Question

A canal is 300cm wide and 120 cm deep. The water in the canal is flowing with a speed of 20 km/h. How much area will it irrigate in 20min, if 8 cm of standing water is desired?

Answer 1

Cross section area of canal = $300\times 120$ = 36000 sq.cm
Speed of water flow = 20 km/h = 20000 m/h = $20\times {10}^3\times {10}^2$ cm/h
$=2\times {10}^{6}cm/hr$
Hence, in the volume of water flowing $36000\times 2\times {10}^6$ cubic cm:
$=7.2\times {10}^{10}cc$
Is to min volume of water = $\frac{7.2\times {10}^{10}}{\text{/}{60}_3}\times \text{/}30=2.4\times {10}^{10}cc$

Let the area of field being irrigated = A sq. cm.
Required studing height of water = 8 cm
$\therefore$ Volume of water in field $=A\times 8$ cc --- (2)
this must be equal to volume of water give by (1)
$\therefore A\times 8=2.4\times {10}^{10}$
$A=\frac{2.4}{8}\times {10}^{10}sq.cm$
$=3\times {10}^{9}sq.cm$
$=\frac{3\times {10}^9}{{10}^4}sq.cm=3\times {10}^{5}sq.cm$