Question

How many cubic metres of Earth must be dug out to form a well $28m$ deep and $2.8m$ in diameter? Also, find the cost of plastering its inner surface at $Rs\,4.5$ per square metre.

Answer 1

Hint:In the given question, we are provided with dimensions of the well that is to be dug. So, we have to find the volume of Earth to be dug out to form a well. We are also required to find the cost of plastering the inner surface of the wall. So, we must remember the formulae for curved surface area and volume of a cylinder to solve the problem. We will substitute the value of known quantities in the formula to get to the required result.

Complete step by step answer:
A well is cylindrical in shape. A cylinder is a three dimensional solid geometrical figure with straight parallel sides and a circular base. In the problem, we are given the depth of well as $28\,m$.
Diameter of well $ = 2.8\,m$
Now, we know that radius is half of the diameter of a shape.
So, we get the radius of well $ = \dfrac{{2.8}}{2}m = 1.4\,m$.
Now, we have to calculate the volume of the Earth dug out to form well. So, Volume of Earth dug out to form a well is equal to the volume of the well itself. We know that the volume of a cylinder is given by $\pi {r^2}h$, where r is the radius of the cylinder and h is the height or depth of a cylinder.

Hence, Volume of Earth dug out $ = \pi {\left( {1.4m} \right)^2}\left( {28m} \right)$
Substituting the value of $\pi $ as $\left( {\dfrac{{22}}{7}} \right)$, we get,
$\dfrac{{22}}{7} \times \left( {1.4m} \right) \times \left( {1.4m} \right) \times \left( {28m} \right)$
Cancelling the common factors in numerator and denominator, we get,
$\left( {22 \times 1.4 \times 1.4 \times 4} \right)\,{m^3}$
Carrying out the calculations, we get,
$\left( {88 \times 1.96} \right)\,{m^3}$
$\Rightarrow 172.48\,{m^3}$
So, the volume of earth dug out to form the well is $172.48\,{m^3}$. Now, we also have to find the cost of plastering the inner surface as well. So, we will have to find the inner curved surface area as well. We know that the curved surface area of a cylinder is $2\pi rh$, where r represents the radius and h represents the height or depth of the cylinder.

So, the inner surface of well $ = 2\pi \left( {1.4m} \right)\left( {28m} \right)$
Putting the value of $\pi $ in the expression, we get,
$2 \times \dfrac{{22}}{7} \times \left( {1.4m} \right)\left( {28m} \right)$
Cancelling the common factors in numerator and denominator, we get,
$2 \times 22 \times \left( {1.4} \right) \times \left( 4 \right){m^2}$
$\Rightarrow 246.4\,{m^2}$
So, the inner surface of the well is $246.4\,{m^2}$.
Now, we multiply the surface area by the rate of plastering to find the total cost.
So, the total cost of plastering inner surface of well $ = Rs\,246.4\, \times 4.5= Rs\,1108.8$

Hence, the cost of plastering the inner surface of a well is $Rs\,1108.8$.

Note:We must know the formulae for area, volume and perimeter of basic shapes like square, rectangle, parallelogram, circle, etc. One must have a strong grip over concepts of transposition in order to solve the equation formed in the problem. We should know to multiply the rate of plastering with the area to find the total cost. We also must take care of the calculations while doing such questions.