Question

An iron pillar has some part in the form of a right circular cylinder and remaining in the form of a right circular cone. The radius of the base of each of cone and cylinder is 8 cm. The cylinderical part is 240 cm high and the conical part in 36 cm high. Find the weight of the pillar if one cu. cm of iron weighs 7.8 grams.​

Answer 1

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We know that

Volume of cylinder = πr²h

Volume of cone = ⅓πr²h

Now,

Volume of cylinder = 3.14 × 8 × 8 × 240

=> 48320.4 cm^3

Now

⅓ × 3.14 × 8 × 8 × 36

1 × 3.14 × 8 × 8 × 12

3.14 × 64 ×12

2411.52 cm^3

Now

Weight of pillar = Volume of cylinder + volume of cone

W = 48320.4 + 2411.52

W = 50730

Now

1kg = 1000gm

7.8/1000 × 50730

0.0078 × 50730

395.4 kg

${\textsf{\textbf{\green{\underline{Weight of pillar is 395 kg}}}}}$

Answer 2

Answer:

$\huge\mathcal{\fcolorbox{cyan}{black}{\pink{ᴀɴsᴡᴇʀ࿐}}}$

We know that

Volume of cylinder = πr²h

Volume of cone = ⅓πr²h

Now,

Volume of cylinder = 3.14 × 8 × 8 × 240

=> 48320.4 cm^3

Now

⅓ × 3.14 × 8 × 8 × 36

1 × 3.14 × 8 × 8 × 12

3.14 × 64 ×12

2411.52 cm^3

Now

Weight of pillar = Volume of cylinder + volume of cone

W = 48320.4 + 2411.52

W = 50730

Now

1kg = 1000gm

7.8/1000 × 50730

0.0078 × 50730

395.4 kg

${\textsf{\textbf{\green{\underline{Weight of pillar is 395 kg}}}}}$