An iron pipe of internal diameter 2.8 cm and uniform thickness 1 mm is melted and a solid cylindrical rod of the same length is formed. Find the diameter of the rod.

Answer 1

The cross-section of the pipe has area
pi/4 (2.9^2-2.8^2) = 0.1425pi cm^2

The rod of radius r has area
pi r^2 cm^2

So, pi r^2 = 0.1425 pi
r=0.3775 cm
So, the diameter is 0.755 cm

Answer 2

Its answer is 1.08 cm diameter

Answer 3

Internal diameter =2.8cm
Radius=1.4cm
Thickness =1mm=0.1cm
External radius =1.4cm+0.1cm=1.5cm
(V1) volume of hollow pipe =pie×h×(R+r)(R-r)=(22÷7 )×h×2.9×0.1=0.9hcm^2
(V) volume of rod with same length=pie×r^2×h=(22÷7)×r^2×h
V1=V
or,0.9h=(22÷7)×r^2×h
or,0.29=r^2
:.r=0.54cm
So,diameter of rod =2r=2×0.54=1.08cm

Answer 4

Is the External Radius = Radius * thickness? How

Answer 5

Thanku very much.And fantastic job!

Answer 6

Thanks a lot to solve this.This is one of the great aid to the students who are taking online class in this lockdown period.

Answer 7

Thanks

Answer 8

Thanks a lot.

Answer 9

Yes I GOT IT , THAKK YOU SO MUCH

Answer 10

Thanks man

Answer 11

Where does this radius 1 .4 come from?

Answer 12

And where does these formulas from e.gPie×h×(R+r)(R-r)?