Question

Two cylinders are $12$ cm and $18$ cm in height. Their diameters are $12$ cm and $16$ cm respectively. Both the cylinders are melted and the material is moulded into a solid right circular cone of height $33$ cm. Find its diameter.

Answer 1

Diameter of cylinder$d_1=12$cm

Height of cylinder$h_1=12$cm

Volume of cylinder$V_1=\pi r_1^2{h}_1=\pi {\left(\frac{d_1}{2}\right)}^2{h}_1=\pi {\left(\frac{12}{2}\right)}^2\times 12=\frac{\pi {12}^2\times 12}{4}$ ...........$\left(1\right)$

since radius $=\frac{1}{2}\times$diameter.

Diameter of cylinder$d_2=18$cm

Height of cylinder$h_2=16$cm

Volume of cylinder$V_2=\pi r_2^2{h}_2=\pi {\left(\frac{d_2}{2}\right)}^2{h}_2=\pi {\left(\frac{18}{2}\right)}^2\times 16=\frac{\pi {18}^2\times 16}{4}$ ...........$\left(2\right)$

since radius$=\frac{1}{2}\times$diameter.

When both the cylinders are melted and moulded into a single cylinder, then the

Volume of single cylinder$V=\pi r^{2}h=\pi {\left(\frac{d}{2}\right)}^{2}h=\frac{\pi d^2\times 33}{4}$ ...........$\left(3\right)$ where $h=33$cm

Now, $\frac{\pi d^2\times 33}{4}=\frac{\pi {12}^2\times 12}{4}+\frac{\pi {18}^2\times 16}{4}$

$\Rightarrow 33d^2={12}^3+{18}^2\times 16$

$\Rightarrow d^2(3\times 11)=12\times 144+18\times 18\times 16$

$\Rightarrow 11d^2=576+1728=2304$ on simplification

$\Rightarrow d^2=\frac{2304}{11}=209.45$

$\therefore d=\sqrt{209.45}=1.316$cm