Question

The inner diameter of a circular well is \[3.5{\text{ }}m\] . It is \[10m\] deep. Find
I.Its inner curved surface area
II.The cost of plastering this curved surface at the rate of \[Rs{\text{ }}70\]

Answer 1

Hint: Here in this question inner diameter is given to be \[3.5{\text{ }}m\] and depth of well is \[10m\] we know the formula of curved surface area \[ = \;2\pi rh\] . Just putting a radius which is half of diameter and depth we will be able to find the inner curved surface area. For the 2nd part of the question we need to plaster the inner curved surface area. If we multiply rate with calculated curved surface area we will surely find the required solution .The outer diameter of the well is not our concern.

Complete step-by-step answer:
In this question the inner diameter of a circular well is \[3.5{\text{ }}m\] and well is \[10m\] deep. The cost of plastering this curved surface area is \[Rs{\text{ }}70\] .

Suppose r is the radius of the given well and h is the depth of the well
Step 1: Firstly, we are going to calculate the curved surface area of the given well. As we know the that the well in circular in shape so its easy for us to calculate it by using the formula \[2\pi rh\]….(1)
By putting the given value in the formula (1) we get \[2 \times \pi \times \dfrac{{3.5}}{2} \times 10\,\,{m^2}\]
\[ = \dfrac{{70}}{2}\pi \,\,{m^2}\]
$ = 110\,{m^{2\,\,\,\,\,\,\,\,\,}}\,\,\,\,\,\,\,\,\,\left[ {taken\,\,\pi = \dfrac{{22}}{7}} \right]$
Yes we have calculated the first part of the problem. Let's Move towards the next part of the question which is as simple as the first part .
Step 2: Here the rate of plastering the well is \[Rs{\text{ }}70\] for per square meter. The cost of plastering for plastering entire inside wall of well is nothing but
\[
   = 110 \times 70 \\
   = 7700\,\,Rs. \\
\]
Hence the inner curved surface area is\[110\pi \,\,{m^2}\]and cost of plastering is \[7700{\text{ }}Rs.\]

Note: In this type of problem student might think about outer surface area which we need not require here. Sometimes students mistakenly write $\pi rh$ instead of $2\pi rh$ which does not justvlead your solution being wrong but it reflects that you don’t understand the basic formula. Always convert all the dimensions in a single unit. It's good if you convert it in standard form like in m instead of cm.