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The largest possible hemisphere is drilled out from a wooden cubical block of side $21$ cm such that the base of the hemisphere is on one of the faces of the cube. Find :
(i) the volume of wood left in the block,
(ii) the total surface area of the remaining solid.
(i) the volume of wood left in the block,
(ii) the total surface area of the remaining solid.
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Diameter of hemisphere $=$ side of the cube $= 21$ cm
$\therefore$ radius of hemisphere $= \frac{21}{2}$ cm
(i) Volume of the wood left $=$ volume of cube $-$ volume of hemisphere
$= 21^3 - \frac{2}{3} \times \frac{22}{7} \times (\frac{21}{2})^3$
$= 6835.5$ cm$^3$
(ii) Total surface area of remaining solid $=$ TSA of cube $-$ base area of hemisphere $+$ CSA of hemisphere
$= 6 \times 21^2 - \frac{22}{7} \times (\frac{21}{2})^2 + 2 \times \frac{22}{7} \times (\frac{21}{2})^2$
$= 2992.5$ cm$^2$
$\therefore$ radius of hemisphere $= \frac{21}{2}$ cm
(i) Volume of the wood left $=$ volume of cube $-$ volume of hemisphere
$= 21^3 - \frac{2}{3} \times \frac{22}{7} \times (\frac{21}{2})^3$
$= 6835.5$ cm$^3$
(ii) Total surface area of remaining solid $=$ TSA of cube $-$ base area of hemisphere $+$ CSA of hemisphere
$= 6 \times 21^2 - \frac{22}{7} \times (\frac{21}{2})^2 + 2 \times \frac{22}{7} \times (\frac{21}{2})^2$
$= 2992.5$ cm$^2$