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Answer :
Inner Radius (r) of sphere = \(\frac{Diameter}{2} = \frac{10.5}{2} = 5.25 cm\)
Surface area of sphere
= \(4 {\pi} {r}^2\)
= [\(4 × \frac{22}{7} × (5.25)^2\) \({cm}^2\)]
= [\(4 × \frac{22}{7} × 5.25 × 5.25\) \({cm}^2\)]
= ((88 × 5.25) \({cm}^2\)
= 173.25 \({cm}^2\)
Therefore, the surface area of a sphere having radius 5.25 cm is 173.25 \({cm}^2\).
Cost of tin-plating 100 \({cm}^{2}\) area = Rs. 16
Cost of tin-plating 173.25 \({cm}^{2}\) area = Rs.\(\frac{16 × 173.25}{100}\) = Rs. 27.72
Therefore, the cost of tin-plating the inner side of the hemispherical bowl is Rs. 27.72.