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 is 36 cm high.find the weight of the pillar if one cu.cm of iron weighs 7.8 grams.​

Answer 1

volume of pillar = volume of cylindrical part + volume of conical part

Volume of cylinder of radius 'R' and height 'h' =

πR^2h,

volume of cone = 1/3 πr^2h

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

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

@${{\underline{\blue{theanuragkumar}}}}$_