Question

It costs ₹ 3300 to paint the inner curved surface of a cylindrical vessel 10 m deep at the rate of Rs 30 per m². Find the
(i) inner curved surface area of the vessel,
(ii) inner radius of the base, and
(iii) capacity of the vessel.

Answer 1

Total cost of paining the inner curved surface of the cylinderical vassel = ₹ 3,300

Rate of painting = ₹ 30 per m²

(i) Inner curved surface area of the vassel

$$\begin{gathered} =\frac{\mathrm{Total}\mathrm{cost}\mathrm{of}\mathrm{painting}\mathrm{the}\mathrm{inner}\mathrm{curved}\mathrm{surface}\mathrm{of}\mathrm{the}\mathrm{cylindrical}\mathrm{vassel}}{\mathrm{Rate}\mathrm{of}\mathrm{painting}} \\ =\frac{3300}{30} \\ =110{\mathrm{m}}^2 \end{gathered}$$
Thus, the inner curved surface area of the vassel is 110 m².

(ii) Depth of the vassel, h = 10 m

Let the inner radius of the base be r m.

∴ Inner curved surface area of the vasssel = $2\mathrm{\pi }rh=2\times \frac{22}{7}\times r\times 10$

$$\begin{gathered} \Rightarrow 2\times \frac{22}{7}\times r\times 10=110 \\ \Rightarrow r=\frac{110\times 7}{2\times 22\times 10} \\ \Rightarrow r=1.75\mathrm{m} \end{gathered}$$
Thus, the inner radius of the base is 1.75 m.

(iii) Capacity of the vassel = $\mathrm{\pi }{r}^{2}h=\frac{22}{7}\times {\left(1.75\right)}^2\times 10=$96.25 m³

Thus, the capacity of the vassel is 96.25 m³.