Question

Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/hrs. How much area will it irrigate in 30 minutes, if 8 cm standing water is needed?

Answer 1

$\boxed{ \boxed{ \overline {\underline{ \bf \red {SOLUTION \: ☻ }}}}}$

$\rm \: Given, \: speed \: of \: flow \: of \: water = 10km/h$

$\rm = 10 \times 1000 \: m/h \: [\because \: 1 \: km = 1000 \: m]$

$\longrightarrow \: \rm \: Length \: of \: water \: flow \: in \: 1 \: h = 0 \times 1000 \: m$

$\longrightarrow \: \rm \: Length \: of \: water \: flow \: in \: 30 \: min \bigg( i.e. \: in \: \frac{1}{2} \: h \bigg)$

$= \rm \frac{1}{2} \times 10 \times 1000 = 5000 \: m$

$\rm \: Now, \: volume \: of \: water \: flowing \: in \: 30 \: min$

$\rm = volume \: of \: cuboid \: of \: length \: 5000 \: m,$

$\rm \: width \: 6 \: m \: and \: depth \: 1.5 \: m$

$\rm = 6 \times 1.5 \times 5000 \: m {}^{3} = 4500 \: m {}^{3}$

$\rm \: Hence, \: the \: required \: area \: covered \: for \: irrigation \: with \:$

$\rm \: 8 \: cm \: or \: \frac{8}{100} \: m \: of \: standing \: water$

$\rm = \frac{4500}{8} \times 100 = 562500 \: m {}^{2}$

$\rm = \: \frac{562500}{10000} \: hec$

$= {\boxed{\underline{\underline{\rm \red{56.25 \: hec}}}}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rm [\because \: 1 \: hec = 10000 \: m {}^{2}]$