Question

36. 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 the each of cone and cylinder is 8 cm. The cylindrical part is 240 cm
high and the conical part is 36cm high. Find the weight of pillar if 1cm³ of irom weighs 8 grams.
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Answer 1

Step-by-step explanation:

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Answer 2

Answer

${\textsf{\textbf{\pink{\underline{Given:− }}}}}$

Radius = 8 cm

Height of cylinder = 240 cm

the conical part in 36 cm high.

${\textsf{\textbf{\purple{\underline{To find:- }}}}}$

Weight of the pillar if one cu. cm of iron weighs 7.8 grams.

${\textsf{\textbf{\orange{\underline{ Solution :- }}}}}$

${\textsf{\textbf{\blue{\underline{ We know that }}}}}$

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

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