Find the change in volume which 1c.c. of water at the surface will undergo, when it is taken to the bottom of the lake 100 m deep, given that volume elasticity is 22000 atmosphere.

Explanation

To find the change in volume of 1 cm³ of water when taken to the bottom of a lake 100 m deep, we can follow these steps:

### Step 1: Calculate the pressure at the bottom of the lake
The pressure at a depth \( h \) in a fluid is given by the formula:
\[ P = \rho g h \]
where:
- \( \rho \) is the density of the fluid (for water, \( \rho \approx 1000 \, \text{kg/m}^3 \)),
- \( g \) is the acceleration due to gravity (\( g \approx 9.81 \, \text{m/s}^2 \)),
- \( h \) is the depth of the fluid (in this case, \( h = 100 \, \text{m} \)).

Substituting the values:
\[ P = 1000 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 100 \, \text{m} = 981000 \, \text{Pa} \, (\text{or} \, 9810 \, \text{kPa}) \]

### Step 2: Convert pressure from pascals to atmospheres
To convert pressure from pascals to atmospheres, we use the conversion factor \( 1 \, \text{atm} \approx 101325 \, \text{Pa} \):
\[ P = \frac{981000 \, \text{Pa}}{101325 \, \text{Pa/atm}} \approx 9.68 \, \text{atm} \]

### Step 3: Use the volume elasticity formula
The change in volume \( \Delta V \) can be calculated using the formula:
\[ \Delta V = -\frac{P \cdot V}{K} \]
where:
- \( P \) is the pressure change,
- \( V \) is the original volume,
- \( K \) is the volume elasticity (bulk modulus).

Given:
- \( V = 1 \, \text{cm}^3 = 1 \times 10^{-6} \, \text{m}^3 \),
- \( K = 22000 \, \text{atm} \).

### Step 4: Convert volume elasticity to pascals
To convert \( K \) from atmospheres to pascals:
\[ K = 22000 \, \text{atm} \times 101325 \, \text{Pa/atm} \approx 2.23 \times 10^6 \, \text{Pa} \]

### Step 5: Calculate the change in volume
Now substituting the values into the volume change formula:
\[ \Delta V = -\frac{(981000 \, \text{Pa}) \cdot (1 \times 10^{-6} \, \text{m}^3)}{2.23 \times 10^6 \, \text{Pa}} \]
Calculating this gives:
\[ \Delta V = -\frac{981000 \times 10^{-6}}{2.23 \times 10^6} \approx -0.00044 \, \text{m}^3 = -0.44 \, \text{cm}^3 \]

### Conclusion
The change in volume of 1 cm³ of water when taken to the bottom of a lake 100 m deep is approximately:
\[ \Delta V \approx -0.00044 \, \text{cm}^3 \]