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Surface Areas and Volumes
RD Sharma
CBSE
Grade 10

Question:

From a solid cylinder whose height is $2.4\text{ cm}$ and diameter $1.4\text{ cm}$, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest $\text{cm}^2$.

Step-by-Step Solution

Key Concept: $r = 0.7\text{ cm}, h = 2.4\text{ cm} \Rightarrow l = \sqrt{0.49 + 5.76} = \sqrt{6.25} = 2.5\text{ cm}$. Total Surface Area $= \text{CSA cylinder} + \text{Base area} + \text{CSA cone} = 2 \pi r h + \pi r^2 + \pi r l = \pi r (2h + r + l) = \dfrac{22}{7} \times 0.7 \times (4.8 + 0.7 + 2.5) = 2.2 \times 8.0 = 17.6\text{ cm}^2 \approx 18\text{ cm}^2$.
$r = 0.7\text{ cm}, h = 2.4\text{ cm} \Rightarrow l = \sqrt{0.49 + 5.76} = 2.5\text{ cm}$. [1.0 Mark]
$\text{Surface Area} = \pi r(2h + r + l) = 2.2 \times (4.8 + 0.7 + 2.5) = 2.2 \times 8 = 17.6\text{ cm}^2$. [1.5 Marks]
Nearest $\text{cm}^2 = 18\text{ cm}^2$. [0.5 Mark]

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🎯 Official CBSE Marking Scheme:
Finding slant height $l = 2.5\text{ cm}$: 1.0 Mark
Calculating total surface area $= 17.6\text{ cm}^2$: 1.5 Marks
Rounding to nearest integer $= 18\text{ cm}^2$: 0.5 Mark

Correct Answer:
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