Question

A solid cuboid of iron with dimensions 33 cm × 40 cm × 15 cm is melted and recast into a cylindrical pipe. The outer and inner diameters of pipe are 8 cm and 7 cm respectively. Find the length of pipe.

Answer 1

Given: Dimensions of cuboid: \[l = 33\ \mathrm{cm},\ b = 40\ \mathrm{cm},\ h = 15\ \mathrm{cm}\]

For cylindrical pipe: Outer diameter = \[8\ \mathrm{cm} \implies R = 4\ \mathrm{cm}\]

Inner diameter = \[7\ \mathrm{cm} \implies r = 3.5\ \mathrm{cm} = \dfrac{7}{2}\ \mathrm{cm}\]

To find: Length (height) of the pipe, [H]

Formula: \[\text{Volume of iron in pipe} = \pi(R^2 - r^2)H\]

\[\text{Volume of cuboid} = l \times b \times h\]

\[\pi(R^2 - r^2)H = l \times b \times h\]

Substitution: \[\dfrac{22}{7} \times \left(4^2 - 3.5^2\right) \times H = 33 \times 40 \times 15\]

Calculation: \[4^2 - 3.5^2 = (4 + 3.5)(4 - 3.5) = 7.5 \times 0.5 = 3.75 = \dfrac{15}{4}\] 

\[\dfrac{22}{7} \times \dfrac{15}{4} \times H = 19800\] 

\[\dfrac{165}{14} \times H = 19800\] 

\[H = \dfrac{19800 \times 14}{165} = 120 \times 14 = 1680\ \mathrm{cm}\]

Answer: \[H = 1680\ \mathrm{cm} = 16.8\ \mathrm{m}\]