Question

The inner diameter of a circular well is 2 m
and its depth is 10.5 cm. Find
(i) the inner curved surface area of the well
(ii) the cost of plastering this curved surface
area at the rate of 2310 per m.​

Answer 1

Answer:

Given:

⇒(i)r= radius=23.5=1.75 , h= depth of the well=10m 

⇒Curved surface area 

=2πrh

=(2×722×1.75×10)m2=110m2

⇒(ii) Cost of plastering = Rs 40 per m2

The cost of plastering the curved surface = Rs (110×40)= Rs 4400.

Answer 2

(i). The inner curved surface area of the well is 0.66 m².

(ii). The cost of plastering the curved surface area of the well at the rate of Rs. 2310 per m² is Rs. 1524.60.

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Let's understand a few concepts:

To solve the given problem we will use the following formula:

$\bigstar\: \boxed{\bold{Curved \:Surface \:Area \:of\:Cylinder = 2\pi rh}}\:\bigstar$

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Let's solve the given problem:

The inner diameter of the circular well (d) = 2 m

∴ The inner radius of the circular well (r) = $\frac{d}{2} = \frac{2}{2} = 1 \:m$

The depth of the circular well (h) = 10.5 cm = $\frac{10.5}{100}\:m = 0.105\:m$

(i). Finding the inner C.S.A. of the well:

Since the opening of the well is circular in shape and has a depth of 10.5 cm so we can say that the well is cylindrical in shape therefore to find its inner CSA we will use the above-mentioned formula.

∴ The inner curved surface area of the circular well is,

= $2\pi rh$

= $2\times \frac{22}{7} \times 1 \times 0.105$

= $2\times 22\times 1 \times 0.015$

= $\boxed{\bold{0.66\:m^2}}$

(ii). Finding the cost of plastering the CSA of the well:

If the cost of plastering 1 m² of curved surface area of the well is Rs. 2310

Then,

The cost of plastering the 0.66 m² curved surface area of the well will be,

=  $Rs. \:2310\:m^2 \times 0.66\:m^2$

= $\boxed{\bold{Rs. \:1524.60}}$

Thus,

(i). The inner curved surface area of the well is 0.66 m².

(ii). The cost of plastering the curved surface area of the well at the rate of Rs. 2310 per m² is Rs. 1524.60.

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