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

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?

2 thoughts on “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”

  1. Answer:

    [tex]volume \: of \: water \: that \: flows \: in \: the \: canal \: in \: one \: hour[/tex]

    [tex] = width \: of \: the \: canal \times dept \: of \: canal \times speed \: of \: the \: canal \: water[/tex]

    [tex] = 3 \times 1.2 \times 20 \times 1000 {m}^{3} = 72000m {}^{3} [/tex]

    [tex] \sf \pink{in \: 20 \: minutes \: the \: volume \: of \: water \: in \: the \: canal \: }[/tex]

    [tex] = 72000 \times \frac{20}{60} {m}^{3} = 24000 {m}^{3} [/tex]

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

    [tex] \sf \pink{ \frac{24000}{0.08} {m}^{2} = 300000 {m}^{2} = 30 \: hectares}[/tex]

  2. [tex]\large\textsf{ }[/tex]

    [tex]\LARGE\mathfrak{\underline\textcolor{aqua}{✯\; Given :-}}[/tex]

    [tex]\large\textsf{ }[/tex]

    [tex]\qquad\tt{:}\longrightarrow\large\textsf{Width of the canal = 300 cm = 3 m}[/tex]

    [tex]\qquad\tt{:}\longrightarrow\large\textsf{Depth of the canal = 120 cm = 1.2 m }[/tex]

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

    [tex]\large\textsf{ }[/tex]

    [tex]\LARGE\mathfrak{\underline\textcolor{aqua}{✯\; To \; \; Find :-}}[/tex]

    [tex]\large\textsf{ }[/tex]

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

    [tex]\large\textsf{ }[/tex]

    [tex]\LARGE\mathfrak{\underline\textcolor{aqua}{✯\; Formula :-}}[/tex]

    [tex]\large\textsf{ }[/tex]

    [tex]\qquad\tt{:}\longrightarrow\boxed{\large\textsf\textcolor{purple}{Distance = Speed × Time }}[/tex]

    [tex]\large\textsf{ }[/tex]

    [tex]\LARGE\mathfrak{\underline\textcolor{aqua}{✯\; Solution :-}}[/tex]

    [tex]\large\textsf{ }[/tex]

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

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

    [tex]\large\textsf{ }[/tex]

    [tex]\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 = $\cfrac{\large\textsf{20000 m}}{\large\textsf{3}}$}}[/tex]

    [tex]\large\textsf{ }[/tex]

    ↦ Amount of water irritated in 20 mins. :-

    [tex]\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}}[/tex]

    [tex]\large\textsf{ }[/tex]

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

    [tex]\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}}[/tex]

    [tex]\large\textsf{ }[/tex]

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

    [tex]\large\textsf{ }[/tex]

    [tex]\large\textsf\textcolor{purple}{ \; \; \; \; \; \; \; \; \; \; \; \; ◈ ━━━━━━━ ✪ ━━━━━━━ ◈}[/tex]

    [tex]\large\textsf{ }[/tex]

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