Question

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

Answer 1

Answer:

$volume \: of \: water \: that \: flows \: in \: the \: canal \: in \: one \: hour$

$= width \: of \: the \: canal \times dept \: of \: canal \times speed \: of \: the \: canal \: water$

$= 3 \times 1.2 \times 20 \times 1000 {m}^{3} = 72000m {}^{3}$

$\sf \pink{in \: 20 \: minutes \: the \: volume \: of \: water \: in \: the \: canal \: }$

$= 72000 \times \frac{20}{60} {m}^{3} = 24000 {m}^{3}$

$\sf \pink{area \: irrigated \: in20 \: minutes \: if \: 8 \: cm \: ie \: 0.08 \: m \: standing \: water \: is \: required \: }$

$\sf \pink{ \frac{24000}{0.08} {m}^{2} = 300000 {m}^{2} = 30 \: hectares}$

Answer 2

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$\LARGE\mathfrak{\underline\textcolor{aqua}{✯\; Given :-}}$

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$\qquad\tt{:}\longrightarrow\large\textsf{Width of the canal = 300 cm = 3 m}$

$\qquad\tt{:}\longrightarrow\large\textsf{Depth of the canal = 120 cm = 1.2 m }$

$\qquad\tt{:}\longrightarrow\large\textsf{Speed of the water flow = 20 km/h = 20000 m/h .}$

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$\LARGE\mathfrak{\underline\textcolor{aqua}{✯\; To \; \; Find :-}}$

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• The total area irrigated in 20 mins. if 8 cm ( 0.08 m ) of standing water is desired = ?

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$\LARGE\mathfrak{\underline\textcolor{aqua}{✯\; Formula :-}}$

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$\qquad\tt{:}\longrightarrow\boxed{\large\textsf\textcolor{purple}{Distance = Speed × Time }}$

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$\LARGE\mathfrak{\underline\textcolor{aqua}{✯\; Solution :-}}$

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↦ Distance covered by water in 1 hr. or 60 mins. = 20000 m .

↦ So , the distance covered by the water in 20 mins. :-

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$\qquad\tt{:}\longrightarrow\large\textsf{Distance = Speed × Time}\\\\\\\qquad\tt{:}\longrightarrow\large\textsf{\; \; \; = \cfrac{\large\textsf{20}}{\large\textsf{60}} ×\large\textsf{20000}}\\\\\\\qquad\tt{:}\longrightarrow\boxed{\large\textsf\textcolor{red}{Distance = \cfrac{\large\textsf{20000 m}}{\large\textsf{3}} }}$

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↦ Amount of water irritated in 20 mins. :-

$\qquad\tt{:}\longrightarrow\large\textsf{ \; \; \; = \cfrac{\large\textsf{3 × 1.2 × 20000}}{\large\textsf{3}} }\\\\\\\qquad\tt{:}\longrightarrow\large\textsf{ \; \; \; = \cfrac{\large\textsf{72000}}{\large\textsf{3}} }\\\\\\\qquad\tt{:}\longrightarrow\boxed{\large\textsf\textcolor{red}{\; \; \; = 24000 m²}}$

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↦ Area irrigated by this water if 8 cm or 0.08 m of standing water is desired will be :-

$\qquad\tt{:}\longrightarrow\large\textsf{\; \; \; = \cfrac{\large\textsf{24000}}{\large\textsf{0.08}} }\\\\\\\qquad\tt{:}\longrightarrow\large\textsf{\; \; \; = \cfrac{\large\textsf{24000 × 100}}{\large\textsf{8}} }\\\\\\\qquad\tt{:}\longrightarrow\large\textsf{\; \; \; = 3000 × 100}\\\\\\\qquad\tt{:}\longrightarrow\boxed{\large\textsf\textcolor{red}{\; \; \; = 300000 m²}}$

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∴ The total area irrigated in 20 mins. if 8 cm ( 0.08 m ) of standing water is desired = 300000 m² = 30 hector .

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$\large\textsf\textcolor{purple}{ \; \; \; \; \; \; \; \; \; \; \; \; ◈ ━━━━━━━ ✪ ━━━━━━━ ◈}$

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