Multiple choice

The surface of water in a swimming pool, when it is full of water, is rectangular with length and breadth $36\ m$ and $10.5\ m$ respectively. The depth of water increases uniformly from $1\ m$ at one end to $1.75\ m$ at the other end. The water in the pool is emptied by a cylindrical pipe of radius $7\ cm$ at the rate of $5\ km/h$. The time (in hours) to empty water in the pool is (take $\pi = \dfrac {22}{7})$.

  1. $6\dfrac {1}{4}$
  2. $6\dfrac {1}{2}$
  3. $6\dfrac {3}{4}$
  4. $6\dfrac {4}{5}$
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C Correct answer
Explanation

Volume of pool = Area * Average depth = (36 * 10.5) * ((1 + 1.75)/2) = 378 * 1.375 = 519.75 m^3. Pipe radius = 0.07m, speed = 5000m/h. Flow rate = Area * speed = pi * r^2 * v = (22/7) * 0.07^2 * 5000 = 77 m^3/h. Time = 519.75 / 77 = 6.75 hours = 6 3/4 hours.

AI explanation

Using the volume flow continuity method, the volume of the pool equals the volume of water exiting the pipe. The volume of the pool is calculated by multiplying the surface area by the average depth, giving 36 m multiplied by 10.5 m multiplied by 1.375 m for a total of 519.75 cubic meters. The emptying rate of the pipe is its cross-sectional area multiplied by the water speed, yielding pi multiplied by 0.07 m squared and then by 5000 m per hour, which equals 77 cubic meters per hour. Dividing the pool volume by this flow rate gives 519.75 divided by 77, resulting in a time of 6 and 3/4 hours.

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