Question

Question 4
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 ₹ 20 per litre. Also find the cost of metal sheet used to make the container, if it costs ₹ 8 per 100 square centimetres.

Answer 1

Height of frustum = 16 cm, R = 20 cm, r = 8 cm
Volume of frustum
$=\frac{1}{3}\pi h(R^2+r^2+Rr)$
$=\frac{1}{3}\pi \times 16({20}^2+8^2+160)$
$=\frac{1}{3}\times \frac{22}{7}\times 16(400+64+160)$
$=\frac{1}{3}\times \frac{22}{7}\times 16\times 624$
$=10459.42c{m}^3$

Cost of milk at ₹ 20 per 1000 cubic cm
$10.45942\times 20=₹209.18$

For calculating surface area, we need to find slant height which can be calculated as follows:
$l=\sqrt{h^2+(R-r{)}^2}$
$=\sqrt{{16}^2+(20-8{)}^2}$
$=\sqrt{256+144}$
$=\sqrt{400}=20cm$

Surface area of frustum
$=\pi (R+r)l+\pi r^2$
$=\pi \left[\right(R+r)l+r^2]$
$=\pi \left[\right(20+8)20+8^2]$
$=\pi (560+64)$
$=\frac{22}{7}\times 624=1961.14c{m}^2$

Cost of metal sheet at ₹ 8 per 100 sq cm
$=₹19.6114\times 8=₹156.89$