A cylindrical tank has a hole of `1cm^(2)` in its bottom. If the water is allowed to flow into the tank from a tube above it at the rate of `70cm^(3)//sec`, then the maximum height up to which water can rise in the tank is

Options

A. 2.5 cm
B. 5 cm
C. 10 cm
D. 0.25 cm

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the problem
We have a cylindrical tank with a hole at the bottom. Water flows into the tank at a rate of 70 cm³/s, and we need to find the maximum height (h) of the water in the tank when the inflow rate equals the outflow rate through the hole.

### Step 2: Define the variables
- Let \( V_1 = 70 \, \text{cm}^3/\text{s} \) (inflow rate)
- Let \( A = 1 \, \text{cm}^2 \) (area of the hole)
- Let \( V_2 \) be the outflow rate through the hole
- Let \( h \) be the height of the water in the tank

### Step 3: Set up the equation
At equilibrium, the inflow rate equals the outflow rate:
\[ V_1 = V_2 \]

### Step 4: Express the outflow rate
The outflow rate \( V_2 \) can be expressed using the area of the hole and the velocity of the water exiting the hole:
\[ V_2 = A \cdot v \]
where \( v \) is the velocity of water exiting the hole.

### Step 5: Use Bernoulli's equation to find the velocity
According to Bernoulli's principle, the pressure difference can be related to the height of the water:
\[ P_0 + \rho g h = P_0 + \frac{1}{2} \rho v^2 \]
Here, \( P_0 \) is the atmospheric pressure, \( \rho \) is the density of water, and \( g \) is the acceleration due to gravity.

By simplifying, we can find:
\[ \rho g h = \frac{1}{2} \rho v^2 \]
Cancelling \( \rho \) from both sides gives:
\[ g h = \frac{1}{2} v^2 \]
Thus, we can express the velocity \( v \) as:
\[ v = \sqrt{2gh} \]

### Step 6: Substitute back into the outflow equation
Now substituting \( v \) back into the outflow rate equation:
\[ V_2 = A \cdot \sqrt{2gh} \]
Substituting the known values:
\[ 70 = 1 \cdot \sqrt{2gh} \]

### Step 7: Solve for height \( h \)
Squaring both sides gives:
\[ 70^2 = 2gh \]
\[ 4900 = 2gh \]
Now, solving for \( h \):
\[ h = \frac{4900}{2g} \]
Using \( g \approx 980 \, \text{cm/s}^2 \):
\[ h = \frac{4900}{2 \cdot 980} \]
\[ h = \frac{4900}{1960} \]
\[ h \approx 2.5 \, \text{cm} \]

### Conclusion
The maximum height up to which water can rise in the tank is approximately **2.5 cm**.