The excess pressure due to surface tension inside a spherical drop is `6 units`. If eight such drops combine, then the excess pressure due to surface tension inside the larger drop is

Options

A. 3 units
B. 6 units
C. 12 units
D. 48 units

Answer

To solve the problem, we need to understand the relationship between the excess pressure inside a spherical drop due to surface tension and the radius of the drop. The formula for excess pressure (ΔP) inside a spherical drop is given by:

\[ \Delta P = \frac{2T}{r} \]

where:
- \( T \) is the surface tension,
- \( r \) is the radius of the drop.

### Step 1: Identify the given data
We know that the excess pressure inside a single drop is 6 units. Therefore, we can write:

\[ \Delta P = 6 \text{ units} \]

### Step 2: Relate excess pressure to radius
From the formula, we can express the radius \( r \) of the smaller drop in terms of surface tension \( T \):

\[ 6 = \frac{2T}{r} \]

Rearranging gives:

\[ r = \frac{2T}{6} = \frac{T}{3} \]

### Step 3: Calculate the volume of the smaller drops
The volume \( V \) of a single spherical drop is given by the formula:

\[ V = \frac{4}{3} \pi r^3 \]

Substituting \( r = \frac{T}{3} \):

\[ V = \frac{4}{3} \pi \left(\frac{T}{3}\right)^3 = \frac{4}{3} \pi \frac{T^3}{27} = \frac{4\pi T^3}{81} \]

### Step 4: Calculate the total volume of 8 drops
If we have 8 such drops, the total volume \( V_{total} \) is:

\[ V_{total} = 8 \times V = 8 \times \frac{4\pi T^3}{81} = \frac{32\pi T^3}{81} \]

### Step 5: Find the radius of the larger drop
Let \( R \) be the radius of the larger drop formed by combining the 8 smaller drops. The volume of the larger drop is:

\[ V_{large} = \frac{4}{3} \pi R^3 \]

Setting the total volume equal to the volume of the larger drop:

\[ \frac{32\pi T^3}{81} = \frac{4}{3} \pi R^3 \]

Cancelling \( \pi \) from both sides and simplifying gives:

\[ \frac{32 T^3}{81} = \frac{4}{3} R^3 \]

Multiplying both sides by \( 3 \):

\[ \frac{96 T^3}{81} = 4 R^3 \]

Dividing both sides by 4:

\[ R^3 = \frac{24 T^3}{81} \]

### Step 6: Calculate the radius \( R \)
Taking the cube root gives:

\[ R = \left(\frac{24 T^3}{81}\right)^{1/3} \]

### Step 7: Calculate the excess pressure in the larger drop
Now, we can find the excess pressure in the larger drop using the formula:

\[ \Delta P_{large} = \frac{2T}{R} \]

Substituting \( R = 2r \) (since the volume of 8 drops implies the radius doubles):

\[ \Delta P_{large} = \frac{2T}{2r} = \frac{T}{r} \]

Since we know \( \Delta P = 6 \) units for the smaller drop, we can substitute:

\[ \Delta P_{large} = \frac{6}{2} = 3 \text{ units} \]

### Final Answer
Thus, the excess pressure due to surface tension inside the larger drop is:

\[ \Delta P_{large} = 3 \text{ units} \]