The cross-section of a railway tunnel is a rectangle 6 m broad and 8 m high surmounted by a semi-circle . The tunnel is 35 m long. Find the cost of plastering the internal surface of the tunnel (excluding the floor) at the rate of Rs 2.25 per `m^(2)`

Answer

To solve the problem, we need to calculate the internal surface area of the railway tunnel (excluding the floor) and then find the cost of plastering that area. The tunnel has a rectangular cross-section with a semi-circle on top.

### Step-by-Step Solution:

1. **Identify the dimensions of the tunnel:**
- The width of the rectangular part is 6 m.
- The height of the rectangular part is 8 m.
- The length of the tunnel is 35 m.

2. **Calculate the radius of the semi-circle:**
- The diameter of the semi-circle is equal to the width of the rectangle, which is 6 m.
- Therefore, the radius \( r \) of the semi-circle is:
\[ r = \frac{6}{2} = 3 \text{ m} \]

3. **Calculate the area of the semi-circle:**
- The area \( A \) of a full circle is given by the formula \( A = \pi r^2 \).
- The area of the semi-circle is half of that:
\[ \text{Area of semi-circle} = \frac{1}{2} \pi r^2 = \frac{1}{2} \times \frac{22}{7} \times (3^2) = \frac{1}{2} \times \frac{22}{7} \times 9 = \frac{198}{14} = \frac{99}{7} \text{ m}^2 \]

4. **Calculate the curved surface area of the semi-circle along the length of the tunnel:**
- The curved surface area (CSA) of the semi-circle is given by the formula \( \text{CSA} = \text{perimeter of semi-circle} \times \text{length} \).
- The perimeter of the semi-circle is \( \pi r \):
\[ \text{Perimeter of semi-circle} = \pi r = \frac{22}{7} \times 3 = \frac{66}{7} \text{ m} \]
- Therefore, the curved surface area of the semi-circle over the length of the tunnel is:
\[ \text{CSA of semi-circle} = \frac{66}{7} \times 35 = \frac{2310}{7} = 330 \text{ m}^2 \]

5. **Calculate the area of the rectangular sides:**
- The area of the two rectangular sides (excluding the floor) is:
\[ \text{Area of two rectangles} = 2 \times \text{height} \times \text{length} = 2 \times 8 \times 35 = 560 \text{ m}^2 \]

6. **Calculate the total internal surface area (excluding the floor):**
- The total internal surface area is the sum of the curved surface area of the semi-circle and the area of the two rectangular sides:
\[ \text{Total Surface Area} = \text{CSA of semi-circle} + \text{Area of two rectangles} = 330 + 560 = 890 \text{ m}^2 \]

7. **Calculate the cost of plastering:**
- The cost of plastering is given as Rs 2.25 per m². Therefore, the total cost is:
\[ \text{Total Cost} = \text{Total Surface Area} \times \text{Cost per m}^2 = 890 \times 2.25 = 2002.50 \text{ Rs} \]

### Final Answer:
The cost of plastering the internal surface of the tunnel (excluding the floor) is **Rs 2002.50**.