Question

An iron pipe 20cm long has exterior diameter 25cm. If the thickness of the pipe is 1 cm , then the whole surface area of the pipe

Answer 1

Length of the pipe = 20 cm

External diameter = 25 cm

External radius, R = 12.5 cm Thickness of the pipe, w = 1 cm Internal radius = 12.5 – 1 = 11.5 cm

Area of the pipe = outer area – inner area Outer area = 2πr(r + h) = 2π×12.5(12.5 + 20) = (812.5)π Inner area = 2π×112.5(11.5 + 20) = (724.5)π

Hence area of the pipe = (812.5)π – (724.5)π = 88π

Answer 2

Answer:

We have,

R = external radius = 12.5 cm

r = Internal radius = ( External radius - thickness ) = (12.5 - 1) = 11.5 cm

h = length of the pipe = 20 cm

______________________

$\therefore$$\sf{Total\:surface\:area\:of\:pipe}$

= (external curved surface) + (internal curved surface) + 2(area of the base of the ring)

= $\sf{2πRh+2πrh+2(πR^2-πr^2)}$

= $\sf{2π(R+r)h+2π(R^2-r^2)}$

= $\sf{2π(R+r)h+2π(R+r)(R-r)}$

= $\sf{2π(R+r)(h+R-r)}$

= $\sf{2×}$$\sf\frac{22}{7}×(12.5+11.5)×(20+12.5-11.5)cm^2$

= $\sf{2×}$$\sf\frac{22}{7}×24×21\:\:cm^2$

= 3168 cm²