Question

HARVEST RAIN WATER
Water is precious, each and every drop of water counts. Keeping this thought in mind Mr. X decided to harvest rain water in his field. For doing so he decided to dig a well of diameter 7m in his field. His field is 150m x 100 m and the well to be dug will be 28m deep.
a)How much water can be stored in the well?
b)If he want to plaster the wall of the well, how much area he need to plaster?
c)Find the cost of plastering at the rate of RS. 15 per square m

Answer 1

${\pmb{\underline{\sf{ Required \ Solution ... }}}}$

Mr. X wanted to harvest Water in his Field for the Futuristic Use and implementation and For this He want to built well in his Field to store water to level up & maintain water table.

As We know that he wanna built well in his Field of Measurement of 150m × 100m of lengths.

$\begin{gathered} \sf 150 \ m\: \: \: \: \: \: \: \: \: \: \: \\ \begin{gathered}\boxed{\begin{array}{}\bf { \red{}}\\{\qquad \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }{}\\ { \sf{ }}\\ { \sf{ }} \\ \\ { \sf{ }}\end{array}}\end{gathered} \sf 100 \: m \end{gathered}$

As We also know that Mr. X decided to dig a well of diameter 7m in his field and 28m deep.

We have to Find the volume of the Digged well that in his Field which is of the Cylindrical shape.

• Radius (r) = Diameter/2 = 7/2 m
• Height (h) = 28 m

$\circ \ {\underline{\boxed{\sf\gray{ Volume_{(Cylinder)} = πr^2h }}}} \\ \\ \colon\implies{\sf{ \dfrac{22}{7} \times \left( \dfrac{7}{2} \right)^2 \times 28 }} \\ \\ \colon\implies{\sf{ 11 \times 7 \times 14 }} \\ \\ \colon\implies{\sf{ 1078 \ m^3 }}$

$\\ ★ {\underline{\sf{Case \ 1^{st} ... }}}$

As We know that we have to find the amount of the water stored in that well.

» 1 m³ = 1000 litres

$\colon\implies{\sf{ 1078 \times 1000}} \\ \\ \colon\implies{\sf{ 1078000 \ Litres }}$

So, 1078000 Litres water can be stored in Digged well.

$\\ ★ {\underline{\sf{Case \ 2^{nd} ... }}}$

As We know that we have to find the Area of well's wall to be plaster.

We got that All the sides wall can be plastered except of Upper side which is open to draw water from the well.

» 5 sides of Well can be Plastered.

» Total Surface Area of the Well = (2πrh + 2πr²) — πr²

$\colon\implies{\sf{ 2πr(h + r) - πr^2 }} \\ \\ \colon\implies{\sf{ 2πr \left(28 + \dfrac{7}{2} \right) - \dfrac{22}{7} \times \left( \dfrac{7}{2} \right)^2 }} \\ \\ \colon\implies{\sf{ 2 \times \dfrac{22}{7} \times \dfrac{7}{2} \left( \dfrac{56+7}{2} \right) - \dfrac{77}{2} }} \\ \\ \colon\implies{\sf{ \cancel{22} \left( \dfrac{63}{ \cancel{2} } \right) - \dfrac{77}{2} }} \\ \\ \colon\implies{\sf{ 693 - \dfrac{77}{2} }} \\ \\ \colon\implies{\sf{ \dfrac{1386-77}{2} }} \\ \\ \colon\implies{\sf{ \cancel{ \dfrac{1309}{2} } }} \\ \\ \colon\implies{\sf{ 654.5 \ m^2 }}$

So, We got that 654.5 m² area of Well can be Plastered.

$\\ ★ {\underline{\sf{Case \ 3^{rd} ... }}}$

As We also know that we have to find the cost of Plastering the walls of the well at the rate of ₹15/m² .

$\colon\implies{\sf{ 654.5 \times 15 }} \\ \\ \colon\implies{\tt{ \₹ \ 9817.5 }}$

Hence,

»The Total cost of Plastering the wall will be ₹9817.5 .