Question

An iron pillar consists of cylindrical portion of 2.8 height and 20 cm in diameter and a cone of 42cm height surmounting it . Find the weight of the pillar if 1cmsquare of iron weighs 7.5 g.

Answer 1

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Answer 2

HERE IS YOUR SOLUTION...⤵⤵
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$\underline{SOLUTION}$ ➡

$\underline{Given :-} \\ \\ \underline{For \: \: the \: \: cylinderical \: \: portion} \\ \\ Height \:, \: h_{1} = 2.8 \: \: m= 280 \: \: cm \\ \\ Diameter \:\: = 20 \: \: cm \\ \\ Radius \:, \: r_{1} = 10 \: \: cm \\ \\ Volumn \: \: = \pi \: r {}^{2} h \\ \\ = [3.14 \times (10) {}^{2} \times 280 ] \: \: cm {}^{3} \\ \\ = 87920 \: \: \: cm {}^{3}$

$\underline{For \: \: the \: \: cone} \\ \\ Height \:, \: h_{2} = 42 \: \: cm \\ \\ Radius \:, \: r_{2} = 10 \: \: cm \: \: \: \: [radius \: \: of \: \: the \: \: cylinderical \: \: portion] \\ \\ Volumn \: = \frac{1}{3} \pi \: r {}^{2} h \\ \\ = [\frac{1}{3} \times 3.14 \times (10) {}^{2} \times 42] \: \: \: cm {}^{3} \\ \\ = 4396 \: \: \: cm {}^{3}$

$Total \: \: volumn \: \: of \: \: the \: \: iron \: \: pillar \: \\ \\ = Volumn \: \: of \: \: the \: \: cylinderical \: \: portion \: + Volumn \: \: of \: \: the \: \: cone \\ \\ =( 87920 + 4396) \: \: \: cm {}^{3} \\ \\ = 92316 \: \: cm {}^{3}$

$Now, \\ \\ Weight \: \: of \: \: 1 \: cm {}^{3} \: \: \: iron = 7.5 \: \: g \\ \\ Weight \: \: of \: \: 92316 \: cm {}^{3} \: \: iron = (7.5 \times 92316) \: g \\ \\ = 692370 \: \: g \\ \\ = 692.37 \: \: kg$

$\underline{ANSWER}$ ➡

$Weight \: \: of \: \: the \: \: iron \: \: pillar \: = 692.37 \: \: kg$

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$\underline{NOTE :-}$

Here , the value of $\pi$ is taken 3.14.

If the value of $\pi$ = ${22}{7}$ is taken, then the answer will be 930 kg.

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