Twenty one cylindrical pillars of the Parliament House are to be cleaned.

Given:

Diameter of the pillars $=0.5 \mathrm{~m}$

Radius of the pillars, $r=0.25 \mathrm{~m}$

Height of the pillars, $h=4 \mathrm{~m}$

Number of pillars $=21$

Rate of cleaning $=$ Rs $2.50$ per square metre

Curved surface area of one pillar $=2 \pi r h$

$=2 \times \frac{22}{7} \times 0.25 \times 4$

$=2 \times \frac{22}{7}$

$=\frac{44}{7} \mathrm{~m}^{2}$

$\therefore$ Curved surface area of one pillar $=\frac{44}{7} \mathrm{~m}^{2}$

Cost of cleaning 21 pillars at the rate of Rs $2.50$ per $\mathrm{m}^{2}=$ Rs $2.5 \times 21 \times \frac{44}{7}$

$=7.5 \times 44$

$\therefore$ Cost of cleaning 21 pillars at the rate of Rs $2.50$ per $\mathrm{m}^{2}=$ Rs 330