Question

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

Answer 1

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-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 2

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Let the height be h

We know that

Radius = 1.4 cm = 14 mm

let radius of new cy = x

Vol of new cyli (V) = Vol of metal initially(v)= vol of outer cy - vol of inner cylinder.

then

V = πh(R²-r²) = πh(14²-1²) = 22/7 ×h×195

= 612.3h mm³

then

V = πx²h = 22/7 ×x²h = 612.3h

x = √612.3×7/22 = 195 ≈ 14 mm

then diameter = 28 mm or 2.8 cm