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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. \( 40 \mathrm{per} \mathrm{m}^{2} \).
Given:
The inner diameter of a circular well is $3.5\ m$. It is $10\ m$ deep.
To do:
We have to find:
(i) Its inner curved surface area.
(ii) the cost of plastering this curved surface at the rate of $Rs.\ 40$ per $m^2$.
Solution:
The inner diameter of the well $= 3.5\ m$
This implies,
Radius $(r)=\frac{3.5}{2}$
$=1.75 \mathrm{~m}$
Depth of the well $(h)=10 \mathrm{~m}$
Therefore,
The inner curved surface area of the well $=2 \pi r h$
$=2 \times \frac{22}{7} \times 1.75 \times 10$
$=440 \times 0.25$
$=110 \mathrm{~m}^{2}$
Rate of plastering the curved surface $=Rs.\ 40$ per $\mathrm{m}^{2}$
Therefore,
Total cost of plastering $=Rs.\ 40 \times 110$
$= Rs.\ 4400$ .