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

Answer:

Answer:

Let R cm and r cm denote the radii of the base of the cylinder and cone respectively. Then,

R = r = 8 cm

Let H and h cm be the height of the cylinder and the cone respectively. Then

H = 240 cm and h = 36 cm

Now, volume of the cylinder = πR²H cm³

= (π × 8 × 8 × 240)cm3

= (π × 64 × 240) cm3

$\sf Volume \: of \: the \: cone = \frac{1}{3} πr {}^{2} h \: cm {}^{3} \\ = \sf \:( \frac{1}{3} π × 8 × 8 × 36) cm {}^{3} \\ = \sf ( \frac{1}{3} π × 64 × 36) cm {}^{3}$

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

$= \sf (π × 64 × 240 + \frac{1}{3} π × 64 × 36) cm {}^{3} \\ \sf= π × 64 × (240 + 12) cm {}^{3} \\ = \sf \: 22/7 × 64 × 252 cm {}^{3} \\ \sf= 22 × 64 × 36 cm {}^{3}$

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

= (22 × 64 × 36) × 7.8 gms

= 395366.4 gms

= 395.3664 kg.

Answer 2

Answer:

${\sf{\purple{\underline{\overline{Given:-}}}}}$

• Radius = 8 cm
• Height of cylinder = 240 cm
• the conical part in 36 cm high.

${\sf{\pink{\underline{\overline{To \: find:-}}}}}$

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

${\sf{\purple{\underline{\overline{Solution :-}}}}}$

$\rm\fbox \red {We know that}$

• 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

Weight of pillar is 395 kg.

${\sf{\purple{\underline{\overline{❀Tʜᴀɴᴋs!!}}}}}$