The inner diameter of a circular well is $ 3.5 \mathrm{~m} $. It is $ 10 \mathrm{~m} $ deep. Find(i) its inner curved surface area,(ii) the cost of plastering this curved surface at the rate of Rs. $ 40 \mathrm{per} \mathrm{m}^{2} $.

Given:

The inner diameter of a circular well is $3.5\ m$. It is $10\ m$ deep.

To do:

We have to find:

(i) Its inner curved surface area.

(ii) the cost of plastering this curved surface at the rate of $Rs.\ 40$ per $m^2$.

Solution:

The inner diameter of the well $= 3.5\ m$

This implies,

Radius $(r)=\frac{3.5}{2}$

$=1.75 \mathrm{~m}$

Depth of the well $(h)=10 \mathrm{~m}$

Therefore,

The inner curved surface area of the well $=2 \pi r h$

$=2 \times \frac{22}{7} \times 1.75 \times 10$

$=440 \times 0.25$

$=110 \mathrm{~m}^{2}$

Rate of plastering the curved surface $=Rs.\ 40$ per $\mathrm{m}^{2}$

Therefore,

Total cost of plastering $=Rs.\ 40 \times 110$

$= Rs.\ 4400$ .

Related Articles The inner diameter of a circular well is $3.5\ m$. It is $10\ m$ deep. Find the cost of plastering this curved surface at the rate of $Rs.\ 40$ per $m^2$.
The inner diameter of a circular well is $3.5\ m$. It is $10\ m$ deep. Find inner curved surface area.
It costs Rs. 2200 to paint the inner curved surface of a cylindrical vessel \( 10 \mathrm{~m} \) decp. If the cost of painting is at the rate of \( Rs. 20 \) per \( \mathrm{m}^{2} \), find(i) inner curved surface area of the vessel,(ii) radius of the base,(iii) capacity of the vessel.
A cylindrical pillar is \( 50 \mathrm{~cm} \) in diameter and \( 3.5 \mathrm{~m} \) in height. Find the cost of painting the curved surface of the pillar at the rate of RS.\( 12.50 \) per \( \mathrm{m}^{2} \).
How many cubic metres of earth must be dug out to sink a well of \( 22.5 \mathrm{~m} \) deep and diameter \( 7 \mathrm{~m} \)? Also, find the cost of plastering the inner curved surface at Rs. 3 per square metre.
The slant height and base diameter of a conical tomb are \( 25 \mathrm{~m} \) and \( 14 \mathrm{~m} \) respectively. Find the cost of white-washing its curved surface at the rate of Rs. 210 per \( 100 \mathrm{~m}^{2} \).
A metal pipe is \( 77 \mathrm{~cm} \) long. The inner diameter of a cross section is \( 4 \mathrm{~cm} \), the outer diameter being \( 4.4 \mathrm{~cm} \) (see below figure). Find its(i) inner curved surface area,(ii) outer curved surface area,(iii) total surface area."\n
Find the surface area of a sphere of diameter(i) \( 14 \mathrm{~cm} \)(ii) \( 21 \mathrm{~cm} \)(iii) \( 3.5 \mathrm{~m} \).
Curved surface area of a right circular cylinder is \( 4.4 \mathrm{~m}^{2} \). If the radius of the base of the cylinder is \( 0.7 \mathrm{~m} \), find its height.
Find the cost of digging a cuboidal pit \( 8 \mathrm{~m} \) long, \( 6 \mathrm{~m} \) broad and \( 3 \mathrm{~m} \) deep at the rate of Rs. 30 per \( \mathrm{m}^{3} \).
Find the cost of sinking a tubewell $280\ m$ deep, having diameter $3\ m$ at the rate of $Rs.\ 3.60$ per cubic metre. Find also the cost of cementing its inner curved surface at $Rs.\ 2.50$ per square metre.
How many cubic metres of earth must be dugout to sink a well $21\ m$ deep and $6\ m$ diameter? Find the cost of plastering the inner surface of the well at Rs. $9.50$ per $m^2$.
The floor of a rectangular hall has a perimeter \( 250 \mathrm{~m} \). If the cost of painting the four walls at the rate of \( Rs. 10 \mathrm{per} \mathrm{m}^{2} \) is \( Rs. 15000 \), find the height of the hall.[Hint: Area of the four walls = Lateral surface area.]
Diameter of the base of a cone is \( 10.5 \mathrm{~cm} \) and its slant height is \( 10 \mathrm{~cm} \). Find its curved surface area.
Find the total surface area of a cone, if its slant height is \( 21 \mathrm{~m} \) and diameter of its base is \( 24 \mathrm{~m} \).
Kickstart Your Career
Get certified by completing the course

Get Started