Verified answer

cylindrical iron pole :

height h = 4 m

diameter d = 28 cm

radius r = 14 cm

cm to m : ÷ 100

[tex] = \frac{14}{100} m[/tex]

volume of cylinder = πr²h

[tex] = \frac{22}{7} \times \frac{14}{100} \times \frac{14}{100} \times 4[/tex]

[tex] = 22 \times \frac{2}{100} \times \frac{14}{100} \times 4[/tex]

[tex] = \frac{2464}{100 \times 100} \: {m}^{3} [/tex]

weight of 1 m³ = 50 kg

[tex]weight \: of \: \frac{2464}{100 \times 100} \: m³ =\frac{2464}{100 \times 100} \times 50[/tex]

[tex] = \frac{2464}{2 \times 100} [/tex]

[tex] = \frac{1232}{100} [/tex]

= 123.2 kg