An air bubble rises the bottom of a lake to the top and increases thereby its volume 15 times . If the atmospheric pressure is 75 cm of hg and density of lake water is `1.02 xx10^(3) kg//m^(3)` then find the depth of lake ( Neglect surface tension).

Explanation

To find the depth of the lake where an air bubble rises and increases its volume by 15 times, we can use Boyle's Law, which states that for a given mass of gas at constant temperature, the product of pressure and volume is constant.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Volume increase factor: \( V_2 = 15V_1 \)
- Atmospheric pressure \( P_1 = 75 \, \text{cm Hg} \)
- Density of lake water \( \rho = 1.02 \times 10^3 \, \text{kg/m}^3 \)
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