Question.
It costs Rs 2200 to paint the inner curved surface of a cylindrical vessel $10 \mathrm{~m}$ deep. If the cost of painting is at the rate of Rs 20 per $\mathrm{m}^{2}$, find
(i) Inner curved surface area of the vessel
(ii) Radius of the base
(iii) Capacity of the vessel
Assume $\left.\pi=\frac{22}{7}\right]$
(i) Inner curved surface area of the vessel
(ii) Radius of the base
(iii) Capacity of the vessel
Assume $\left.\pi=\frac{22}{7}\right]$
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.
(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.