An air bubble doubles its radius on raising from the bottom of water reservoir to the surface of water in it. If the atmosphetic pressure is equal to 10 m of water, the height of water in the reservoir will be

Options

A. 10 m
B. 20 m
C. 70 m
D. 80 m

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Understand the problem
We have an air bubble that doubles its radius when it rises from the bottom of a water reservoir to the surface. We need to find the height of the water in the reservoir given that the atmospheric pressure is equivalent to 10 meters of water.

### Step 2: Relate the volumes of the bubble
Let the initial radius of the bubble at the bottom be \( R_1 \) and the radius at the surface be \( R_2 \). According to the problem, \( R_2 = 2R_1 \).

The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi R^3 \]

Thus, the volumes at the two positions are:
\[ V_1 = \frac{4}{3} \pi R_1^3 \]
\[ V_2 = \frac{4}{3} \pi R_2^3 = \frac{4}{3} \pi (2R_1)^3 = \frac{4}{3} \pi (8R_1^3) = 8V_1 \]

### Step 3: Apply the ideal gas law
Using the ideal gas law, we know that for a gas, the product of pressure and volume is constant when temperature and the number of moles are constant:
\[ P_1 V_1 = P_2 V_2 \]

### Step 4: Express pressures
Let \( P_0 \) be the atmospheric pressure at the surface (10 m of water), and \( H \) be the height of the water in the reservoir. The pressure at the bottom of the reservoir is:
\[ P_1 = P_0 + \rho g H \]
where \( \rho \) is the density of water and \( g \) is the acceleration due to gravity.

### Step 5: Substitute the volumes and pressures
Substituting the values into the ideal gas law:
\[ (P_0 + \rho g H) V_1 = P_2 (8V_1) \]

Dividing both sides by \( V_1 \):
\[ P_0 + \rho g H = 8P_2 \]

### Step 6: Substitute \( P_2 \)
At the surface, the pressure \( P_2 \) is equal to the atmospheric pressure:
\[ P_2 = P_0 \]
Thus, we can substitute \( P_2 \) into the equation:
\[ P_0 + \rho g H = 8P_0 \]

### Step 7: Solve for \( H \)
Rearranging gives:
\[ \rho g H = 8P_0 - P_0 \]
\[ \rho g H = 7P_0 \]
Substituting \( P_0 = \rho g \times 10 \) (since \( P_0 \) is equivalent to 10 m of water):
\[ \rho g H = 7(\rho g \times 10) \]
Cancelling \( \rho g \) from both sides:
\[ H = 70 \text{ m} \]

### Conclusion
The height of the water in the reservoir is \( H = 70 \) meters.

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