$A$ 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 container,at the rate of $Rs. \, 25$ per $litre$ (use $\pi = 3.14$).

(N/A) The capacity (volume) of the bucket is given by the formula $V = \frac{\pi h}{3} [r_1^2 + r_2^2 + r_1 r_2]$.
Given $h = 30 \, cm$,$r_1 = 20 \, cm$,and $r_2 = 10 \, cm$.
Capacity $= \frac{3.14 \times 30}{3} [20^2 + 10^2 + 20 \times 10] = 31.4 [400 + 100 + 200] = 31.4 \times 700 = 21980 \, cm^3$.
Since $1000 \, cm^3 = 1 \, litre$,the capacity is $21.98 \, litres$.
The cost of milk at $Rs. \, 25$ per $litre$ is $21.98 \times 25 = Rs. \, 549.50$.
The slant height $l = \sqrt{h^2 + (r_1 - r_2)^2} = \sqrt{30^2 + (20 - 10)^2} = \sqrt{900 + 100} = \sqrt{1000} \approx 31.62 \, cm$.
The surface area of the bucket (open at the top) is the curved surface area plus the area of the base: $A = \pi l (r_1 + r_2) + \pi r_2^2$.
$A = 3.14 \times 31.62 \times (20 + 10) + 3.14 \times 10^2 = 3.14 \times 31.62 \times 30 + 314 = 2978.56 + 314 = 3292.56 \, cm^2 \approx 3292.6 \, cm^2$.