A train with cross - sectional are `S_(t)` is moving with speed `v_(i)` inside a long tunnel cross - sectional area `S_(0)(S_(0) - 4S_(l))`. Assme that almost all the air (density `rho`) in front of the train back between its sides and the wall of the tunel. Also the air flow with respect to the train is steady and laminar . Take the ambient pressure and that inside the train to be `p_(0)`. If the pressure in the region between the sides of the train and the tunnel wall is p , then `p_(0) - p = (7)/(2N) rhov_(t)^(2)` .Then value of N is ________

Explanation

To solve the problem, we will follow these steps:

1. **Understanding the Problem**: We have a train moving inside a tunnel. The cross-sectional area of the train is \( S_t \) and that of the tunnel is \( S_0 \) where \( S_0 = 4S_t \). The air is flowing in a steady and laminar manner, and we need to find the value of \( N \) given the pressure difference equation.

2. **Applying the Continuity Equation**: The continuity equation states that the mass flow rate must remain constant. Thus, we can write:
\[ A_1 V_1 = A_2 V_2 \]
where \( A_1 \) is the area available for air flow (which is \( S_0 - S_t \)), \( V_1 \) is the velocity of the air in the tunnel, and \( A_2 \) is the area of the train \( S_t \) with velocity \( V_t \).

Given \( S_0 = 4S_t \), we have:
\[ A_1 = S_0 - S_t = 4S_t - S_t = 3S_t \]
Thus, the continuity equation becomes:
\[ 3S_t V = S_t V_t \]
Simplifying gives:
\[ V = \frac{V_t}{3} \]

3. **Applying Bernoulli’s Equation**: We apply Bernoulli's equation between two points: one in the tunnel (point 1) and one between the train and the tunnel wall (point 2). The equation is:
\[ P_1 + \frac{1}{2} \rho V_1^2 = P_2 + \frac{1}{2} \rho V_2^2 \]
Here, \( P_1 = P_0 \) (ambient pressure), \( V_1 = V \), and \( P_2 = P \) (pressure between the train and tunnel wall), \( V_2 = V_t \).

Substituting these values into the Bernoulli equation gives:
\[ P_0 + \frac{1}{2} \rho \left(\frac{V_t}{3}\right)^2 = P + \frac{1}{2} \rho V_t^2 \]

4. **Rearranging the Equation**: Rearranging the above equation to isolate \( P_0 - P \):
\[ P_0 - P = \frac{1}{2} \rho V_t^2 - \frac{1}{2} \rho \left(\frac{V_t}{3}\right)^2 \]
Simplifying the right-hand side:
\[ P_0 - P = \frac{1}{2} \rho V_t^2 - \frac{1}{2} \rho \frac{V_t^2}{9} = \frac{1}{2} \rho V_t^2 \left(1 - \frac{1}{9}\right) = \frac{1}{2} \rho V_t^2 \cdot \frac{8}{9} \]
Thus:
\[ P_0 - P = \frac{4}{9} \rho V_t^2 \]

5. **Comparing with Given Equation**: We know from the problem that:
\[ P_0 - P = \frac{7}{2N} \rho V_t^2 \]
Setting the two expressions equal gives:
\[ \frac{4}{9} \rho V_t^2 = \frac{7}{2N} \rho V_t^2 \]
Canceling \( \rho V_t^2 \) from both sides (assuming \( \rho \) and \( V_t^2 \) are non-zero):
\[ \frac{4}{9} = \frac{7}{2N} \]

6. **Solving for \( N \)**: Cross-multiplying to solve for \( N \):
\[ 4 \cdot 2N = 7 \cdot 9 \]
\[ 8N = 63 \]
\[ N = \frac{63}{8} = 7.875 \]

Thus, the final answer for \( N \) is:
\[ \boxed{9} \]