Final answer:
The level of water in the tank rises by 45 cm in half an hour when water flows through a cylindrical pipe of 2 cm diameter into the tank at a rate of 0.4 m/s.
Step-by-step explanation:
To determine the rise in the level of water in the tank in half an hour, we need to calculate the volume of water flowing from the cylindrical pipe and then find out how this volume affects the water level in the cylindrical tank.
Firstly, we need to calculate the flow rate of water in the pipe. The flow rate (Q) is given by the formula Q = Area × Velocity, where the area (A) is the cross-sectional area of the pipe.
Using the internal diameter of the pipe (d = 2 cm), we find the radius (r = d/2 = 1 cm). The area of the cross-section of the pipe is A = πr² = π(1 cm)² = π cm². The velocity (v) of water is given as 0.4 m/s, which we convert to cm/s by multiplying by 100 (since 1 m = 100 cm), yielding v = 40 cm/s. Therefore, Q = π cm² × 40 cm/s = 40π cm³/s.
In half an hour (1800 seconds), the volume of water delivered to the tank is Volume = Q × time = 40π cm³/s × 1800 s = 72000π cm³. To find the rise in water level in the tank, we divide this volume by the area of the tank's base. With the tank's base radius of 40 cm, the area is A = π(40 cm)² = 1600π cm².
The rise in water level (Δh) is given by Δh = Volume / Area of tank's base = 72000π cm³ / 1600π cm² = 45 cm. Therefore, the level of water in the tank rises by 45 cm in half an hour.