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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 $=Rs.\ 40 \times 110$

$= Rs.\ 4400$ .

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