A hemispherical bowl made of brass has inner diameter 10.5 cm.
Question. A hemispherical bowl made of brass has inner diameter $10.5 \mathrm{~cm}$. Find the cost of tin-plating it on the inside at the rate of Rs 16 per $100 \mathrm{~cm}^{2} .\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$

Solution:

Inner radius $(r)$ of hemispherical bowl $=\left(\frac{10.5}{2}\right) \mathrm{cm}=5.25 \mathrm{~cm}$

Surface area of hemispherical bowl $=2 \pi r^{2}$

$=\left[2 \times \frac{22}{7} \times(5.25)^{2}\right] \mathrm{cm}^{2}$

$=173.25 \mathrm{~cm}^{2}$

Cost of tin-plating $100 \mathrm{~cm}^{2}$ area $=$ Rs 16

Cost of tin-plating $173.25 \mathrm{~cm}^{2}$ area $=\operatorname{Rs}\left(\frac{16 \times 173.25}{100}\right)=\operatorname{Rs} 27.72$

Therefore, the cost of tin-plating the inner side of the hemispherical bowl is Rs $27.72$.
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