The excess pressure inside a spherical drop of water is four times that of another drop. Then, their respective mass ratio is

Options

A. `1:16`
B. `8:1`
C. `1:4`
D. `1:64`

Explanation

To solve the problem of finding the mass ratio of two spherical drops of water, given that the excess pressure inside one drop is four times that of the other, we can follow these steps:

### Step-by-Step Solution:

1. **Understand Excess Pressure in Spherical Drops**:
The excess pressure (\(P\)) inside a spherical drop of liquid is given by the formula:
\[ P = \frac{2S}{R} \]
where \(S\) is the surface tension of the liquid and \(R\) is the radius of the drop.

2. **Set Up the Relationship Between the Two Drops**:
Let the radius of the first drop be \(R_1\) and the radius of the second drop be \(R_2\). According to the problem, the excess pressure in the first drop is four times that in the second drop:
\[ P_1 = 4P_2 \]
Substituting the formula for excess pressure, we have:
\[ \frac{2S}{R_1} = 4 \cdot \frac{2S}{R_2} \]

3. **Cancel Common Terms**:
Since the surface tension \(S\) is the same for both drops (as they are both water), we can cancel \(2S\) from both sides:
\[ \frac{1}{R_1} = 4 \cdot \frac{1}{R_2} \]

4. **Rearrange to Find the Relationship Between Radii**:
Rearranging the equation gives:
\[ R_1 = \frac{R_2}{4} \]

5. **Calculate the Mass of Each Drop**:
The mass (\(m\)) of a spherical drop can be expressed as:
\[ m = \rho \cdot V \]
where \(\rho\) is the density and \(V\) is the volume. The volume \(V\) of a sphere is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Therefore, the masses of the two drops can be expressed as:
\[ m_1 = \rho \cdot \frac{4}{3} \pi R_1^3 \]
\[ m_2 = \rho \cdot \frac{4}{3} \pi R_2^3 \]

6. **Find the Mass Ratio**:
The mass ratio \( \frac{m_1}{m_2} \) can be calculated as follows:
\[ \frac{m_1}{m_2} = \frac{\frac{4}{3} \pi R_1^3}{\frac{4}{3} \pi R_2^3} \]
The \( \frac{4}{3} \pi \) cancels out:
\[ \frac{m_1}{m_2} = \frac{R_1^3}{R_2^3} \]

7. **Substitute the Relationship of Radii**:
We already established that \( R_1 = \frac{R_2}{4} \). Therefore:
\[ \frac{m_1}{m_2} = \left(\frac{R_2/4}{R_2}\right)^3 = \left(\frac{1}{4}\right)^3 = \frac{1}{64} \]

8. **Final Result**:
Thus, the mass ratio of the two drops is:
\[ m_1 : m_2 = 1 : 64 \]