A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is Rs. 12 per\( m^2 \) , what will be the cost of painting all these cones? (Use \({\pi}\) = 3.14 and take 1.04 = 1.02)

Given :
Radius (r) of conical cap = \(\frac{40}{2} cm = 20 cm = 0.2 m\)
Height (h) of conical cap = 1 m

Slant height (l) of the conical cap
\(\Rightarrow \) \({l}^2 = {h}^2 + {r}^2\)
\(\Rightarrow \) \(l^2= {1}^2 + {0.2}^2 {m}^2\)
\(\Rightarrow \) \(l^2= 1.04 {m}^2\)
\(\Rightarrow \) \({l}^2 = {1.02}^2 m\)
\(\Rightarrow \) l = 1.02 m

\(\therefore \) the slant height Of the tent is 1.02 m.

We know, CSA of tent
= \({\pi}rl\)
= [\(3.14 × 0.2 × 1.02\) \({m}^2\)]
= \(0.64056\) \({m}^2\)

So, CSA of 50 such cones
= \(50 × 0.64056\) \({m}^2\)
= 32.028 \({m}^2\).

Now, we have,

Cost of painting 1 \({m}^2\) area = Rs. 12
Cost of painting 32.02 \({m}^2\) area = Rs. (32.028 x 12)
= Rs. 384.336
= Rs 384.34 (approximately)

Therefore, it Will cost Rs. 384.34 in painting 50 such hollow cones.