Question

A container 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. 15 per litre and the cost of the metal sheet used, if it costs Rs. 5 per 100 $c{m}^2$. (Take $\pi$= 3·14)

• A. Rs. 156.75 , Rs. 92.45
• B. Rs. 155.75, Rs. 95.63
• C. Rs. 156.75 , Rs 97.97
• D. Rs. 150, Rs. 100

Answer 1

The correct option is C

Rs. 156.75 , Rs 97.97

R = 20 cm, r = 8cm, h = 16 cm

l = $\sqrt{h^2+(R-r{)}^2}=\sqrt{256+144}cm=20cm$

Volume of container = $\frac{1}{3}\pi h\left(R^2+r^2+Rr\right)$

$=\frac{1}{3}\times \left(3.14\right)\times 16(400+64+160)c{m}^3$

$=\frac{1}{3}\times 3.14\times 16\times 624c{m}^3$

$=3.14\times 16\times 208c{m}^3$

$=10449.93c{m}^3$

Therefore, the quantity of milk in the container = $\frac{10449.92}{1000}$ = 10.45 liters

Cost of milk at the rate of Rs.15 per liters = Rs.{10.45 × 15 } = Rs. 156.75

Surface area of the metal sheet used to make the container

$=\pi l(R+r)+\pi r^2=\pi \left[l(R+r)+r^2\right]$

$=3.14\times \left[20\times (20+8)+8^2\right]c{m}^2$

$=3.14\times \left[20\times 28+64\right]c{m}^2$

$=3.14\times 624c{m}^2=1959.36c{m}^2$


Therefore, the cost of the metal sheet at rate of Rs.5 per 100 $c{m}^2$

= Rs. 1959.36 $\times$ $\frac{5}{100}$ = Rs.97.97 approx.