Surface Areas and Volumes — Question 9

Given,

Inner diameter of hemispherical bowl (d) = 10.5 cm

Inner radius of hemispherical bowl (r) = $\dfrac{\text{Diameter}}{2} = \dfrac{10.5}{2}$ = 5.25 cm

Curved surface area of hemispherical bowl = 2πr²

$= 2 \times \dfrac{22}{7} \times (5.25)^2 \\[1em] = 2 \times \dfrac{22}{7} \times 27.5625 \\[1em] = 2 \times 22 \times 3.9375 \\[1em] = 173.25 \text{ cm}^2.$

Given,

The cost of tin-plating 100 cm² of the bowl = ₹ 16

∴ The cost of tin-plating 1 cm² of the bowl = ₹ $\dfrac{16}{100}$

∴ The cost of tin-plating 173.25 cm² area of the bowl = ₹ $\dfrac{16}{100} \times 173.25$ = ₹ 27.72

Hence, the cost of tin-plating the hemispherical bowl is ₹ 27.72.