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

Explanation

To solve the problem, we need to find the mass ratio of two spherical drops of water given that the excess pressure inside one drop is four times that of the other.

### Step-by-Step Solution:

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

2. **Setting Up the Problem**:
Let the radius of the first drop be \(r\) and the radius of the second drop be \(R\). According to the problem, the excess pressure in the first drop is four times that in the second drop:
\[ P_1 = 4P_2 \]

3. **Expressing the Excess Pressures**:
Using the formula for excess pressure, we can write:
\[ P_1 = \frac{2T}{r} \quad \text{and} \quad P_2 = \frac{2T}{R} \]
Substituting these into the equation gives:
\[ \frac{2T}{r} = 4 \left(\frac{2T}{R}\right) \]

4. **Simplifying the Equation**:
Canceling \(2T\) from both sides (assuming \(T \neq 0\)):
\[ \frac{1}{r} = \frac{8}{R} \]
Rearranging this gives:
\[ R = 8r \]

5. **Finding the Volume Ratio**:
The volume \(V\) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Therefore, the volumes of the two drops are:
\[ V_1 = \frac{4}{3} \pi r^3 \quad \text{and} \quad V_2 = \frac{4}{3} \pi R^3 \]
The mass \(m\) of the drops can be expressed as:
\[ m_1 = \rho V_1 = \rho \left(\frac{4}{3} \pi r^3\right) \quad \text{and} \quad m_2 = \rho V_2 = \rho \left(\frac{4}{3} \pi R^3\right) \]
where \(\rho\) is the density of water.

6. **Calculating the Mass Ratio**:
The mass ratio \( \frac{m_1}{m_2} \) is:
\[ \frac{m_1}{m_2} = \frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3} = \frac{r^3}{R^3} \]
Substituting \(R = 8r\):
\[ \frac{m_1}{m_2} = \frac{r^3}{(8r)^3} = \frac{r^3}{512r^3} = \frac{1}{512} \]

7. **Final Result**:
The mass ratio of the two drops is:
\[ \frac{m_1}{m_2} = \frac{1}{512} \]