Question

A hollow cylindrical iron pipe with external and internal radii 8 cm and 6 cm
respectively and length 35 cm is melted and recast into a solid wire of thickness
2.8 cm. Find the length of the wire.

Step by step plzz!!​

Answer 1

The external radius of the cylinder is :- 8cm

The internal radius of the cylinder is :- 6cm

Height or length of the cylinder is :- 35 cm

Thickness of wire = diameter of the wire = 2.8 cm

Therefore , Radius = 2.8 / 2 = 1.9 cm

Volume of cylinder = Volume of solid wire

[ Since, the cylinder is melted and recast into the solid wire ]

Volume of cylinder = π( R² - r² ) h

➡ Volume of cylinder =22 / 7 × ( 8² - 6² )× 35

➡Volume of cylinder =22 / 7 × ( 64 - 36 )× 35

➡ Volume of cylinder = 22 / 7 × 28 × 35

➡ Volume of cylinder = 22 × 5 × 28

➡ Volume of cylinder = 110 × 28

➡ Volume of cylinder = 3080 cm³

Volume of cylinder = Volume of wire = 3080 cm³

Therefore, Volume of wire = πr²l

3080 = 22 / 7 × 1.9 × l

➡ ( 3080 × 7 ) / ( 22 × 1.9 ) = l

➡ l = 515.78 cm

The length of the wire is 515.78 cm.

Answer 2

GIVEN :

• External radii of a hollow cylindrical pipe = 8 cm.
• Internal radii of a hollow cylindrical pipe = 6 cm.
• Length = 35 cm is melted.
• Thickness of a solid wire = 2.8 cm, Radius = $\sf \dfrac {2.8}{2} \ = \ 1.9 \ cm$

TO FIND :

• The length of the wire = ?

SOLUTION :

To find the length of the wire.

Volume of cylinder = volume of wire.

$\large \red {\underline {\boxed {\sf Volume \ of \ cylinder \ = \ \pi (R^2 \ - \ r^2) \times h}}}$

Put the given values in the above formula,

:$\implies \sf \dfrac {22}{7} \times (8^2 - 6^2) \times 35$

:$\implies \sf \dfrac {22}{7} \times (64 - 36) \times 35$

:$\implies \sf \dfrac {22}{7} \times 28 \times 35$

:$\implies \sf 22 \times 5 \times 28$

:$\implies \sf 110 \times 28$

$\therefore$ Volume of cylinder = 3080 cm³.

Now,

Volume of cylinder = volume of wire = 3080 cm³.

$\sf Volume \ of \ wire \ = \ \pi r^2 l$

:$\implies \sf 3080 \ = \ \dfrac {22}{7} \times 1.9 \times l$

:$\implies \sf \dfrac {(3080 \times 7)}{(22 \times 1.9)} \ = \ l$

$\qquad \large \red {\underline {\boxed {\sf l \ = \ 515.78 \ cm}}}$

$\therefore$ The length of the wire is 515.78 cm.