CBSE 2025 · Region 1 · Set 1 · Q30 · 3 marks
A room is in the form of a cylinder surmounted by a hemispherical dome. The base radius of the hemisphere is half of the height of the cylindrical part. If the room contains $\displaystyle \frac{1408}{21} \mathrm{~m}^{3}$ of air, find the height of the cylindrical part. (Use $\displaystyle \pi=\frac{22}{7}$ ).
Marking-scheme solution
Let r is the radius of hemisphere and cylinder and h is the height of cylinder
\[h=2 r
\]
Volume of air in room \(\displaystyle =\frac{2}{3} \pi \mathrm{r}^{3}+\pi \mathrm{r}^{2} \mathrm{~h}\)
\[\begin{aligned}
\frac{1408}{21} & =\frac{2}{3} \pi r^{3}+\pi r^{2}(2 r) \\
\frac{1408}{21} & =\frac{8}{3} \times \frac{22}{7} \times r^{3} \\
r^{3} & =8 \\
\therefore \quad r & =2 \mathrm{~m} \\
\text { and } h & =4 \mathrm{~m}
\end{aligned}
\]
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CBSE Class 10 Mathematics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.