Question

a cyndrical iron pipe has an Outer diameter 28cm and inner diameter of 14 cm its length is 1 m find the cost of painting at that 10 rupees per centimetre cube.please help me

Answer 1

Solution:

Since it is a hollow cylinder,and both the ends of a pipe are open

Area to be painted is outer curved surface area,and ring in both the ends

Outer radius R =14 cm

Inner radius r = 7 cm

length /height h = 1 m = 100 cm

CSA of Cylinder =
$2 \times \pi \times R \times h \\ \\ = 2 \times \frac{22}{7} \times 14 \times 100 \\ \\ = 4 \times 2200 \\ \\ = 8800 \: {cm}^{2}$
Surface area of one end ring:
$= \pi {(R - r)}^{2} \\ \\ = \frac{22}{7} \times {(14 - 7)}^{2} \\ \\ = \frac{22}{7} \times 49 \\ \\ = 22 \times 7 \\ \\ = 154 \: {cm}^{2}$
Surface area of two circular rings=
$154 \times 2 \\ \\ = 308 \: {cm}^{2}$
Total area to be painted = CSA of pipe + 2 circular rings at end

$=8800 + 308 \\ \\ = 9108 \: {cm}^{2}$
cost of painting at the rate 10 rupees per cm-sq

Total cost

$9108 \times 10 \\ \\ = 91080 \: Rs$