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The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find
i) its inner curved surface area
ii) the cost of plastering this curved surface at the rate of Rs. 40 per \({m}^2\).


Answer :

Given :
Inner radius (r) of circular well = \(\frac{3.5}{2}\) m = 1.75 m
Depth (h) of circular well = 10 m

i) We know that,

Inner curved surface area
= \(2{\pi}rh\)
= \(2 × \frac{22}{7} × 1.75 × 10\) \({m}^2\)
= (44 × 0.25 × 10) \({m}^2\)
= 110 \({m}^2\)

Therefore, the inner curved surface area of the circular well is 110 \({m}^2\)

ii) We have, Cost of plastering 1 \({m}^2\) area = Rs. 40
Cost of plastering 100 \({m}^2\) area = Rs. (110 x 40) = Rs. 4400

Therefore, the cost of plastering the CSA of this well is Rs. 4400.

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