Question
From a solid cylinder of height $14\ cm$ and base diameter $7\cm,$ two equal conical holes each of radius 2.1cm and height 4cm are cut: off. Find the volume of the remaining solid.
✓
Answer
Height of the cylinder $(\mathrm{h})=14 \mathrm{~cm}$,
Base diameter $=7 \mathrm{~cm}$
$\Rightarrow$ Radius of the base of the cylinder $(r)=3.5 \mathrm{~cm}$
Volume of the cylinder $\pi \mathrm{r}^2 \mathrm{~h}$
$ =\frac{22}{7} \times 3.5 \times 3.5 \times 14 $
$ =22 \times 0.5 \times 3.5 \times 14 $
$ =539 \mathrm{~cm}^3$
Radius of the conical holes $\left(r_1\right)=2.1 \mathrm{~cm}$,
Height of the conical holes $\left(h_1\right)=4 \mathrm{~cm}$
Volume of the conical hole $=\frac{1}{3} \pi \mathrm{r}_1{ }^2 \mathrm{~h}_1$
$ =\frac{1}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 4 $
$ =18.48 \mathrm{~cm}^3$
Volume of the two conical hole $=2 \times 18.48$
$=36.96 \mathrm{~cm}^3$
Volume of the remaining solid = volume of the cylinder - Volume of two conical hole
$ =539-36.96 $
$ =502.04 \mathrm{~cm}^3$
Base diameter $=7 \mathrm{~cm}$
$\Rightarrow$ Radius of the base of the cylinder $(r)=3.5 \mathrm{~cm}$
Volume of the cylinder $\pi \mathrm{r}^2 \mathrm{~h}$
$ =\frac{22}{7} \times 3.5 \times 3.5 \times 14 $
$ =22 \times 0.5 \times 3.5 \times 14 $
$ =539 \mathrm{~cm}^3$
Radius of the conical holes $\left(r_1\right)=2.1 \mathrm{~cm}$,
Height of the conical holes $\left(h_1\right)=4 \mathrm{~cm}$
Volume of the conical hole $=\frac{1}{3} \pi \mathrm{r}_1{ }^2 \mathrm{~h}_1$
$ =\frac{1}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 4 $
$ =18.48 \mathrm{~cm}^3$
Volume of the two conical hole $=2 \times 18.48$
$=36.96 \mathrm{~cm}^3$
Volume of the remaining solid = volume of the cylinder - Volume of two conical hole
$ =539-36.96 $
$ =502.04 \mathrm{~cm}^3$
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