It costs Rs 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep.

Solution:

(i) Rs 20 is the cost of painting $1 \mathrm{~m}^{2}$ area.

Rs 2200 is the cost of painting $=\left(\frac{1}{20} \times 2200\right) \mathrm{m}^{2}$ area

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

Therefore, the inner surface area of the vessel is $110 \mathrm{~m}^{2}$.

(ii) Let the radius of the base of the vessel be $r$.

Height (h) of vessel = 10 m

Surface area $=2 \pi r h=110 \mathrm{~m}^{2}$

$\Rightarrow\left(2 \times \frac{22}{7} \times r \times 10\right) \mathrm{m}=110 \mathrm{~m}^{2}$

$\Rightarrow r=\left(\frac{7}{4}\right) \mathrm{m}=1.75 \mathrm{~m}$

(iii) Volume of vessel $=\pi r^{2} h$

$=\left[\frac{22}{7} \times(1.75)^{2} \times 10\right] \mathrm{m}^{3}$

$=96.25 \mathrm{~m}^{3}$

Therefore, the capacity of the vessel is $96.25 \mathrm{~m}^{3}$ or 96250 litres.