B. `(4piR^(2)-4pir^(2))T`
C. `[4/3pir^(3)-4/3piR^(3)]T`
D. `(4pir^(2)-n4pir^(2))//T`

Answer

To find the energy needed for breaking a drop of radius \( R \) into \( n \) drops each of radius \( r \), we can follow these steps:

### Step 1: Understand the concept of surface tension
Surface tension (\( T \)) is the energy required to create a unit area of surface. When a drop is broken into smaller drops, the surface area increases, and thus energy is required to create this new surface area.

### Step 2: Calculate the initial energy
The initial drop has a radius \( R \). The surface area \( A_i \) of the initial drop is given by the formula for the surface area of a sphere:
\[ A_i = 4\pi R^2 \]
The initial energy (\( E_i \)) associated with this surface area is:
\[ E_i = T \times A_i = T \times 4\pi R^2 \]

### Step 3: Calculate the final energy
When the drop is broken into \( n \) smaller drops, each of radius \( r \), the surface area \( A_f \) of one small drop is:
\[ A_f = 4\pi r^2 \]
Since there are \( n \) such drops, the total surface area for all \( n \) drops is:
\[ A_f = n \times A_f = n \times 4\pi r^2 \]
The final energy (\( E_f \)) associated with this total surface area is:
\[ E_f = T \times A_f = T \times n \times 4\pi r^2 \]

### Step 4: Calculate the energy required for breaking the drop
The energy required (\( E \)) to break the original drop into \( n \) smaller drops is the difference between the final energy and the initial energy:
\[ E = E_f - E_i \]
Substituting the expressions for \( E_f \) and \( E_i \):
\[ E = (T \times n \times 4\pi r^2) - (T \times 4\pi R^2) \]
Factoring out \( T \):
\[ E = T \times [n \times 4\pi r^2 - 4\pi R^2] \]
This can be simplified to:
\[ E = 4\pi T \times (n r^2 - R^2) \]

### Final Answer
Thus, the energy needed for breaking a drop of radius \( R \) into \( n \) drops each of radius \( r \) is:
\[ E = 4\pi T (n r^2 - R^2) \]

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