Question

A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the cost of the milk which can completely fill the container, at the rate of Rs.20 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs.8 per 100 cm². [Take it = 3.14]​

Answer 1

$\green{\fbox{\tt{࿐answer \: ࿐}}}$

R = 20 cm

r = 8 cm

h = 20 cm

$l = \sqrt{h ^{2} } + (r \: + \: r \: ) ^{2}$

$= \sqrt{16} ^{2} + (20 + 8) ^{2}$

∴ Slant height, Quantity of milk in the container = 10449.82100010449.821000

$= \sqrt{256} + 144$

$\sqrt{400}$

$l = 20cm$

$volume \: of \: the \: metallic \: sheet$

$\frac{1}{3} \pi \: h \: ( {r}^{2} + {r}^{2} + rr \:$

$\frac{1}{3} \pi \times 16( {20}^{2} + {8}^{2} + 20 \times 8$

Cost of 1 litre of milk is Rs. 20,

$\frac{1}{3} \pi \: \times 16 \times 624$

$\frac{1}{3} \pi \times 16 \times 624$

$\frac{1}{3} \times 3.14 \times 16 \times 624$

$10449.92 {cm}^{3}$

Cost of 10.45 litres of milk ……. ??

∴ 20 × 10.45 = Rs. 209

. Cost of metal sheet = π(R + r) + πr2

= π {20 × (20 + 8) + (8)2}

= 3.14 × 624 = 1959.36 cm2 .

∴ Cost of preparing metallic container :

For 100 cm2 Rs. 8,

∴ For 1959.36 cm2 …… ??

= 8 x 1159.361001159.36100

= Rs. 156.75.

$\green{\fbox{\tt{࿐i \: hope \: it \: helps \: \: ࿐}}}$