Question

Friends helps this math question.​

Answer 1

Hey mate ✋✋

✔✔Thanks for the question ✔✔

Here's your answer ➡➡

Given,

h = 16 cm
$r_{1}$= 20 cm
$r_{2}$= 8 cm

$\therefore$ volume = $\frac{1}{3} \pi{h}( { r_{1} }^{2} + r_{1} × r_{2} + {r_{2}}^{2}) \\ \\ = \frac{1 \times 314 \times 16}{3 \times 100} ( {20}^{2} + 20 \times 8 + {8}^{2} ) \\ \\ = \frac{314 \times 16}{3 \times 100} (400 + 160 + 64) \\ \\ = \frac{314 \times 16}{3 \times 100} \times 624 \\ \\ = \frac{104492}{100} \\ \\ = 1044.92 \: {cm}^{3}$

Now, milk in container =$\frac{1044.92}{1000}$ l
=10.449 l

$\therefore$ cost = ₹ 20 × 10.449
= ₹ 209

Area of metal used =$\pi( r_{1} + r_{2}) l + \pi { r_{2}}^{2} \\ \\ = \frac{314}{100} (8 + 20) \sqrt{ {16}^{2} + {(20 - 8)}^{2} } + \frac{314 \times 64}{100} \\ \\ = \frac{314 \times 28}{100} \times \sqrt{256 + 144} \: + \frac{314 \times 64}{100} \\ \\ = \frac{314 \times 28}{100} \times \sqrt{400} + \frac{314 \times 64}{100} \\ \\ = \frac{314 \times 28 \times 20}{100} + \frac{314 \times 64}{100} \\ \\ = \frac{314 \times 28 \times 20 + 314 \times 64}{100} \\ \\ = \frac{195936}{100} \\ \\ = 1959.36 \: {cm}^{2}$

Again,

Cost of 100${cm}^{2}$ metal = ₹ 8

cost of 1${cm}^{2}$ metal = ₹ $\frac{8}{100}$

cost of 1959.36${cm}^{2}$ metal = ₹ $\frac{8}{100}$ × 1959.36

= ₹156.74