Answer:
Given, the radius of hemisphere is 3.5 cm and let the height of the cone be h cm. Now, Volume of wood \[=166\frac{5}{6}c{{m}^{3}}\] \[\frac{2}{3}\pi {{r}^{3}}+\frac{1}{3}\pi {{r}^{2}}h=166\frac{5}{6}\] \[\frac{2}{3}\times \frac{22}{7}\times 3.5\times 3.5\times 3.5+\frac{1}{3}\times \frac{22}{7}\times 3.5\times 3.5\times h\] \[=\frac{1001}{6}\] \[\frac{22}{7}\times 3.5\times 3.5\left( \frac{2}{3}\times \frac{7}{2}+\frac{1}{3}\times h \right)=\frac{1001}{6}\] \[\frac{22}{7}\times \frac{7}{2}\times \frac{7}{2}\left( \frac{7+h}{3} \right)=\frac{1001}{6}\] \[7+h=\frac{1001\times 7\times 2\times 2\times 3}{6\times 7\times 7\times 22}\] \[h=\frac{1001}{77}-7\] \[h=6\,cm\]. Area of hemispherical part of the toy \[=2\pi {{r}^{2}}\] \[=2\times \frac{22}{7}\times \frac{7}{2}\times \frac{7}{2}\] \[=77\,\,c{{m}^{2}}\] \[\therefore \] The cost of painting the hemispherical part of the toy \[=Rs.\,(77\times 10)\] \[=Rs.\,770\]
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