Answer 1

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.