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Surface Areas and Volumes
NCERT Exemplar Ch 11
CBSE_NCERT_EXEMPLAR_CH11
Grade 10

Question:

A solid piece of iron in the form of a cuboid of dimensions $49\text{ cm} \times 33\text{ cm} \times 24\text{ cm}$, is remelted to form a solid sphere. The radius of the sphere is:

$21\text{ cm}$
$28\text{ cm}$
$14\text{ cm}$
$35\text{ cm}$

Step-by-Step Solution

Key Concept: Volume of sphere $\dfrac{4}{3} \pi r^3 = l \times b \times h$.
Stepwise Solution:

$\dfrac{4}{3} \times \dfrac{22}{7} \times r^3 = 49 \times 33 \times 24$. [0.5 Mark]

$\dfrac{88}{21} r^3 = 38808 \Rightarrow r^3 = \dfrac{38808 \times 21}{88} = 441 \times 21 = 9261 = 21^3 \Rightarrow r = 21\text{ cm}$. [0.5 Mark]

Marking Scheme:

• Volume equality setup: 0.5 Mark
• Solving for radius $r = 21\text{ cm}$: 0.5 Mark

Correct Answer: $21\text{ cm}$
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