The width of a rectangular sewer is twice its depth while discharging 1.5 m/sec. The width of the sewer is

Solution (By Examveda Team)

First, let's understand the problem:
We have a rectangular sewer (like a rectangular channel).
We know the width is twice the depth.
We also know the discharge (flow rate) is 1.5 m³/sec.
We need to find the width.

Here's how we can approach it:
1. Area and Velocity: Remember the relationship: Discharge (Q) = Area (A) * Velocity (V). We know Q (1.5 m³/sec) and V (1.5 m/sec). So, we can find A.

2. Calculate Area: Area (A) = Discharge (Q) / Velocity (V) = 1.5 m³/sec / 1.5 m/sec = 1 m²

3. Relate Width and Depth: Let's say the depth = d. The problem tells us that the width = 2d.

4. Area in terms of 'd': The area of a rectangle is width * depth. So, Area (A) = (2d) * d = 2d²

5. Solve for 'd': We know A = 1 m², and A = 2d². Therefore, 2d² = 1
d² = 1/2 = 0.5
d = √0.5 ≈ 0.707 m

6. Find the Width: Remember, width = 2d. So, Width = 2 * 0.707 m ≈ 1.414 m

7. Choose the closest answer: The closest option to 1.414 m is Option C: 1.36 m. This is because in real-world scenarios, there might be some approximations or rounding involved. So, C is the best choice.

Therefore, the correct answer is Option C: 1.36 m