A rectangular reservoir is 120m long and 75m wide. At what speed per h...
To solve this problem, we need to calculate the speed at which water must flow into the reservoir through the square pipe in order for the water level to rise by 2.4 meters in 18 hours.
We are given the following information:
- Length of the reservoir = 120 m
- Width of the reservoir = 75 m
- Rise in water level = 2.4 m
- Time taken for the rise = 18 hours
- Width of the square pipe = 20 cm
First, let's convert the width of the square pipe from centimeters to meters:
Width of the square pipe = 20 cm = 0.2 m
Next, let's calculate the volume of water that needs to flow into the reservoir to raise the water level by 2.4 meters:
Volume = Length × Width × Height
Volume = 120 m × 75 m × 2.4 m
Volume = 21600 m³
Now, let's calculate the flow rate of water required to fill this volume in 18 hours:
Flow rate = Volume / Time
Flow rate = 21600 m³ / 18 hours
Flow rate = 1200 m³/hour
Since the width of the square pipe is 0.2 m, we can calculate the speed at which the water must flow through the pipe:
Speed = Flow rate / Area of the pipe
Speed = 1200 m³/hour / (0.2 m × 0.2 m)
Speed = 1200 m³/hour / 0.04 m²
Speed = 30000 m/hour
Finally, let's convert the speed from meters per hour to kilometers per hour:
Speed = 30000 m/hour × (1 km / 1000 m)
Speed = 30 km/hour
Therefore, the water must flow into the reservoir through the square pipe at a speed of 30 km/hour for the water level to rise by 2.4 meters in 18 hours. Hence, option B is the correct answer.
A rectangular reservoir is 120m long and 75m wide. At what speed per h...
Volume of the water accumulated the reservoir 18 hours = (120 × 75 × 2.4) m
3Let speed of water = v km/hour.
The width of cuboid = b = 20/100 = 1/5m
Height = h = 20/100 = 1/5 m
Length of water cuboid formed in 18 hours
= 18 v km = 18 × 1000 v m
= 18000 v m
Volume of the water accumulated in reservoir in 18 hours

⇒ 720v = 120 × 75 × 2.4
