Question

A container, open from the top and made 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 (in rupees) of the milk which can completely fill the container, at the rate of ₹ 20 per litre.

Also, find the cost (in rupees) of metal sheet used to make the container, if it costs ₹ 8 per 100$c{m}^2$. (Take $\pi$ = 3.14).

• A. 209; 156.75
• B. 221; 156.75
• C. 209; 162.5
• D. 221; 162.5

Answer 1

The correct option is A

209; 156.75

Height of frustum of cone = h = 16 cm

Radius of lower end $=r_1=8cm$

Radius of upper end $=r_2=20cm$

Volume of container =V $=\frac{1}{3}\pi h\left(\right(r_1{)}^2+(r_2{)}^2+(r_1\left)\right(r_2)$

$=\frac{1}{3}\times 3.14\times 16\left(\right(8{)}^2+(20{)}^2+(8\left)\right(20\left)\right)$

$=10449.92c{m}^3$

10449.92 $c{m}^3$ = 10.44992 liters

Cost of milk to fill the container at the rate of ₹ 20 per liter = 20 × 10.44992 = ₹ 209

Slant height of frustum of cone(l)$=\sqrt{(h^2+(r_2-r_1{)}^2)}=\sqrt{({16}^2+(20-8{)}^2)}=\sqrt{(256+144)}=\sqrt{400}$ = 20 cm

Total surface area of frustum $=\pi l(r_1+r_2)+\pi (r_1{)}^2=3.14\times 20(8+20)+(3.14\times 64)=1758.40+200.96=1959.36c{m}^2$

100 $c{m}^2$ of metal sheet costs = ₹ 8

1 $c{m}^2$ of metal sheet would cost = ₹$\frac{8}{100}$

1959.36 $c{m}^2$ of metal sheet would cost = $\frac{8}{100}\times 1959.36=₹156.75$