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 $\left.\pi=\frac{22}{7}\right]$
(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 .
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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