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.
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19
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![\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}](https://tex.z-dn.net/?f=%5Cunderline%7BGiven+%3A-%7D+%5C%5C+%5C%5C+%5Cunderline%7BFor+%5C%3A+%5C%3A+the+%5C%3A+%5C%3A+cylinderical+%5C%3A+%5C%3A+portion%7D+%5C%5C+%5C%5C+Height+%5C%3A%2C+%5C%3A+h_%7B1%7D+%3D+2.8+%5C%3A+%5C%3A+m%3D+280+%5C%3A+%5C%3A+cm+%5C%5C+%5C%5C+Diameter+%5C%3A%5C%3A+%3D+20+%5C%3A+%5C%3A+cm+%5C%5C+%5C%5C+Radius+%5C%3A%2C+%5C%3A+r_%7B1%7D+%3D+10+%5C%3A+%5C%3A+cm+%5C%5C+%5C%5C+Volumn+%5C%3A+%5C%3A+%3D+%5Cpi+%5C%3A+r+%7B%7D%5E%7B2%7D+h+%5C%5C+%5C%5C+%3D+%5B3.14+%5Ctimes+%2810%29+%7B%7D%5E%7B2%7D+%5Ctimes+280+%5D+%5C%3A+%5C%3A+cm+%7B%7D%5E%7B3%7D+%5C%5C+%5C%5C+%3D+87920+%5C%3A+%5C%3A+%5C%3A+cm+%7B%7D%5E%7B3%7D+ "\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}](https://tex.z-dn.net/?f=%5Cunderline%7BFor+%5C%3A+%5C%3A+the+%5C%3A+%5C%3A+cone%7D+%5C%5C+%5C%5C+Height+%5C%3A%2C+%5C%3A+h_%7B2%7D+%3D+42+%5C%3A+%5C%3A+cm+%5C%5C+%5C%5C+Radius+%5C%3A%2C+%5C%3A+r_%7B2%7D+%3D+10+%5C%3A+%5C%3A+cm+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%5Bradius+%5C%3A+%5C%3A+of+%5C%3A+%5C%3A+the+%5C%3A+%5C%3A+cylinderical+%5C%3A+%5C%3A+portion%5D+%5C%5C+%5C%5C+Volumn+%5C%3A+%3D+%5Cfrac%7B1%7D%7B3%7D+%5Cpi+%5C%3A+r+%7B%7D%5E%7B2%7D+h+%5C%5C+%5C%5C+%3D+%5B%5Cfrac%7B1%7D%7B3%7D+%5Ctimes+3.14+%5Ctimes+%2810%29+%7B%7D%5E%7B2%7D+%5Ctimes+42%5D+%5C%3A+%5C%3A+%5C%3A+cm+%7B%7D%5E%7B3%7D+%5C%5C+%5C%5C+%3D+4396+%5C%3A+%5C%3A+%5C%3A+cm+%7B%7D%5E%7B3%7D+ "\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}")
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Here , the value of
is taken 3.14.
If the value of = 
is taken, then the answer will be 930 kg.
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Here , the value of
If the value of
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✝✝…HOPE…IT…HELPS…YOU…✝✝
Answered by
26
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!
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