Question

An iron pillar has some part in the form of a right circular cylinder and the remaining in the form of a right circular cone. The radius of the base of each of the cone and cylinder is 8 cm. The cylinder part is 240 cm high and the conical part is 36 cm high. Find the weight of the pillar if 1 cubic centimetre of iron with 7.5 g.​

Answer 1

Qᴜᴇsᴛɪᴏɴ :

➥ An iron pillar has some part in the form of a right circular cylinder and the remaining in the form of a right circular cone. The radius of the base of each of the cone and cylinder is 8 cm. The cylinder part is 240 cm high and the conical part is 36 cm high. Find the weight of the pillar if 1 cubic centimetre of iron with 7.5 g.

Aɴsᴡᴇʀ :

➥ The weight of the iron pillar = 380.16 kg

Gɪᴠᴇɴ :

➤ Radius of the cylinder (r) = 8 cm

➤ Height of the cone (r) = 8 cm

➤ Height of the cylinder (h) = 240 cm

➤ Height of the cone (H) = 36 cm

Tᴏ Fɪɴᴅ :

➤ The weight of the iron pillar = ?

Sᴏʟᴜᴛɪᴏɴ :

Total volume of the iron pillar = volume of the cylinder + volume of the cone

➩ πr²h + $\sf{\dfrac{1}{3}}$πr²H

➩ πr²h $\sf{\left( h + \dfrac{1}{3}H\right)}$

☛ On putting values

$\sf{: \implies \left[\dfrac{22}{7} \times 8 \times 8 \left( 240 + \dfrac{1}{3} \times 36 \right) \right] }$

$\sf{:\implies \underline{ \overline{ \boxed{ \green{ \bf{\:\:50688 \: {cm}^{3}\:\:}}}}} }$

$\therefore$ weight of the pillar = volume in cm³ × weight per cm³

$\sf{: \implies \left( \dfrac{50688 \times 7.5}{1000} \right) kg}$

$:\implies\sf \left({\cancel{\dfrac{380160}{1000}}}\right)kg$

$: \implies \underline{ \overline{ \boxed{ \purple{ \bf{\:\:380.16 \: kg\:\:}}}}}$

Hence, the weight of the pillar is 380.16 kg.

Answer 2

The volume of the cylindrical part = πr²h

= 22/7 x 8² x 240

= 48274 2/7 cm³

The volume of the conical part = 1/3 x base area x height

= 1/3 x 22/7 x 8² x 36

= 2413 5/7 cm³

Total volume = 48274 2/7 + 2413 5/7

= 50688 cm³

Mass = Density x Volume

= 7.5 x 50688

= 380160g

= 380.16 kg