Theory:

1. A metallic cylinder of height \(36 \ cm\) and radius of base \(8 \ cm\) is melted and reshapes into a form of a sphere. Find the diameter of the sphere.
 
Solution:
 
Height of the cylinder \(=\) \(36 \ cm\)
 
The radius of the base of the cylinder \(=\) \(8 \ cm\)
 
Volume of the cylinder \(=\) \(\pi r^2 h\) cu. units
 
\(=\) \(\pi \times 8^2 \times 36\)
 
\(= 2304 \pi \ cm^3\)
 
Let '\(r\)' be the radius of the sphere.
 
Volume of sphere \(=\) \(\frac{4}{3} \pi r^3\) cu. units
 
Since the cylinder reshapes into a sphere, the volume remains unchanged.
 
Volume of sphere \(=\) Volume of the cylinder
 
\(\frac{4}{3} \pi r^3 = 2304 \pi\)
 
\(r^3\) \(=\) 2304π×3π×4
 
\(r^3\) \(=\) \(1728\)
 
\(r = 12\)
 
\(d = r \times 2 = 12 \times 2 = 24\)
 
Therefore, the diameter of the sphere is \(24 \ cm\).
 
 
2. How many spherical lead shots of radius \(3.5 \ cm\) can be made out of a solid rectangular lead piece with dimensions \(33 \ cm\), \(21 \ cm\) and \(14 \ cm\).
 
Solution:
 
The radius of the spherical lead shot \(=\) \(3.5 \ cm\)
 
Volume of spherical lead shot \(=\) \(\frac{4}{3} \pi r^3\) cu. units
 
\(=\) \(\frac{4}{3} \times \frac{22}{7} \times (3.5)^3\)
 
\(=\) \(\frac{539}{3}\) \(cm^3\)
 
Length of the rectangular lead piece \(=\) \(33 \ cm\)
 
Breadth of the rectangular lead piece \(=\) \(21 \ cm\)
 
Height of the rectangular lead piece \(=\) \(14 \ cm\)
 
Volume of the rectangular lead piece \(=\) \(lbh\) cu. units
 
\(= 33 \times 21 \times 14\)
 
\(= 9702 \ cm^3\)
 
\(\text{Number of spherical lead shots} = \frac{\text{Volume of rectangular lead piece}}{\text{Volume of spherical lead shots}}\)
 
\(=\) 97025393
 
\(=\) 9702×3539
 
\(=\) \(54\)
 
Therefore, \(54\) spherical leads can be made from the rectangular lead piece.