Question

A metallic bucket open at the top of height $24cm$ is $I$ the form of frustum of a cone the radii of whose lower and upper ends are $7cm$ and $14cm$ respectively find
volume of the water that can fill the bucket

Answer 1

Volume of feustum= volume of one $OCD-OAB$

$\Rightarrow \frac{1}{3}\pi r_2^2{h}_2-\frac{1}{3}\pi r_1^2{h}_1$

$=\frac{\pi }{3}({14}^2{h}_2-7^2{h}_1)$

$\frac{\pi }{3}\times 7^2(4h_2-h_1)$

Now, In $△OMB,\angle OBM=\angle ODN\left(MB||ND\right),\angle OMB-\angle OND={90}^o,\angle O=\angle O$ By $AA$ simplify $△OMB\sim △ONB$.

$\Rightarrow \frac{h_1}{h_2}=\frac{r_1}{r_2}\Rightarrow \frac{h_1}{h_2}=\frac{7}{14}=\frac{1}{2}\Rightarrow 2h_1$

Now, $h_2-h_1=24\Rightarrow 2h_1-h_1=24\Rightarrow h_1=24cm,h_2=48cm$

$\Rightarrow$ volume $=\frac{49\pi }{3}(8h_1-h_1)=\frac{49\pi }{3}\times 7\times 248$

$=49\times 56\pi =8616.16c{m}^3$