Question

An iron pillar consists of a cylindrical portion of 2.8 m. height and 20 cm. in diameter and a cone of 42 cm. height surmounting it. Find the weight of the pillar if 1cm3 of iron weighs 7.5 g.

Answer 1

HERE IS YOUR SOLUTION...⤵⤵
===================================

$\underline{SOLUTION}$ ➡

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

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

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

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

$\underline{ANSWER}$ ➡

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

_________________________________

$\underline{NOTE :-}$

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.

_________________________________
✝✝…HOPE…IT…HELPS…YOU…✝✝

Answer 2

Answer:

693 kg

Step-by-step explanation:

Volume of cylinder:

Height = 2.8 m = 280 cm³.

Diameter d = 20 cm.

Radius = d/2 = 10 cm.

Volume of cylinder = πr²h

                                = (22/7) * (10)² * 2.8

                                = 88000 cm³.

Volume of cone:

Height h = 42 cm.

Radius r = 10 cm.

Volume of cone = (1/3) πr²h

                          = (1/3) * (22/7) * (10)² * 42

                          = 4400 cm³.

Volume of pillar:

Volume of cylinder + Volume of Cone

= 88000 + 4400

= 92400 cm³

Given that Weight of 1 cm³ of iron = 7.5 g.

Weight of pillar = 7.5 * 92400

                         = 693000 g

                        = 693 kg

Hope it helps!