Question

A bucket is in the form of a frustum of a cone of height 30 cm with radii of its lower and upper ends as 10 cm and 20 cm respectively. Find the capacity and surface area of the bucket. Also, find the cost of milk which can completely fill the cointainer, at the rate of Rs.25 per litre. (Use $\pi$ = 3.14)

Answer 1

Solution:-

Radius of the upper end of the frustum of cone = R = 20 cm
radius of the lower end of the frustum of cone = r = 10 cm
H = 30 cm
Volume

$=\frac{1}{3}\pi h(R^2+r^2+Rr)$

$=\frac{1}{3}\times \frac{22}{7}\times 30({20}^2+{10}^2+20\times 10)=\frac{660}{21}(400+100+200)=22000c{m}^3=22litres$

Now, Slant height

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

$l=\sqrt{(20-10{)}^2+{30}^2}=\sqrt{1000}=31.62cm$

Slant height is 31.62 cm

Surface area =

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

$=\frac{22}{7}({20}^2+{10}^2+(20+10)\times 30)=\frac{22}{7}(400+100+30\times 30)=\frac{22\times 1400}{7}=4400c{m}^2$

Cost of 1 liter milk = Rs. 25
Total cost of 22 liter milk = 22$\times$25
= Rs. 550