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- Surface Areas and Volumes - Exercise 13.1
- Surface Areas and Volumes - Exercise 13.2
- Surface Areas and Volumes - Exercise 13.3
- Surface Areas and Volumes - Exercise 13.4
- Surface Areas and Volumes - Exercise 13.5
- Surface Areas and Volumes - Exercise 13.6
- Surface Areas and Volumes - Exercise 13.7
- Surface Areas and Volumes - Exercise 13.8
- Surface Areas and Volumes - Exercise 13.9

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

(i) its inner curved surface area,

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

(Assume π=22/7)

Answer:

Inner radius of circular well, r = 3.5/2m = 1.75m

Depth of circular well, say h = 10m

(i)Inner curved surface area = 2πrh

= ( 2 x 22/7 x 1.75 x 10)

= 110

Therefore, the inner curved surface area of the circular well is 110 \mathrm{m}^{2}

(ii)Cost of plastering 1 m area = Rs.40

Cost of plastering 110 m area = Rs (110 x 40) = Rs.4400

Therefore, the cost of plastering the curved surface of the well is Rs. 4400.

"hello students welcome to lido q a video
session
i am your math tutor and question for
today is
the inner diameter of the circular well
is 3.5 meter
and it's the 10 meter deep find
first its inner cow surface area second
the cost of plastering this crowd
surface area
at the rate of rupees 40 per meter
square
assume pi is equal to 22 by 7 now here
to begin with let us see what is given
the inner radius of the circular wall is
given
in the form of diameter so here radius
can be found with the help of diameter
3.5 meter diameter by 2
will be radius so radius
is equal to
diameter upon 2 that is 3.5
upon 2 and it is 1.75
meter now depth of the circular well
say h the h will be
10 meter which is given so height
of well h
emitter now the first thing we need to
find is the inner curve surface area
formula to find the inner curve surface
area of the cylinder
is 2 pi rh val is cylindrical in shape
basically
so 2 pi r h
2 into 22 by 7 into
1.75 into 10
so curves inner curve surface area is
110
meter square therefore
this is the inner curve surface area
now the next thing we need to find is
the cost of plastering this curve
service area
at the rate of rupees 40 per meter
square
so here
rate of fast plastering
one meter square
is rupees 40 so how much will be the
rate of plastering
40 meter square
so for this we have to multiply one
meter cost
with the amount of area to be plastered
so 40 into 110
so rate will be getting
as 4400 rupees
that is our final answer so if you have
any query regarding this
you can post it in our comment section
and subscribe to lido for more such q a
thank you for watching
"

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