Question

The inner diameter of a circular well is 0 .7 m and it is 27 m deep.
(a) Find the inner curved surface area of the well. ( Take π = 22/7. )
(b) Determine the cost of plastering the inner curved surface area of the well at the rate of Rs 80 per square metre..ye sang lavkar pls​

Answer 1

Answer:

a )The inner curved surface area of the circular well = 59.4 cm²

b) The cost of plastering at the rate of Rs.80 per square = Rs.4752

Step-by-step explanation:

Given:

The inner diameter of the circular well = 0.7 m

The height of the circular well = 27 m

To find out:

The curved surface area and the cost of plastering the inner curved surface area of the well.

Solution:

From the given, we can understand that the well is in the shape of a cylinder with a diameter of 0.7 m and a height of 27 m.

a) The curved surface area does not include the bottom and upper parts.

Hence the formula to find the curved surface area (A) of a cylinder is given as,

      A = 2πrh

Where 'r' is the radius and 'h' is the height of the cylinder.

Radius = Diameter ÷ 2

          = 0.7 ÷ 2 = 0.35 m

    $\pi = \dfrac{22}{7}$

By substituting the values we get,

⇒     A = (2 × $\frac{22}{7}$ × 0.35 × 27)

         = 59.4 cm²

b) The cost of plastering at the rate of Rs.80 per square = 59.4 × 80

                                                                           = Rs.4752

           

Answer 2

Step-by-step explanation:

a )The inner curved surface area of the circular well = 59.4 cm²

b) The cost of plastering at the rate of Rs.80 per square = Rs.4752

Step-by-step explanation:

Given:

The inner diameter of the circular well = 0.7 m

The height of the circular well = 27 m

To find out:

The curved surface area and the cost of plastering the inner curved surface area of the well.

Solution:

From the given, we can understand that the well is in the shape of a cylinder with a diameter of 0.7 m and a height of 27 m.

a) The curved surface area does not include the bottom and upper parts.

Hence the formula to find the curved surface area (A) of a cylinder is given as,

      A = 2πrh

Where 'r' is the radius and 'h' is the height of the cylinder.

Radius = Diameter ÷ 2

          = 0.7 ÷ 2 = 0.35 m

    \pi = \dfrac{22}{7}π=722

By substituting the values we get,

⇒     A = (2 × \frac{22}{7}722 × 0.35 × 27)

         = 59.4 cm²

b) The cost of plastering at the rate of Rs.80 per square = 59.4 × 80

                                                                           = Rs.4752