A solid metallic sphere of radius $12 ext{ cm}$ is melted and recast into $27$ identical spheres. What is th

Question

A solid metallic sphere of radius $12 	ext{ cm}$ is melted and recast into $27$ identical spheres. What is the ratio of the surface area of the original sphere to the total surface area of $8$ smaller spheres so formed?

Examoogle · Accepted Answer

The correct answer is Option B: 9 : 8. Ratio of radii $R/r = n^{1/3}$. For $n=27$, $R/r = 3$. Surface area ratio is $A_{large} / (k 	imes A_{small}) = (R^2) / (k 	imes r^2) = (3^2) / k = 9/k$. For $k=8$, the ratio is $9/8$.

Step-by-Step Solution & Concepts:
Shortcut Trick: Ratio of radii $R/r = n^{1/3}$. For $n=27$, $R/r = 3$. Surface area ratio is $A_{large} / (k 	imes A_{small}) = (R^2) / (k 	imes r^2) = (3^2) / k = 9/k$. For $k=8$, the ratio is $9/8$.

Formula Used: $$V_{original} = n 	imes V_{small} \implies \frac{4}{3}\pi R^3 = n 	imes \frac{4}{3}\pi r^3$$ and $$SA = 4\pi r^2$$

Given: Original sphere radius R = 12 cm. Recast into 27 smaller spheres of radius r. Comparing surface area of the original sphere to 8 smaller spheres.

Steps: Find radius of small spheres: Let $R = 12$ cm be the radius of the large sphere and $r$ be the radius of the $n=27$ smaller spheres. Using volume conservation: $V_{large} = 27 	imes V_{small}$. Thus, $\frac{4}{3}\pi R^3 = 27 	imes \frac{4}{3}\pi r^3$. Simplifying gives $(\frac{R}{r})^3 = 27$, so $\frac{R}{r} = 3$.. Calculate surface areas: Surface area of original sphere $SA_{orig} = 4\pi R^2 = 4\pi(12)^2 = 576\pi$. Surface area of one small sphere $SA_{small} = 4\pi r^2 = 4\pi(4)^2 = 64\pi$.. Determine required ratio: We need the ratio of the surface area of the original sphere to the total surface area of 8 smaller spheres: Ratio = $\frac{SA_{orig}}{8 	imes SA_{small}} = \frac{576\pi}{8 	imes 64\pi}$.

Answer: B is correct because the ratio of the surface area of the original sphere to 8 smaller spheres is 9:8.

Exam Tip: This problem uses the relationship between scale factor ($k$), area ratio ($k^2$), and volume ratio ($k^3$). If the linear dimension ratio is $3:1$, the area ratio is $9:1$, which simplifies solving such geometry problems.

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