Question

A milk container of height 16 cm is made of metal sheet in the form of frustum of cone with radii of its lower and upper ends as 8 cm and 20cm respectively. Find the cost of milk at the rate of rupees 22 per litre which the container can hold

Answer 1

Answer:

Step-by-step explanation:

volume of a frustum= 1/3 × 22/7 × h (1r2+ 2r2+ r1r2)

=1/3×22/7×16 (8'2+20'2+8×20)

=1/3×22/7×16(64+400+160)

=1/3×22/7×16 (624)

=22/7×16 (208)

=352×29.71

=10459.42 Ltrs

now,if 1 litre costs 22 rupees

therefore the container holds the milk of cost 230107.24 rupees

Answer 2

Given that , the height of milk container (h) = 16 cm

Radius of lower end of milk container (r) = 8 cm

and radius of upper end of milk container (R) = 20 cm

∴Volume of the milk container

$= \frac{\pi h}{3} (R^{2} +r^{2} +Rr)\\\\= \frac{22}{7} \times\frac{16}{3} (20^{2} +8^{2} +20\times8 )\\\\= \frac{22\times16}{21} (400+64+160)\\\\= \frac{22\times16\times624}{21}= \frac{219648}{21} \\\\= 10459.42cm^{3} = 10.45942 L$

Cost of one L milk = 22/-

Cost of 10.49542L milk = 22*10.49542 = 230.12 Rs