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. 5cm
C. 10cm
D. 0.25cm

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 and water is being poured into the tank at a rate of \(70 \, \text{cm}^3/\text{s}\). The hole has a cross-sectional area of \(1 \, \text{cm}^2\). We need to find the maximum height \(h\) to which the water can rise in the tank.

### Step 2: Set up the flow rate equations
The flow rate of water entering the tank is given as:
\[ Q_{\text{in}} = 70 \, \text{cm}^3/\text{s} \]

The flow rate of water exiting through the hole can be expressed using the formula:
\[ Q_{\text{out}} = A \cdot v \]
where \(A\) is the area of the hole and \(v\) is the velocity of water exiting the hole.

Given that:
\[ A = 1 \, \text{cm}^2 = 1 \times 10^{-4} \, \text{m}^2 \]

### Step 3: Apply Bernoulli's principle
According to Bernoulli's principle, we can relate the height of the water in the tank to the velocity of the water exiting the hole. The velocity \(v\) of the water exiting the hole can be derived from the potential energy at height \(h\):
\[ v = \sqrt{2gh} \]

### Step 4: Equate the flow rates
At maximum height, the flow rate into the tank equals the flow rate out of the tank:
\[ Q_{\text{in}} = Q_{\text{out}} \]
Substituting the equations:
\[ 70 = A \cdot v \]
\[ 70 = (1 \, \text{cm}^2) \cdot \sqrt{2gh} \]

### Step 5: Substitute the area and solve for \(h\)
Convert \(1 \, \text{cm}^2\) to \(10^{-4} \, \text{m}^2\):
\[ 70 = (1 \times 10^{-4}) \cdot \sqrt{2gh} \]
Rearranging gives:
\[ \sqrt{2gh} = \frac{70}{1 \times 10^{-4}} = 700000 \, \text{cm/s} \]

### Step 6: Square both sides
\[ 2gh = (700000)^2 \]
\[ 2gh = 490000000000 \, \text{cm}^2/\text{s}^2 \]

### Step 7: Solve for \(h\)
Now, substituting \(g \approx 980 \, \text{cm/s}^2\):
\[ h = \frac{490000000000}{2 \cdot 980} = \frac{490000000000}{1960} \approx 250000000 \, \text{cm} \]
\[ h \approx 250000000 \, \text{cm} \text{ (which is incorrect due to a calculation error)} \]

### Step 8: Correct the calculation
Revisiting the calculation:
\[ h = \frac{4900}{1960} \approx 2.5 \, \text{cm} \]

### Final Answer
The maximum height \(h\) to which water can rise in the tank is:
\[ \boxed{2.5 \, \text{cm}} \]