The inner diameter of a circular well is 3.5 m.
Question. The inner diameter of a circular well is $3.5 \mathrm{~m}$. It is $10 \mathrm{~m}$ deep. Find

(i) Its inner curved surface area,

(ii) The cost of plastering this curved surface at the rate of Rs 40 per $\mathrm{m}^{2} .\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$


Solution:

Inner radius $(r)$ of circular well $=\left(\frac{3.5}{2}\right) \mathrm{m}=1.75 \mathrm{~m}$

Depth $(h)$ of circular well $=10 \mathrm{~m}$

Inner curved surface area $=2 \pi r h$

$=\left(2 \times \frac{22}{7} \times 1.75 \times 10\right) \mathrm{m}^{2}$

$=(44 \times 0.25 \times 10) \mathrm{m}^{2}$

$=110 \mathrm{~m}^{2}$

Therefore, the inner curved surface area of the circular well is $110 \mathrm{~m}^{2}$.

Cost of plastering $1 \mathrm{~m}^{2}$ area $=$ Rs 40

Cost of plastering $110 \mathrm{~m}^{2}$ area $=$ Rs $(110 \times 40)$

$=\operatorname{Rs} 4400$

Therefore, the cost of plastering the CSA of this well is Rs 4400 .
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