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.