The length of a metallic pipe is 7.56 m. Its external and internal radii are 2.5 cm and 1.5 cm respectively. If 1 cm$$^3$$ of the metal weigh 7.5 g, then the weight of the pipe is:(Take $$\pi = \frac{22}{7}$$)

Given that,

The length of a metallic pipe is 7.56 m

External radius and internal radius of the pipe$$(R)=2.5cm=2.5\times 10^{-2}$$m and $$r=1.5cm=1.5\times 10^{-2}$$m

The density of the metal $$\rho=\dfrac{7.5gm}{1cm^3}=\dfrac{7.5 \times 10^{-3}}{1\times10^{-6}m^3}=7500\dfrac{kg}{m^3}$$

We know that the volume of the cylinder $$=\pi r^2 h $$

So, Volume of the pipe $$(V)=\pi (R^2-r^2) h=\dfrac{22\times[(2.5\times 10^{-2})^2-(1.5\times 10^{-2})^2]\times 7.56}{7}$$

$$\Rightarrow (V)=\dfrac{22\times[(6.25\times 10^{-4})-(2.25\times 10^{-4})]\times 7.56}{7}$$

$$\Rightarrow (V)=\dfrac{22\times[4\times 10^{-4}]\times 7.56}{7}$$

$$\Rightarrow (V)=22\times4\times 10^{-4}\times 1.08=95.04\times 10^{-4}m^3$$

$$\Rightarrow (V)=95.04\times 10^{-4}m^3$$

We know that mass (m) = density x Volume $$=\rho \times V$$

$$\Rightarrow m=7500 \times 95.04\times 10^{-4}=71.28 Kg$$