Question

A milk container is in the form of frustum of a cone, whose volume is $10459c{m}^3$. The radii of lower and upper circular ends at 8 cm and 20 cm respectively. Find the cost of metal sheet used in making the container at the rate of Rs. 2/- per square centimeter.

Answer 1

For frustum,

Volume $\left(V\right)=10459{cm}^3$

Radius of lower circular end $\left(r_1\right)=8cm$

Radius of upper circular nd $\left(r_2\right)=20cm$

TNow,

Volume of frustum $=\frac{\pi h}{3}({r_1}^2+{r_2}^2+r_1{r}_2)$

$10459=\frac{22\times h}{7}(64+400+160)$

$\Rightarrow h=\frac{10459\times 7}{22\times 624}=5.33cm$

Now, slant height $\left(l\right)$ is given as-

$l=\sqrt{h^2+{(r_1-r_2)}^2}$

$\Rightarrow l=\sqrt{28.40+144}=13.13cm$

Therefore,

Total area of the frustum $\left(A\right)=\pi ({r_1}^2+{r_2}^2+(r_1+r_2)l)$

$\Rightarrow A=\frac{22}{7}(64+400+367.64)$

$\Rightarrow A=2613.725{cm}^2$

Given that the rate of making sheet is $Rs.2\text{per sq. cm}$

Cost of metal sheet $=area\times rate$

$\Rightarrow$ Cost of metal sheet $=3218.285\times 2=Rs.5227.45$

Hence the cost of metal sheet is $Rs.5227.45$.

Hence the correct answer is $5227.45$.