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i). its inner curved surface area

ii). the cost of plastering the curved surface at the rate of Rs 40 per m2.

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Hint- Here, we will be proceeding by using the formula for curved surface area of the cylinder.

Given, the inner diameter of a circular well $d = 3.5$ m

Depth or height of the circular well $h = 10$ m

Radius of the circular well $r = \dfrac{d}{2} = \dfrac{{3.5}}{2} = 1.75$ m

Since, a circular well will be cylindrical in shape.

As we know that the formula for the inner curved surface area of a cylinder with inner radius $r$ and height $h$ is given by ${\text{Inner curved surface area}} = 2\pi rh$

Using above formula, we get

i. ${\text{Inner curved surface area}} = 2\left( {\dfrac{{22}}{7}} \right)\left( {1.75} \right)\left( {10} \right) = 110$ m2.

ii. Also, given \[{\text{cost of plastering the curved surface}} = {\text{Rs 40 per }}{{\text{m}}^2}\]

\[ \Rightarrow {\text{Cost of plastering the curved surface}} = 40 \times \left( {{\text{Curved surface area}}} \right) = 40 \times 110 = {\text{Rs }}4400\]

Note- For this particular problem, in the cost of plastering the curved surface of the circular well we are considering the inner surface area of the circular well (inner curved surface of cylinder obtained in the first part of the problem) because no information is given for the outer curved surface area.

Given, the inner diameter of a circular well $d = 3.5$ m

Depth or height of the circular well $h = 10$ m

Radius of the circular well $r = \dfrac{d}{2} = \dfrac{{3.5}}{2} = 1.75$ m

Since, a circular well will be cylindrical in shape.

As we know that the formula for the inner curved surface area of a cylinder with inner radius $r$ and height $h$ is given by ${\text{Inner curved surface area}} = 2\pi rh$

Using above formula, we get

i. ${\text{Inner curved surface area}} = 2\left( {\dfrac{{22}}{7}} \right)\left( {1.75} \right)\left( {10} \right) = 110$ m2.

ii. Also, given \[{\text{cost of plastering the curved surface}} = {\text{Rs 40 per }}{{\text{m}}^2}\]

\[ \Rightarrow {\text{Cost of plastering the curved surface}} = 40 \times \left( {{\text{Curved surface area}}} \right) = 40 \times 110 = {\text{Rs }}4400\]

Note- For this particular problem, in the cost of plastering the curved surface of the circular well we are considering the inner surface area of the circular well (inner curved surface of cylinder obtained in the first part of the problem) because no information is given for the outer curved surface area.