# 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.

Result

l =  3.478 m

#### Solution:

$D_{1}=4 \ cm=4 / 100 \ m=0.04 \ m \ \\ D_{2}=4.8 \ cm=4.8 / 100 \ m=0.048 \ m \ \\ \ \\ r_{1}=D_{1}/2=0.04/2=\dfrac{ 1 }{ 50 }=0.02 \ \text{m} \ \\ r_{2}=D_{2}/2=0.048/2=\dfrac{ 3 }{ 125 }=0.024 \ \text{m} \ \\ \ \\ S=\pi \cdot \ r_{2}^2 - \pi \cdot \ r_{1}^2=3.1416 \cdot \ 0.024^2 - 3.1416 \cdot \ 0.02^2 \doteq 0.0006 \ \text{m}^2 \ \\ \ \\ m=15 \ \text{kg} \ \\ h=7800 \ \text{kg/m}^3 \ \\ \ \\ V=m/h=15/7800 \doteq \dfrac{ 1 }{ 520 } \doteq 0.0019 \ \text{m}^3 \ \\ \ \\ l=V/S=0.0019/0.0006 \doteq 3.478 \doteq 3.478 \ \text{m}$

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