Question

The inner diameter of a circular well is 3.5m. It is 10m deep. Find the cost of plastering this curved surface at the rate of Rs. 40 per ${m^2}$.

Answer 1

Hint: In the question we have given the diameter of a circular well and their depth is also given so we draw a diagram. And then we find the diagram of a cylinder, then we use the volume of the cylinder and find the required value.

Complete step-by-step answer:

According to the question the dimensions of the well given is ${\text{Diameter}} = 3.5m$ and ${\text{Height}} = 10m$
Now first we find the radius of the inner well, we get
Inner radius (r) of circular well=$\dfrac{{Diameter}}{2}$
 $ \Rightarrow \left( {\dfrac{{3.5}}{2}} \right)m$
$ \Rightarrow 1.75m$
As Depth (h) of circular well $ = 10cm$,Here we will apply formula of curved surface area as well is of cylindrical shaped
$ \Rightarrow Inner{\text{ }}curved{\text{ }}surface{\text{ }}area = 2\pi rh$
Now we put the values of r , h and $\pi = \dfrac{{22}}{7}$ in the formula we get,
\[ \Rightarrow \left( {2 \times \dfrac{{22}}{7} \times 1.75 \times 10} \right){m^2}\]
By solving the above equation we get
\[ \Rightarrow \left( {2 \times \dfrac{{22}}{7} \times \dfrac{{175}}{{100}} \times 10} \right){m^2}\]
We simplify the terms and we get
\[ \Rightarrow \left( {2 \times 22 \times \dfrac{{25}}{{10}}} \right){m^2}\]
Now we multiply the terms we get
\[ \Rightarrow Inner{\text{ }}curved{\text{ }}surface{\text{ }}area = 110{m^2}\]
As We have find area of inner surface, now we will find its cost of plastering the inner surface area
$\because $Cost of plastering $1{m^2}$ area $ = Rs.40$
So, Cost of plastering${\text{110}}{{\text{m}}^{\text{2}}}{\text{ area }}$, we multiply this area by $Rs.40$, we get
${\text{ = Rs}}\left( {{\text{110 \times 40}}} \right){\text{ = Rs}}{\text{.4400}}$

Therefore, the cost of plastering the Curved Surface Area of this well is Rs 4400.

Note: In these types of we have to take care of units of parameters. We should take the value of \[\pi = \dfrac{{22}}{7}\] for calculation. As we have to plaster the inner surface of the well so we consider it as the curved surface area of the well because it is cylindrical in shape. We should remember that radius is always half of the diameter. While solving we should take care of the dimension that diameter is given so we have to find radius.