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Mathematics · 2026 · 4 marks
CBSE 2026 · Region 5 · Set 2 · Q38
A wall mounted lamp, made of fabric, is shown below. Lamp has cuboidal shape, open from top and bottom. A spherical bulb of diameter $\displaystyle 7$ cm is latched with a very thin rod. (Ignore the rod while making calculations.)
Dimensions of the cuboid are $\displaystyle 24$ cm × $\displaystyle 12$ cm × $\displaystyle 17$ cm.(i)Find the surface area of the bulb.(ii)What could be the maximum diameter of the bulb if at least $\displaystyle 1$ cm space is left from each side ?(iii)Find the area of the fabric used if there is a fold of $\displaystyle 2$ cm on top and bottom edges.Find the space available inside the lamp. $\displaystyle 1172$-$\displaystyle 2$ { } of $\displaystyle 24$
A wall mounted lamp, made of fabric, is shown below. Lamp has cuboidal shape, open from top and bottom. A spherical bulb of diameter $\displaystyle 7$ cm is latched with a very thin rod. (Ignore the rod while making calculations.)
Dimensions of the cuboid are $\displaystyle 24$ cm × $\displaystyle 12$ cm × $\displaystyle 17$ cm.
(i)
Find the surface area of the bulb.
(ii)
What could be the maximum diameter of the bulb if at least $\displaystyle 1$ cm space is left from each side ?
(iii)
Find the area of the fabric used if there is a fold of $\displaystyle 2$ cm on top and bottom edges.
Find the space available inside the lamp. $\displaystyle 1172$-$\displaystyle 2$ { } of $\displaystyle 24$
Marking-scheme solution
(i) Surface area of the bulb \(\displaystyle =4 \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}=154 \mathrm{~cm}^{2}\)
(ii) Maximum diameter of the bulb = Minimum side length - $\displaystyle 2$ cm
\[\text { = } 12-2=10 \mathrm{~cm}
\]
(a)
With $\displaystyle 2$ cm extra cloth for top and bottom edges,
new dimensions are $\displaystyle 24$ cm × $\displaystyle 12$ cm × $\displaystyle 21$ cm
Area of fabric used \(\displaystyle =2 \times 21 \times(24+12)=1512 \mathrm{~cm}^{2}\)
\[\text { (iii) (b)Space available } \begin{aligned}
= & 24 \times 12 \times 17-\frac{4}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2} \\
& =4896-\frac{539}{3} \\
& =\frac{14149}{3} \mathrm{~cm}^{3} \text { or } 4716.3 \mathrm{~cm}^{3} \text { (approx.) }
\end{aligned}
\]
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CBSE Class 10 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.