Steel tube

The steel tube has an inner diameter of 4 cm and an outer diameter of 4.8 cm. The density of the steel is 7800 kg/m3. Calculate its length if it weighs 15 kg.

Correct result:

l =  3.478 m

Solution:

$$\begin{gathered} D_1=4\text{ cm}\to \text{ m}=4\mathrm{/}100\text{ m}=0.04\text{ m } \\ {D}_2=4.8\text{ cm}\to \text{ m}=4.8\mathrm{/}100\text{ m}=0.048\text{ m } \\ {r}_1=D_1\mathrm{/}2=0.04\mathrm{/}2={\displaystyle \frac{1}{50}}=0.02\text{ m } \\ {r}_2=D_2\mathrm{/}2=0.048\mathrm{/}2={\displaystyle \frac{3}{125}}=0.024\text{ m } \\ S=\pi \cdot r_2^2-\pi \cdot r_1^2=3.1416\cdot 0.024^2-3.1416\cdot 0.02^2\doteq 0.0006{\text{m}}^2 \\ m=15\text{ kg } \\ h=7800{\text{kg/m}}^3 \\ V=m\mathrm{/}h=15\mathrm{/}7800={\displaystyle \frac{1}{520}}\doteq 0.0019{\text{m}}^3 \\ l=V\mathrm{/}S=0.0019\mathrm{/}0.0006=3.478\text{ m} \end{gathered}$$

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