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

Let r1 cm and r2 cm denote the radii of the base of the cylinder and cone respectively. Then,

r1 = r2 = 8 cm

Let h1 and h2 cm be the height of the cylinder and the cone respectively. Then

h1 = 240 cm and h2 = 36 cm

Now, volume of the cylinder = πr12h1 cm3

= (π × 8 × 8 × 240)cm3

= (π × 64 × 240) cm3

Volume of the cone = 1/3πr22h2 cm3

= (1/3π × 8 × 8 × 36) cm3

= (1/3π × 64 × 36) cm3

∴ Total volume of iron = Volume of the cylinder + Volume of the cone

= (π × 64 × 240 + 1/3π × 64 × 36) cm3

= π × 64 × (240 + 12) cm3

= 22/7 × 64 × 252 cm3

= 22 × 64 × 36 cm3

Hence, total weight of the pillar = Volume Weight per cm3

= (22 × 64 × 36) × 7.8 gms

= 395366.4 gms

= 395.3664 kg.

hope it helps..

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

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