Question

A solid cube of iron 53cm×40cm×15cm is melted and recast in cylindrical pipe and outter and inner radius of pipe is 8and 7 cm respectively find lenght of pipe

Answer 1

Volume of the Cuboid= l×b×h,

= 53×40×15,

=31800 cm³.

So, volume of the iron which makes the rod= 31800 cm³,

Inner radius, r = 7 cm,

Outer Radius, R= 8 cm.

Then, volume of the pipe= Area of the ring × height.

= (π R²-π r²)× h,

= π (R²-r²)× h = 53× 40× 15,

(8²-7²)× h= 53× 40× 15 × 7/22,

(64-49)× h= 53× 40× 15× 7/22,

h = 53× 40× 15× 7/22× 15,

So, Height= 53× 40× 7/22,

= 674.54 cm (approximately).

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