Question

The inner diameter of a circular well is 3.5 m and it is 10 m deep. Find the
inner curved surface area of the well. Find the cost of plastering this curved surface at the rate of 10 per m². (Take π=22/7)​

Answer 1

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Answer 2

$\bf \: \underline{Given} :$ ↴

$\sf \: The \: inner \: diameter \: of \: a \: circular \: well \: is \: 3.5 \: meter.$

$\sf \: It \: is \: 10 \: meter \: deep.$

$\bf \: \underline{To \: find }:$ ↴

$\sf \: The \: inner \: curved \: surface \: area.$

$\sf \: The \: cost \: of \: plastering \: this \: curved \: surface \: at \: the \: rate \: of \: Rs. \: 10/m^{2}$

$\sf \: \underline{\underline{Solution }}$

$\sf \: As \: we \: know \: that,$

$\boxed{ \pink{\sf \: Diameter \: of \: circle \: = 2 \times Radius}}$

$\sf \: \implies \: 3.5 \: = 2 \times Radius$

$\sf \: \implies \: \dfrac{3.5}{2} = Radius$

$\sf \: \implies \: 1.75 \: meter = Radius$

$\bf \: Now,$

$\sf \longrightarrow \: Height = 10 \: meter$

$\sf \longrightarrow \: Radius = 1.75 \: meter$

$\bf \: So,$

$\boxed{ \pink{\sf \: Curved \: surface \: area \: of \: circular \: well = 2\pi \times r \times h}}$

$\sf \implies \: 2 \times \dfrac{22}{7} \times 1.75 \times 10$

$\sf \implies \: \dfrac{44}{17} \times 17.5$

$\sf \implies \: \dfrac{770}{7} \: m ^{2}$

$\sf \implies \: 110 \: m ^{2}$

$\sf \: \underline{ Curved \: surface \: area \: of \: the \: circular \: well = 110 \: m ^{2}}$

$\bf \: Hence,$

$\sf \: The \: cost \: of \: plastering \: this \: curved \: surface \: at \: the \: rate \: of \: Rs. \: 10/m^{2} =$

$\sf \: \underline{ 110 \times 10 = Rs. \: 1100}$