Water is flowing through a horizontal tube having crosssectional areas of its two ends being A and A' such that the ratio All' is 5. If the pressure difference of water between the two ends is `3 xx 10^(5) Nm^(-2)` the velocity of water with which it enters the tube will be (neglect gravity effects)

Options

A. `5m s ^(-1)`
B. `10 ms ^(-1)`
C. `25 ms ^(-1)`
D. `50 sqrt10 ms ^(-1)`

Explanation

To solve the problem, we will use the principles of fluid dynamics, particularly the continuity equation and Bernoulli's equation.

### Step-by-step Solution:

1. **Identify Given Data:**
- Cross-sectional areas: \( A \) and \( A' \) such that \( \frac{A}{A'} = 5 \).
- Pressure difference: \( P_1 - P_2 = 3 \times 10^5 \, \text{N/m}^2 \).
- Density of water, \( \rho = 1000 \, \text{kg/m}^3 \).

2. **Express Areas:**
- Let \( A' = A \). Then, \( A = 5A' \) implies \( A = 5A \) and \( A' = A \).

3. **Apply the Continuity Equation:**
- The continuity equation states that the mass flow rate must be constant. Hence,
\[ A V_1 = A' V_2 \]
- Substituting \( A' = \frac{A}{5} \):
\[ A V_1 = \frac{A}{5} V_2 \implies V_2 = 5 V_1 \]

4. **Apply Bernoulli's Equation:**
- According to Bernoulli's principle, we have:
\[ P_1 + \frac{1}{2} \rho V_1^2 = P_2 + \frac{1}{2} \rho V_2^2 \]
- Rearranging gives:
\[ P_1 - P_2 = \frac{1}{2} \rho V_2^2 - \frac{1}{2} \rho V_1^2 \]
- Substituting the pressure difference:
\[ 3 \times 10^5 = \frac{1}{2} \rho (V_2^2 - V_1^2) \]

5. **Substitute for \( V_2 \):**
- Substitute \( V_2 = 5 V_1 \):
\[ 3 \times 10^5 = \frac{1}{2} \rho ((5 V_1)^2 - V_1^2) \]
- Simplifying gives:
\[ 3 \times 10^5 = \frac{1}{2} \rho (25 V_1^2 - V_1^2) = \frac{1}{2} \rho (24 V_1^2) \]
- Therefore:
\[ 3 \times 10^5 = 12 \rho V_1^2 \]

6. **Substitute the Density of Water:**
- Using \( \rho = 1000 \, \text{kg/m}^3 \):
\[ 3 \times 10^5 = 12 \times 1000 \times V_1^2 \]
\[ 3 \times 10^5 = 12000 V_1^2 \]

7. **Solve for \( V_1^2 \):**
- Rearranging gives:
\[ V_1^2 = \frac{3 \times 10^5}{12000} = 25 \]

8. **Calculate \( V_1 \):**
- Taking the square root:
\[ V_1 = \sqrt{25} = 5 \, \text{m/s} \]

### Final Answer:
The velocity of water with which it enters the tube is \( V_1 = 5 \, \text{m/s} \).