The pressure at the bottom of an open tank of water is 3p where p is the atmospheric pressure. If the water is drawn out till the level of water remains one fifth, the pressure at the bottom of the tank will now be

Options

A. 2p
B. `(13//5)p`
C. (7/5)p
D. (4/5)p

Answer

To solve the problem, we need to understand how pressure at the bottom of a tank changes with the height of the water column.

1. **Understanding Initial Conditions**:
- We know that the pressure at the bottom of the tank is given as \(3p\), where \(p\) is the atmospheric pressure.
- This means that the pressure due to the water column plus the atmospheric pressure equals \(3p\).

2. **Relating Pressure to Height**:
- The pressure at the bottom of the tank due to the water column can be expressed as:
\[ P = P_0 + \rho g h \]
where \(P_0\) is the atmospheric pressure, \(\rho\) is the density of water, \(g\) is the acceleration due to gravity, and \(h\) is the height of the water column.
- Given that the total pressure at the bottom is \(3p\), we can write:
\[ 3p = p + \rho g h \]
- Simplifying this gives:
\[ \rho g h = 2p \]

3. **Finding the New Height**:
- Now, when the water is drawn out until the level of water remains one fifth of the original height, the new height \(h' = \frac{h}{5}\).

4. **Calculating New Pressure**:
- The new pressure at the bottom of the tank can be calculated using the same pressure formula:
\[ P' = P_0 + \rho g h' \]
- Substituting \(h' = \frac{h}{5}\):
\[ P' = p + \rho g \left(\frac{h}{5}\right) \]
- We already know that \(\rho g h = 2p\), so we can substitute for \(\rho g h\):
\[ P' = p + \frac{2p}{5} \]

5. **Simplifying the Expression**:
- Now, we simplify:
\[ P' = p + \frac{2p}{5} = \frac{5p}{5} + \frac{2p}{5} = \frac{7p}{5} \]

6. **Final Result**:
- Therefore, the pressure at the bottom of the tank after drawing out the water to one fifth of the original height is:
\[ P' = \frac{7p}{5} \]

### Final Answer:
The pressure at the bottom of the tank will now be \(\frac{7p}{5}\).