Question

# A hemispherical bowl made of brass has inner diameter $10.5cm$. Find the cost of tin-plating it on the inside at the rate of $Rs16per100c{m}^{2}$. (Assume $\mathrm{\pi }=\frac{22}{7}$)

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Solution

## Step 1: Find the inner curved surface areaThe inner radius of the hemispherical bowl$=\frac{10.5}{2}=5.25cm$(Radius is half of the diameter)To find the cost of tin-plating inside the bowl we have to find the inner curved surface area of the hemispherical bowl,The inner curved surface area of the hemispherical bowl$=2{\mathrm{\pi r}}^{2}$$2{\mathrm{\pi r}}^{2}=2×\frac{22}{7}×5.25×5.25=173.25{\mathrm{cm}}^{2}$Step 2: Find the Cost of tin-platingCost of tin-plating $100c{m}^{2}$ area = $Rs16$So, the Cost of tin-plating $1c{m}^{2}$ area = $Rs\frac{16}{100}$Cost of tin-plating $173.25c{m}^{2}$ area $=Rs\frac{16}{100}×173.25=Rs27.72$Hence, the cost of tin-plating the inner side of the hemispherical bowl at the rate of $Rs16per100c{m}^{2}$ is $Rs27.72$.

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