CBSE 2025 · Region 5 · Set 1 · Q34 · 5 marks
From one of the faces of a solid wooden cube of side $\displaystyle 14$ cm, maximum number of hemispheres of diameter $\displaystyle 1.4$ cm are scooped out. Find the total number of hemispheres that can be scooped out. Also, find the total surface area of the remaining solid.From a solid cylinder of height $\displaystyle 24$ cm and radius $\displaystyle 5$ cm, two cones of height $\displaystyle 12$ cm and radius $\displaystyle 5$ cm are hollowed out. Find the volume and surface area of the remaining solid.
From one of the faces of a solid wooden cube of side $\displaystyle 14$ cm, maximum number of hemispheres of diameter $\displaystyle 1.4$ cm are scooped out. Find the total number of hemispheres that can be scooped out. Also, find the total surface area of the remaining solid.
From a solid cylinder of height $\displaystyle 24$ cm and radius $\displaystyle 5$ cm, two cones of height $\displaystyle 12$ cm and radius $\displaystyle 5$ cm are hollowed out. Find the volume and surface area of the remaining solid.
Marking-scheme solution
Total number of hemispheres \(\displaystyle =\frac{14 \times 14}{1.4 \times 1.4}\)
\[\text { = } 100
\]
Total Surface Area of remaining solid = Surface Area of Cube + Curved Surface Area of $\displaystyle 100$ hemispheres - Area of $\displaystyle 100$ circles
\[=6 \times 14 \times 14+100 \times 2 \times \frac{22}{7} \times 0.7 \times 0.7-100 \times \frac{22}{7} \times 0.7 \times 0.7
\]
\(\displaystyle =1330\)
∴ Total surface area of remaining solid is \(\displaystyle 1330 \mathrm{~cm}^{2}\).
Volume of remaining solid = Volume of cylinder -Volume of two cones
\[\begin{aligned}
& =\frac{22}{7} \times 5 \times 5 \times 24-2 \times \frac{1}{3} \times \frac{22}{7} \times 5 \times 5 \times 12 \\
& =\frac{8800}{7} \text { or } 1257.14 \mathrm{~cm}^{3} \text { approx. }
\end{aligned}
\]
\(\displaystyle l=\sqrt{(12)^{2}+(5)^{2}}=13 \mathrm{~cm}\)
Surface Area of remaining solid = Curved Surface Area of cylinder + Curved Surface Area of two cones
\[\begin{aligned}
& =2 \times \frac{22}{7} \times 5 \times 24+2 \times \frac{22}{7} \times 5 \times 13 \\
& =\frac{8140}{7} \text { or } 1162.85 \mathrm{~cm}^{2} \text { approx. }
\end{aligned}
\]
Surface Areas and VolumesVolume of a Combination of SolidsApplylong_answerhard
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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.