Q.A hollow cylindrical iron pipe is 6 meters long. Its internal and external diameters are 3.5 cm and 4.2 cm respectively. Calculate the volume of iron in the pipe. If the weight of iron is 5 kg per cubic decimeter, then find the total weight of the pipe.

The internal diameter of the pipe = 3.5 cm = 0.35 decimeters ∴ Internal radius of the pipe = \(\cfrac{0.35}{2}\) decimeters The external diameter of the pipe = 4.2 cm = 0.42 decimeters ∴ External radius of the pipe = \(\cfrac{0.42}{2}\) decimeters Length of the cylindrical pipe = 6 meters = 60 decimeters Volume of iron in the hollow cylindrical pipe = \(\pi \left\{ \left(\cfrac{0.42}{2}\right)^2 - \left(\cfrac{0.35}{2}\right)^2 \right\} \times 60\) cubic decimeters = \(\cfrac{22}{7} \times \left(\cfrac{0.42}{2} + \cfrac{0.35}{2}\right) \times \left(\cfrac{0.42}{2} - \cfrac{0.35}{2}\right) \times 60\) cubic decimeters = \(\cfrac{22}{7} \times \cfrac{0.77}{2} \times \cfrac{0.07}{2} \times 60\) cubic decimeters = 2.541 cubic decimeters If the weight of iron is 5 kg per cubic decimeter, then the weight of the pipe = \(2.541 \times 5\) kg = 12.705 kg
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