Question

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

Answer 1

$\huge\mathfb{Solution :-}$

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

the length of wire(h) =5m

Step-by-step explanation:

A hollow cylindrical iron pipe external (R) =8cm

A hollow cylindrical iron pipe internal radii (r) =6cm

A hollow cylindrical iron pipe length (h) = 35cm

solid wire of thickness = 2.8cm

the length of wire(h) = ?

A hollow cylindrical iron pipe (v) = πh(R+r) (R-r)

= π x35 (8+6) (8-6)

=π x35 x28

= 980πcm^3

Again,

solid wire of thickness (r) = 2.8\2 =1.4cm

now,

solid wire of thickness of cylindrical (v)=hollow cylindrical iron pipe

(v)

so,

πr^2h = πh(R+r) (R-r)

or, (1.4) ^2h = 980 [π = 22\7 ]

1.96 h =980π

h = 980π\ 1.6π

h= 500cm

h = 500\100 [ 1m = 100cm]

h = 5m.

the length of wire(h) =5m