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

volume of cubical iron =volume of cylindrical pipe
53 x 40 x 15=pi. {R^2-r^2} L

53 x 40 x 15=22/7 x {(8/2)^2-(7/2)^2} L

53 x 40 x 15=22/7 x 15/4 x L

L=(53 x 40 x 15)/(22/7 x 15/4)cm

Answer 2

Answer:-

Let the length of the pipe be h cm.

Then, Volume of cuboid = (53 x 40 x 15) cm3

Internal radius of the pipe = 7/2 cm = r

External radius of the pipe = 8/2 = 4 cm = R

So, the volume of iron in the pipe = External Volume – Internal Volume

= πR2²h – πr²h

= πh(R²– r²)

= πh(R – r) (R + r)

= π(4 – 7/2) (4 + 7/2) x h

= π(1/2) (15/2) x h

The volume of iron in the pipe = volume of iron in cuboid

π(1/2) (15/2) x h = 53 x 40 x 15

h = (53 x 40 x 15 x 7/22 x 2/15 x 2) cm

h = 2698 cm

Therefore, the length of the pipe is 2698 cm.