Question

A milk container of height 16 cm is made of metal sheet in the form of a frustum of a cone with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the cost of milk at the rate of Rs.44 per litre which the container can hold.

Answer 1

We are given,
$h=16cm,r_1=20cm,r_2=8cm$
Volume of frustum of the cone

$=\frac{1}{3}\pi h(r_1^2+r_2^2+r_1{r}_2)$

$=\frac{1}{3}\times 3.14\times 16\times ({20}^2+8^2+20\times 8)=10450c{m}^2$

Since,

$1000c{m}^3=1L;10450c{m}^3=\frac{1}{1000}\times 10450=10.45L$
Cost of 1l of milk= Rs20
Cost of 10.45l of milk= Rs(20$\times$10.45) = Rs 209


Total surface area of the frustum of the cone

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

$=3.14(20+8)\sqrt{{16}^2+(20-8{)}^2}+\pi {20}^2+\pi 8^2=3215c{m}^2$

Cost of$100c{m}^2$of metal sheet =Rs8
Cost of$3215c{m}^2$of metal sheet

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

$rs\frac{8}{100}\times 3215=Rs.250$