Question

A container opened at the top and made ip of a metal sheet ,is in the form of a frustum of a cone of height 16cm with radii of its lower and upper ends as 8 and 20 cm find the cost of milk can fill the container at the rate of rs50 per liture

Answer 1

it is Rs 523.8

it is equal to Rs 524

hope it helps u....

Answer 2

Given:

r1 = 20cm,

r2 = 8cm,

h = 16cm

Slant height =

$\longrightarrow$$\sf\red{l = \sqrt{ {h}^{2} + {(r1 - r2)}^{2} }}$

$\longrightarrow$$\sf\red{ \sqrt{ {(16)}^{2} + {(20 - 8)}^{2} } cm}$

$\longrightarrow$$\sf\red{\sqrt{256 + 144} cm}$

$\longrightarrow$$\sf\red{\sqrt{400} cm}$

$\longrightarrow$$\sf\red{20cm}$

Volume of the container =

$\longrightarrow$$\sf\red{v = \frac{1}{3} \pi \: h( {r1}^{2} + {r2}^{2} + r1r2}$

$\longrightarrow$$\sf\red{\frac{1}{3} \times 3.14 \times 16( {(20)}^{2} + {8}^{2} + 20 \times 8) {cm}^{3}}$

$\longrightarrow$$\sf\red{\frac{1}{3} \times 3.14 \times 16(400 + 64 + 16) {cm}^{3}}$

$\longrightarrow$$\sf\red{\frac{1}{3} \times 3.14 \times 16 \times 624 {cm}^{3}}$

$\longrightarrow$$\sf\red{10449.92 {cm}^{3}}$

$\longrightarrow$$\sf\red{ \frac{10449.92}{1000} litres = 10.45litres(approx.)}$

Cost of milk that fills the container at the rates of Rs. 20 per litre

= $\sf\red{Rs.20×10.45=Rs.209}$

TSA of container = CSA of the container + Base area

$\longrightarrow$$\sf\red{\pi(r1 + r2) l + \pi {r2}^{2}}$

$\longrightarrow$$\sf\red{(3.14 \times (20 + 8) \times 20 + 3.14 \times 8 \times 8) {cm}^{2} }$

$\longrightarrow$$\sf\red{(3.14 \times 28 \times 20 + 3.14 \times 64) {cm}^{2}}$

$\longrightarrow$$\sf\red{3.14 \times 624 {cm}^{2}}$

$\longrightarrow$$\sf\red{1959.36 {cm}^{2} }$

Cost of the metal sheet required at the rate of Rs. 8 per 100 cm square

= $\sf\red{Rs.\frac{8}{100} \times 1959.36 = Rs.156.75(approx.}$