# A Solid Cuboid of Iron with Dimensions 53 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 - Mathematics

A solid cuboid of iron with dimensions 53 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.

#### Solution

Volume of solid cuboid of iron = Volume of cylindrical pipe

$\Rightarrow lbh = \pi h\left( R^2 - r^2 \right)$

$\Rightarrow 53 \times 40 \times 15 = \frac{22}{7} \times h\left[ \left( \frac{8}{2} \right)^2 - \left( \frac{7}{2} \right)^2 \right]$

$\Rightarrow 53 \times 40 \times 15 = \frac{22}{7} \times h\left[ 4^2 - \left( 3 . 5 \right)^2 \right]$

$\Rightarrow 53 \times 40 \times 15 = \frac{22}{7} \times h \times 3 . 75$

$\Rightarrow h = 2698 . 18 cm$

Disclaimer: The answer given in the book is not correct.

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#### APPEARS IN

RD Sharma Class 10 Maths
Chapter 14 Surface Areas and Volumes
Exercise 14.1 | Q 23 | Page 29