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To make the teaching-learning process easier, creative and innovative, a teacher brings clay in the classroom to teach the topic of mensuration. She forms a cylinder of radius $2.1$ cm and height $5$ cm with the clay and put a hemisphere of same radius on its top in such a way that the base of hemisphere covers the top of cylinder. Using the above information, and $\pi = \frac{22}{7}$, find : (a) The volume of cylinder so formed. (b) The volume of hemispherical part. (c) (i) The surface area of the complete solid. OR (c) (ii) The surface area of the cylindrical part, if hemisphere is not put on it.
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(a) Volume of cylinder = $\frac{22}{7} \times \frac{21}{10} \times \frac{21}{10} \times 5 = 69.3$ cm$^3$
(b) Volume of hemispherical part = $\frac{2}{3} \times \frac{22}{7} \times \frac{21}{10} \times \frac{21}{10} \times \frac{21}{10} = 19.404$ cm$^3$
(c) (i) Surface area of the complete solid = $2 \times \frac{22}{7} \times \frac{21}{10} \times 5 + 3 \times \frac{22}{7} \times \frac{21}{10} \times \frac{21}{10} = 107.58$ cm$^2$
OR
(c) (ii) Required surface area = $2 \times \frac{22}{7} \times \frac{21}{10} \times 5 + 2 \times \frac{22}{7} \times \frac{21}{10} \times \frac{21}{10} = 93.72$ cm$^2$
(b) Volume of hemispherical part = $\frac{2}{3} \times \frac{22}{7} \times \frac{21}{10} \times \frac{21}{10} \times \frac{21}{10} = 19.404$ cm$^3$
(c) (i) Surface area of the complete solid = $2 \times \frac{22}{7} \times \frac{21}{10} \times 5 + 3 \times \frac{22}{7} \times \frac{21}{10} \times \frac{21}{10} = 107.58$ cm$^2$
OR
(c) (ii) Required surface area = $2 \times \frac{22}{7} \times \frac{21}{10} \times 5 + 2 \times \frac{22}{7} \times \frac{21}{10} \times \frac{21}{10} = 93.72$ cm$^2$