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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 \({m}^2\) = \(2 × \frac{22}{7} × r × 10 m\)
\(\Rightarrow \) 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.