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Mathematics — CBSE Class 10 board question

Q1. [4]
A wall mounted lamp, made of fabric, is shown below. Lamp has cuboidal shape, open from top and bottom. A spherical bulb of diameter 7 cm is latched with a very thin rod. (Ignore the rod while making calculations.) Dimensions of the cuboid are 24 cm × 12 cm × 17 cm.
Based on the above information, answer the following questions:
  1. (i) Find the surface area of the bulb. [1]
  2. (ii) What could be the maximum diameter of the bulb if at least 1 cm space is left from each side? [1]
  3. (iii) Find the area of the fabric used if there is a fold of 2 cm on top and bottom edges. OR Find the space available inside the lamp. [2]
Previously asked in CBSE board exam
2026 30/5/1 Q37
Generated by claude-sonnet-4-6 · 2026-06-15 10:36 · grounding stimulus
Model Answer

(i) Surface area of the bulb:

Diameter = 7 cm, so radius = 3.5 cm

Surface area = $4\pi r^2 = 4 \times \dfrac{22}{7} \times 3.5 \times 3.5 = 154 \text{ cm}^2$

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(ii) Maximum diameter of bulb leaving 1 cm from each side:

The smallest dimension of the base = 12 cm.
Maximum diameter = 12 − 1 − 1 = 10 cm

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(iii) Area of fabric used (with 2 cm fold on top and bottom):

The fold adds 2 cm on each of the two open ends, so effective height = 17 + 2 + 2 = 21 cm.

Lateral surface area of cuboid (open top & bottom) = Perimeter of base × height
= 2(24 + 12) × 21
= 2 × 36 × 21
= 1512 cm²

OR

Space available inside the lamp:

Volume of cuboid = 24 × 12 × 17 = 4896 cm³

Volume of bulb = $\dfrac{4}{3}\pi r^3 = \dfrac{4}{3} \times \dfrac{22}{7} \times 3.5^3 = \dfrac{4}{3} \times \dfrac{22}{7} \times 42.875 = 179.67 \approx 179.67 \text{ cm}^3$

Space available = 4896 − 179.67 = 4716.33 cm³

Source: Mensuration, Surface Areas and Volumes

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Explanation
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