Question

A metallic cylinder has radius $3$cm and height $5$cm. It is made of metal A. To reduce its weight, a conical hole is drilled in the cylinder, as shown and it is completely filled with a lighter metal B. The conical hole has a radius of $3/2$cm and its depth is $8/9$cm. Calculate the ratio of the volume of the metal A to the volume of the metal B in the solid.

Answer 1

Volume of metal $B=\frac{1}{3}\pi r^{1^2}{h}^1=\frac{1}{3}\pi {\left(\frac{3}{2}\right)}^2\left(\frac{8}{9}\right)=\frac{2\pi }{3}$

Volume of metal $A=\pi r^{2}h-\frac{1}{3}\pi r^{1^2}{h}^1=\frac{\pi }{3}(3r^{2}h-r^{1^2}{h}^1)$

$=\frac{22}{7\times 3}(3\times 9\times 5-\frac{9}{4}\times 89)=\frac{22}{21}(135-2)=\frac{22\times 133}{21}$

$=139.33c{m}^3$

ratio $=\frac{139.33\times 3}{2\pi }=\frac{209}{3.14}=\frac{209}{\pi }=\frac{209\times 7}{22}=\frac{133}{2}$