Question

A milk tanker cylindrical in shape having diameter 2 m and length 4.2 m supplies milk to the two booths in the ratio 3:2. One of the milk booths has cuboid vessels having base area 3.96 sq. m. and the other has a cylindrical vessel having radius 1 m. Find the level of milk in each of the vessels.
$\sf \:( use \: \pi = \frac{22}{7})$
​

Answer 1

$\huge\large{\boxed{\underline{\underline{\tt{\pink{Solution:-}}}}}}$

$\bf \: For \: cylindrical \: milk \: tank$

$\rm \: Radius \: (r) = \frac22 \: m = 1m$

$\rm \: Length \: (h) = 4.2 \: m = \frac{42}{10} \: m$

$\sf \red{ \therefore} \: \ \: \: \: \: \: \: \: \: \: \: \rm \: Volume$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rm = \pi r {}^{2} h = \frac{22}{7}(1) {}^{2} \frac{42}{10}$

$\: \: \: \: \: \: \rm = 22 \times \frac{6}{10} = 22 \times \frac{3}{5} = \frac{66}{5} \: m {}^{3}$

$\sf \: Given \: ratio = \: \bf\red {3 \ratio2}$

$\red{\therefore} \rm \: \: \: \: \: Sum \: of \: ratios = 3 + 2 = 5$

$\red{\therefore} \: \rm \: Volume \: of \: milk \: supplied \: to \: first \: booth$

$\: \: \: \: \: \: \: \: \: \rm = \frac35 \times \frac{66}{5} = \frac{198}{25} \: m {}^{3} \: \: \: \: \: \: \: \:... \bf(1)$

$\rm \: Volume \: of \: milk \: supplied \: to \: second \: booth$

$\: \: \: \rm = \ \frac{2}{5} \times \frac{66}{5} \: cm {}^{3} = \frac{132}{25} \: m {}^{3} \: \: \: \: \: \: \: \: \: \: \: \: ... \bf(2)$

$\bf \: For \: first \: booth \: ( \sf \red{cuboidal \: vessel})$

$\rm \: Let \: the \: level \: of \: milk \: be \: \sf \red{h} \rm \: m.$

$\rm \: Then, \: Volume \: of \: milk.$

$\: \: \: \: \: \: \tt \red{ = Area \: of \: base \times Height}$

$\: \: \: \: \: \: \: \: \: \rm = 3.96h \: m {}^{3} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ... \bf{(3)}$

$\bf \: From \: (3) \: and \: (1),$

$\: \: \: \: \: \: \rm \: 3.96h = \frac{198}{25} \implies \frac{396 \: h}{100} = \frac{198}{25}$

$\implies \: \: \: \: \: \: \rm \: h = \frac{198}{25} \times \frac{100}{396} = \frac{4}{2} = 2 \: m$

$\bf \: For \: second \: booth \: (\sf \red{cylindrical \: vessel})$

$\rm \: Let \: the \: level \: of \: milk \: be \: \bf \red{h} \rm \: m.$

$\rm \: Then,$

$\rm \: Volume \: of \: milk$

$\: \: \: \: \: \rm = \: \pi R {}^{2} H = \pi(1) {}^{2} H = \frac{22}{7}H \: \: \: \: \: \: ... \bf{(4)}$

$\rm \: From \: (4) \: and \: (2),$

$\: \: \: \: \: \: \frac{22}{7} \: H = \frac{132}{25}$

$\red \therefore \: \: \: \: \: \: \rm \: H = \frac{132}{25} \times \frac{7}{22} = \frac{6 \times 7}{25} = \frac{42}{25} \: m$

$\: \: \: \: \: \: \: \: \rm = \frac{42 \times 4}{25 \times 4} = \frac{168}{100} = \tt \red{1.68 \: m}$