Multiple choice

Water flows out through a circular pipe, whose internal diameter is $\displaystyle {1} \frac{1}{3}\, cm$, at the rate of $0.63$ m per second into a cylindrical tank, the radius of whose base is $0.2$ m. By how much will the level of water rise in one hour?

  1. $2.32 m$
  2. $2.52 m$
  3. $2.72 m$
  4. $2.92 m$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Volume of water flowing per second = Area of pipe * velocity = pi * (r_pipe)^2 * v. Diameter = 4/3 cm, so radius = 2/3 cm = 2/300 m. Area = pi * (2/300)^2. Volume per hour = Area * 0.63 * 3600. This volume fills the tank to height h: pi * (0.2)^2 * h = Volume per hour. Solving for h gives 2.52 m.

AI explanation

Using the volume of a cylinder formula, the cross-sectional area of the pipe is pi multiplied by the radius squared, where the diameter is 4/3 cm (1/67 m), so the radius is 2/3 cm (1/150 m) and the area is pi divided by 22500 square metres. In one hour, the length of water flowing is 0.63 m/s multiplied by 3600 s, which equals 2268 m. Therefore, the total volume of water flowing into the cylindrical tank is pi/22500 multiplied by 2268, resulting in 0.1008 pi cubic metres. Since the base area of the cylindrical tank is pi multiplied by 0.4, the rise in water level is the volume divided by the base area, which is 0.1008 divided by 0.4 to equal 2.52 m.

Keep practicing — related questions