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 cm3 of wood has a mass of 0.6 g.
Inner radius of the wooden pipe, $r=\frac{24}{2}=12 \mathrm{~cm}$
Outer radius of the wooden pipe, $R=\frac{28}{2}=14 \mathrm{~cm}$
Length of the wooden pipe, h = 35 cm
$\therefore$ Volume of wood in the pipe $=\pi\left(R^{2}-r^{2}\right) h=\frac{22}{7} \times\left(14^{2}-12^{2}\right) \times 35=5720 \mathrm{~cm}^{3}$
It is given that 1 cm3 of wood has a mass of 0.6 g.
$\therefore$ Mass of the pipe $=$ Volume of wood in the pipe $\times 0.6=5720 \times 0.6=3432 \mathrm{~g}=\frac{3432}{1000}=3.432 \mathrm{~kg}$
Thus, the mass of the pipe is 3.432 kg.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.