A solid iron ball is melted and 64 smaller solid balls of equal aize are made uaing the entire volum...........

Quantitative Aptitude (CDS) Mensuration 2D

A solid iron ball is melted and 64 smaller solid balls of equal aize are made uaing the entire volume of iron. What is the ratio of the surface area of the larger ball to the aum of the surface areas of all the smaller balls ?

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2023 CDS -1 MATH

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0.25

0.50

0.75

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Explanation:

- When a solid iron ball is melted into smaller balls, the total volume remains constant.

- Let the radius of the large ball be \( R \). The volume is \( \frac{4}{3} \pi R^3 \).

- Let the radius of each smaller ball be \( r \). The volume is \( \frac{4}{3} \pi r^3 \).

- Given 64 smaller balls, total volume is \( 64 \times \frac{4}{3} \pi r^3 \).

- Equating volumes: \( \frac{4}{3} \pi R^3 = 64 \times \frac{4}{3} \pi r^3 \).

- Solving, \( R^3 = 64r^3 \), thus \( R = 4r \).

- Surface area of large ball: \( 4 \pi R^2 \).

- Surface area of one small ball: \( 4 \pi r^2 \).

- Total surface area of 64 small balls: \( 64 \times 4 \pi r^2 \).

- Ratio of surface areas: \( \frac{4 \pi R^2}{64 \times 4 \pi r^2} = \frac{R^2}{64r^2} = \frac{(4r)^2}{64r^2} = \frac{16r^2}{64r^2} = \frac{1}{4} = 0.25 \).

Option 1: 0.25 is the correct answer.