Question

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

Answer 1

Hint: We are asked to find the area irrigated by the flowing water in $20$ minutes, given $8\,cm$ of standing or still water is required. For this, firstly calculate the volume of water that flows in the canal in one hour then using this calculate the amount of water irrigated in $20$ minutes.

Complete step by step solution:
According to the question we are given,
A canal of $width=300\,cm=3m$ (as $1m\,=\,100\,cm$ )
and $depth\left( height \right)=120cm=1.2\,m$
and water is flowing in the canal with speed $=20\,km/hr$
Now, as all our units is in metres we will convert the speed also from $km/hr$ to $m/hr$,
Thus $\text{speed}=20km/hr=20,000m/hr$ (as $1\,km=\,1000\,m$ )
Now, we have to calculate the volume of water that flows in the canal in one hour,
So we have the formula to calculate volume as,
$Volume=Area~\times speed$
$\Rightarrow $\[Volume\text{ }=\text{ }width\text{ }of\text{ }canal\times height\text{ }of\text{ }canal\times speed\text{ }of\text{ }water\text{ }in\text{ }canal\]
Substituting the values of width, height and speed we get,
\[Volume=3m\times 1.2m\times 20,000m\]
$\Rightarrow \,$$Volume=72,000{{m}^{3}}$
Thus we got the volume of water flowing in the canal in $1$ hour or $60$ minutes as $72,000\,{{m}^{3}}$
Now, we have to calculate volume of water flowing in $20$ minutes,
 So, first calculate volume of water flowing in $1$ minute $=\dfrac{72,000}{60}$ (as $1\,hour=\,60\,\min $ )
$\therefore $ Volume of water flowing in $20$ minutes $=\dfrac{72,000}{60}\times 20=24,000{{m}^{3}}$
Thus we got the volume of water flowing in the canal in $20$ minutes as $24,000\,{{m}^{3}}$
Now, we have to find the area covered in $20$ minutes when $8\,cm$ of standing water is required,
So $8cm\,=\,0.08\,m$ (as $1\,m\,=\,100\,cm$ )
So, now area covered $20\,$minutes with $0.08\,m$ of standing water $=\dfrac{24,000{{m}^{3}}}{0.08m}$
$\Rightarrow \,3,00,000\,{{m}^{2}}$
$=\,\,30\,hectares$ (as $1\,hectare\,=\,10,000\,{{m}^{2}}$ )

Thus we got the area covered in $20$ minutes with $8\,cm$ of standing water as $30\,hectares$.

Note: The most important thing to keep in mind while solving such problems is the units. Carefully examine the units of the quantities mentioned in the question and then accordingly convert all the quantities in the same unit. While solving the problem all the quantities should be in the same unit.