A canal is 300 ,cm wide and 120 ,cm deep. The | vedclass

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(A) First,convert all dimensions to meters: Width $= 300 \,cm = 3 \,m$,Depth $= 120 \,cm = 1.2 \,m$,Speed $= 20 \,km/h = 20000 \,m/h$.Volume of water

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$30$

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(A) First,convert all dimensions to meters: Width $= 300 \,cm = 3 \,m$,Depth $= 120 \,cm = 1.2 \,m$,Speed $= 20 \,km/h = 20000 \,m/h$.Volume of water flowing through the canal in $1 \,hour$ $= 	ext{Width} 	imes 	ext{Depth} 	imes 	ext{Speed} = 3 	imes 1.2 	imes 20000 = 72000 \,m^3$.Volume of water flowing in $20 \,minutes$ $= \frac{72000 	imes 20}{60} = 24000 \,m^3$.Let the area to be irrigated be $A \,m^2$. The volume of water required for irrigation is $	ext{Area} 	imes 	ext{Height of standing water}$.Given height $= 8 \,cm = 0.08 \,m$.So,$A 	imes 0.08 = 24000$.$A = \frac{24000}{0.08} = 300000 \,m^2$.Since $1 \,hectare = 10000 \,m^2$,the area in hectares $= \frac{300000}{10000} = 30 \,hectares$.

$A$ canal is $300 \,cm$ 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 $20 \,minutes$ if $8 \,cm$ of standing water is desired? (in $hectares$)

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