Question

A solid cuboid of iron with dimensions 53 cm × 40 cm × 15 cm is melted and recast into a cylindrical pipe. The outer and inner diameters of pipe are 8 cm and 7 cm respectively. Find the length of the pipe.

Answer 1

Hi,

1) Dimensions of the solid cubiod iron ,

Length = l = 53 cm

Breadth = b = 40 cm

Height = h = 15 cm

Volume = v1 = lbh

2) dimensions of the cyliderical pipe,

Outer diameter = D = 8 cm

Outer radius = R = D / 2 = 8/2 = 4 cm

Inner diameter = d = 7 cm

Inner radius = r = d/ 2 = 7 / 2 = 3.5 cm

Let height = H cm

Volume of the pipe = v2 = pi × R ^2 × H - pi × r ^2 × H

= pi × H ( R ^2 - r ^2 )

= pi × H × ( R + r ) ( R - r )

According to the problem,

Solid cubiod melted and recast into a cyliderical pipe.

Therefore,

Volume of the pipe = volume of the solid cubiod

Pi × H × ( R + r ) ( R - r ) = l × b × h

22 / 7 × H × ( 4 + 3.5 ) ( 4 - 3.5 ) = 53 × 40 × 15

22/7 × H × 7.5 × 0.5 = 53 × 40 × 15

H = ( 53 × 40 × 15 × 7 ) / ( 22 × 7.5 × 0.5 )

After cancellation

H = 2698.18 cm

I hope this will usful to you.

*****

Answer 2

Answer:

Volume of cuboid = 53 × 40 × 15

                            = 31800 cm³

Volume of hollow-cylinder = πh (R² - r²)

                                         = 22/7 × h (4² - 3.5²)

                                         = 22/7 × h (16 - 12.25)

                                         = 22/7 × h * 3.75

31800 = 22/7 × h × 3.75

222600 = 22 × h × 3.75

222600 = 82.5 × h

h = 222600/82.5

  = 2698.18 cm