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Mathematics · 2023 · 3 marks
CBSE 2023 · Region 1 · Set 1 · Q31
(a) A room is in the form of cylinder surmounted by a hemi-spherical dome. The base radius of hemisphere is one-half the height of cylindrical part.…
A room is in the form of cylinder surmounted by a hemi-spherical dome. The base radius of hemisphere is one-half the height of cylindrical part. Find total height of the room if it contains $\displaystyle \left(\frac{1408}{21}\right) \mathrm{m}^{3}$ of air. $\displaystyle \left(\right.$ Take $\displaystyle \left.\pi=\frac{22}{7}\right)$
An empty cone is of radius $\displaystyle 3$ cm and height $\displaystyle 12$ cm. Ice-cream is filled in it so that lower part of the cone which is $\displaystyle \left(\frac{1}{6}\right)^{\text {th }}$ of the volume of the cone is unfilled but hemisphere is formed on the top. Find volume of the ice-cream. (Take $\displaystyle \pi=3.14$ )
Marking-scheme solution
Let h be height of cylindrical part and r be radius of hemisphere
\[\begin{aligned}
& \text { Volume of room }=2 \pi r^{3}+\frac{2}{3} \pi r^{3}=\frac{1408}{21} \\
& \Rightarrow r=2
\end{aligned}
\]
Therefore, \(\displaystyle \mathrm{h}=4\)
Height of the room is \(\displaystyle =6 \mathrm{~m}\)
Volume of the cone \(\displaystyle =\frac{1}{3} \times \pi \times 9 \times 12=36 \pi c m^{3}\)
Volume of ice-cream in the cone \(\displaystyle =\frac{5}{6} \times 36 \times \pi=30 \pi c m^{3}\)
Volume of ice-cream on top \(\displaystyle =\frac{2}{3} \times 27 \times \pi=18 \pi c m^{3}\)
Total volume of the ice-cream \(\displaystyle =(30 \pi+18 \pi)=48 \pi c m^{3}\)
\[=48 \times 3.14=150.72 \mathrm{~cm}^{3}
\]
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CBSE Class 10 Mathematics past-paper question from the 2023 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.