Question

It costs Rs. 2200 to paint the inner curved of a cylindrical vessel 10m deep. If the cost of painting is at the rate 20 per m2.. Find:
Inner curved surface area of the vessel
Radius of the base
Capacity of the vessel.

Answer 1

Hint: The cost of painting the curved surface (CSA) area is given, using this we can find the CSA which will lead us for finding the radius (by applying formula). Capacity of the vessel can be calculated by substituting the value of radius in the formula.
Use the following formulas to calculate area and capacity of the cylinder:
Curved Surface area (CSA) of Cylinder of cylinder = 2πrh
Capacity = volume of cylinder = πr2h

Complete step-by-step answer:

Given: Height (h) = 10 m
Cost to paint inner curved surface area = Rs.2200
Cost to paint per m2 = Rs. 20
Let CSA = x
We know
Cost of painting (inner CSA) = Inner Curved surface area X Cost of painting per m2
Substituting the given values, we get:$$
$$$2200 = x \times 20$
$x = 2200/20$
∴ $x = 110$
Therefore, the inner Curved surface area of the vessel is 110 m²
Let the radius of the cylinder be r.
Height= h= 10 m (given)
We have,
Curved Surface area= 110 (calculated)
$2\pi rh = 110$ [as CSA of cylinder is 2πrh]
Substituting the values, we get:
$2 \times 22/7 \times r \times 10 = 110$ [π=22/7]
Calculating the values of r:
$r = (110 \times 7)/(2 \times 22 \times 10)$
$r = 7/4$
$r = 1.75$
Therefore, the radius of the vbase of the vessel is 1.75 m.

We have,
Height = h= 10 m (given)
Radius =r =1.75 m or 7/4 m (calculated)
Now,
Capacity= Volume of vessel = $\pi {r^2}h$ [as volume of cylinder =$\pi {r^2}h$]
=$22/7 \times 7/4 \times 7/4 \times 10$
$ = 1540/16$
$ = 96.25$m³
(r is taken as 7/4 instead of 1.75 for easier calculation)
Therefore the capacity of the vessel is 96.25 m³.

Note: Always remember to complete the answer with respective units at the end (like m2 for area) and use the given values, appropriate formulas to find the unknown value.
Curved surface area (CSA) can also be called as lateral surface area (LSA) and the circular bases of the cylinder are always parallel and congruent.