Solve this

Inner radius of the bowl, $r=\frac{10.5}{2}=5.25 \mathrm{~cm}$

$\therefore$ Inner curved surface area of the bowl $=2 \pi r^{2}=2 \times \frac{22}{7} \times(5.25)^{2}=173.25 \mathrm{~cm}^{2}$

Rate of tin-plating $=₹ 32$ per $100 \mathrm{~cm}^{2}$

∴ Cost of tin-plating the bowl on the inside

= Inner curved surface area of the bowl × Rate of tin-plating

$=173.25 \times \frac{32}{100}$

$=₹ 55.44$

Thus, the cost of tin-plating the bowl on the inside is ₹ $55.44$.