A container opened at 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 milk which can completely fill the container, at t

#School

A container opened at 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 milk which can completely fill the container, at the rate of Rs 50 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs10 per100 cm². (Take $\pi$ = 3·14)

Answers (1)

Given $\rightarrow$ height of frustrum, h= 16 cm
upper end radius, R = 20 cm
Lower end radius, r = 8 cm
slant height frustrum = l

$l= \sqrt{h^{2}+\left ( R-r \right )^{2}}$
$l= \sqrt{16^{2}+\left ( 20-8 \right )^{2}}= \sqrt{256+144}= \sqrt{400}$
$l= 20\, cm$
Volume of milk = volume of the frustum
$= \frac{1}{3}\pi h\left [ R^{2}+r^{2}+Rr \right ]$
$= \frac{1}{3}\times 3\cdot 14\times 16\left [ 20^{2}+8^{2}+20\times 8 \right ]$
$= \frac{1}{3}\times 3\cdot 14\times 16\times 624$
$= 3\cdot 14\times 16\times 208= 10450\, cm^{3}$
$= 10450 \, cm^{3}= \frac{10450}{100}\, litres$
Vol of milk $= 10\cdot 4\, litres$
Cost of 1 litre milk = 50 Rs
Cost of 10.4 litre milk = $50\times 10\cdot 4$ Rs
                                   $\Rightarrow 520$ Rupees
Surface are of frustum $= \pi \left ( R+r \right )l+\pi r^{2}$
                                     $= \pi \left [ \left ( 20+8 \right )\times 20+8^{2} \right ]$
                                     $= 3\cdot 14\times \left ( 650+64 \right )$
                                     $= 1959 \, cm^{2}$
Cost of 100 cm² sheet = 10 Rs
Cost of 1 cm² sheet $= \frac{10}{100}\, Rs$
Cost of 1959 cm ² sheet = $1959\times \frac{10}{100}$
                                      = $195\cdot 9\, Rs$
                                       $\simeq \, 196$ Rupees