The cross section of a swimming pool is a trapezium as shown n the adjacent diagram. The length of the pool is $200 m$ . Water is pumped into it through a pipe of cross section $1.2 m^{2}$ at a speed of $0.5$ km/hr. Find the time needed to fill the pool completely.

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Solution

Volume of pool=(Area of C.S ) $\times L$
$= \frac{1}{2} (0.8 + 1.2) \times 2 \times 200$
$= 400 ㎥$
Vol of water in $1$ hr $= 0.5 ㎞ \times 1.2 ㎡$
$= 500 \times 1.2 ㎥$
$= 600 ㎥$
Time $= \frac{400}{600} = \frac{2}{3}$ hrs $= 40$ minutes

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Question 104

Below are the drawings of cross sections of two different pipes used to fill swimming pools. Figure A is a combination of 2 pipes each having a radius of 8 cm. Figure B is a pipe having a radius of 15 cm. If the force of the flow of water coming out of the pipes is the same in both the cases, which will fill the swimming pool faster?

Question 104

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The cross section of a swimming pool is a trapezium as shown n the adjacent diagram. The length of the pool is 200 m. Water is pumped into it through a pipe of cross section 1.2 m2 at a speed of 0.5 km/hr. Find the time needed to fill the pool completely.

Volume of pool=(Area of C.S )×L=12(0.8+1.2)×2×200=400㎥Vol of water in 1 hr=0.5㎞×1.2㎡=500×1.2㎥=600㎥Time=400600=23 hrs =40 minutes

The cross section of a swimming pool is a trapezium as shown n the adjacent diagram. The length of the pool is $200 m$ . Water is pumped into it through a pipe of cross section $1.2 m^{2}$ at a speed of $0.5$ km/hr. Find the time needed to fill the pool completely.

Volume of pool=(Area of C.S ) $\times L$
$= \frac{1}{2} (0.8 + 1.2) \times 2 \times 200$
$= 400 ㎥$
Vol of water in $1$ hr $= 0.5 ㎞ \times 1.2 ㎡$
$= 500 \times 1.2 ㎥$
$= 600 ㎥$
Time $= \frac{400}{600} = \frac{2}{3}$ hrs $= 40$ minutes