Question

A castle has a pillar, which is a cylinder in shape with a cone of same base radius surmounted on it. The total height of the pillar is 60 metres and radius of the pillar is 5 m. Find the height of the conical part, if the volume of the material used in making of pillar is 4400 m³.

Answer 1

$\large \underline{ \purple{\underline{ \sf Solution.}}}$

$\sf Let \: the \: height \: of \: the \: conical \: part \: of \: \\ \sf pillar \: be \: h \: m.$

$\sf Then, \: height \: of \: the \: cylindrical \: part$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf = (60 - h) \: m$

$\sf Clearly \: radius \: of \: cylinder$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf = \: Radius \: of \: cone$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf = 5 \: m$

$\sf Volume \: of \: the \: pillar = 4400 \: m {}^{3}$

$\sf Volume \: of \: cylinder$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf = \pi \ r {}^{2} h$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf = \pi(5) {}^{2} (60 - h) \: m {}^{3} \: \: ...(1)$

$\sf Volume \: of \: cone$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf = \frac{1}{3}\pi \: r {}^{2} h = \frac{1}{3}(5) {}^{2} h \: m {}^{3} \: \: ...(2)$

$\sf According \: to \: the \: question,$

$\sf(1) + (2)$

$\: \: \sf = \pi(5) {}^{2} (60 - h) + \frac{1}{3}\pi(5) {}^{2} h = 4400$

$\implies \: \: \: \: \: \: \sf\pi25(60 - h) + \frac{25\pi \: h}{3} = 4400$

$\implies \: \: \: \: \: \: \: \: \: \sf \frac{25\pi}{3} [3(60 - h) + h] = 4400$

$\implies \: \: \: \: \: \: \sf \frac{25\pi}{3}(180 - 3h + h) = 4400$

$\implies \: \: \: \: \: \: \: \sf \frac{25\pi}{3}(180 - 2h) = 4400$

$\implies \: \: \: \: \: \sf \frac{25}{3}. \frac{22}{7}(180 - 2h) = 4400$

$\implies \: \: \: \sf22 \times 22 \times (180 - 2h) \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: = 4400 \times 3 \times 7$

$\implies \: \: \: \: \: \: \: \: \: 180 - 2h = \frac{4400 \times 3 \times 7}{25 \times 22}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf = \frac{200 \times 3 \times 7}{25}$

$\implies \: \: \: \: \: \: \sf 18 - 2 \: h = 8 \times 3 \times 7 = 168$

$\implies \: \: \: \: \: \: \: \: \sf2h = 180 - 168 = 12$

$\implies \: \: \: \: \: \: \: \: \: \: \: \sf \: h = \frac{12}{2} = 6 \: m$

$\sf Hence, \: the \:height \: of \: the \: conical \: part \\ \sf \: is \: \red{6 \: m.}$