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i) inner curved surface area of the vessel,

ii) radius of the base,

iii) capacity of the vessel.

Answer :

i)Given :

Rs. 20 is the cost Of painting 1 \({m}^2\) area.

\(\therefore\) when Rs. 2200 is the cost of painting, area is

= \(\frac{1}{20} × 2200\) \({m}^2\)

= 110 \({m}^2\) area

Therefore, the inner curved surface area of the vessel is 110 \({m}^2\).

ii)Height (h) of vessel = 10 m

Surface area = 110 \({m}^2\)

Let the radius of the base of the vessel be r.

We know that,

Surface area = \(2 {\pi} r h\)

But, CSA = 110 \({m}^2\) ...(from i))

\(\Rightarrow \) 110 \({m}^2\) = \(2 x \frac{22}{7} x r x 10 m\)

\(\Rightarrow\) r = \(\frac{7}{4} m = 1.75 m\)

iii) Now, volume of vessel

= \({\pi} {r}^2 h\)

= \(22 × (1.75)^2 × 10 {m}^3\)

= 96.25 \({m}^3\)

Therefore, the capacity of the vessel is 96.25 \({m}^3\) or 96250 litres.

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