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# The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 $${cm}^{3}$$ of wood has a mass of 0.6 g.

Given :
Inner radius of cylindrical pipe (r) = $$\frac{24}{2} cm$$ = 12 cm
Outer radius of cylindrical pipe (R) = $$\frac{28}{2} cm$$ = 14 cm
Height (h) of pipe = Length of pipe = 35 cm

We know that,

Volume of pipe
= $${\pi}({R}^2 - {r}^2)h$$
= $$\frac{22}{7} × ({14}^2 - {12}^2) × 35$$ $${cm}^{3}$$
= 110 x 52 $${cm}^{3}$$
= 5720 $${cm}^{3}$$

Now, Mass of 1 $${cm}^{3}$$ wood = 0.6 g

So, Mass of 5720 $${cm}^{3}$$ wood br> = (5720 x 0.6) g
= 3432 g
= 3.432 kg

Therefore, mass of the pipe is 3.432 kg.