Question

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 ₹40 m per m²​

Answer 1

SOLUTION:-

Given:

•The inner diameter of a circular well is 3.5m.

•It is 10m deep.

To find:

•It's inner curved surface area.

•The cost of plastering this curved surface at rate of Rs.40/m².

Explanation:

We know that, diameter of circle: 2r

$2 \times Radius = 3.5 \\ \\ Radius = \frac{3.5}{2} \\ \\ Radius = 1.75m$

We have, height= 10m

Formula of the curved surface area of the circular well: 2πrh

$Curved \: surface \: area = 2 \times \frac{22}{7} \times 1.75 \times 10 \\ \\ Curved \: surface \: area = \frac{44}{7} \times 17.5 \\ \\ Curved \: surface \: area = \frac{770}{7} {m}^{2} \\ \\ Curved \: surface \: area = 110 {m}^{2}$

&

Cost of plastering the curved surface at rate of Rs.40/m².

Cost of plastering the curved surface area Rs.[110 × 40]

Cost of plastering= Rs.4400.

Answer 2

Appropriate Question :-

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• $\textsf{The Inner Radius of a Circular Well is 3.5 m. It is 10 m Deep. Find The Cost of Plastering the Inner Curved Surface at the Rate of Rs. 20/ m²}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Required Solution :-

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• $\textsf{Inner Radius (r) = 3.5 m and Height/ Depth (h) = 10 m}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀_______________________________________________

⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\begin {aligned} \red {\bold {☯ \;Inner\: Curved\: Surface\: Area\: of\: the\: Well}} & = \blue {\bold {(2 \pi rh)\: sq.\: units}} \\\\& \Rightarrow 2 \times \frac{22}{7} \times 3.5 \times 10\\\\& \Rightarrow \frac{44}{7} \times 35 = \red {\bold {220\: m^2\;☯}} \end {aligned}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\textsf{The Inner Curved Surface Area of the Well = 220 m²}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\begin {aligned} \green{\bold {⚛ \;Cost\: of\: Painting\: the\: Inner\: Curved\: Surface\: Area\: of\: the\: Well}} & = Rs. (20 \times 220)\\\\& = \red{\bold {Rs.\: 4,400\; ⚛}}\end {aligned}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Hence,

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• $\textsf{The Cost of Painting the Inner Curved Surface Area of the Well is Rs. 4,400 }$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀___________________________________________________ ⠀⠀⠀⠀⠀⠀⠀⠀⠀