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 need to find the maximum height (h) of water that can rise in a cylindrical tank with a hole at the bottom. The water is flowing into the tank at a rate of 70 cm³/sec, and the hole has an area of 1 cm².

### Step-by-Step Solution:

1. **Identify the Rates**:
- The rate of water flowing into the tank (Vin) is given as 70 cm³/sec.
- The area of the hole (A) is given as 1 cm².

2. **Determine the Efflux Speed**:
- The efflux speed (v) of water exiting through the hole can be calculated using Torricelli’s theorem, which states:
\[ v = \sqrt{2gh} \]
where \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)) and \( h \) is the height of the water column.

3. **Calculate the Rate of Water Exiting**:
- The rate of water exiting the tank (Vout) can be expressed as:
\[ V_{out} = A \cdot v = A \cdot \sqrt{2gh} \]
- Substituting the area of the hole:
\[ V_{out} = 1 \, \text{cm}^2 \cdot \sqrt{2gh} \]

4. **Set the Rates Equal for Maximum Height**:
- At maximum height, the rate of water flowing in equals the rate of water flowing out:
\[ V_{in} = V_{out} \]
- Therefore:
\[ 70 \, \text{cm}^3/\text{s} = 1 \cdot \sqrt{2gh} \]

5. **Solve for h**:
- Rearranging the equation gives:
\[ \sqrt{2gh} = 70 \]
- Squaring both sides:
\[ 2gh = 4900 \]
- Now, substituting \( g = 981 \, \text{cm/s}^2 \) (since we are using cm for height):
\[ 2 \cdot 981 \cdot h = 4900 \]
- Simplifying:
\[ 1962h = 4900 \]
- Thus:
\[ h = \frac{4900}{1962} \approx 2.5 \, \text{cm} \]

### Final Answer:
The maximum height up to which water can rise in the tank is **2.5 cm**.