Question : A solid metallic sphere of radius 4 cm is melted and recast into spheres of 2 cm each. What is the ratio of the surface area of the original sphere to the sum of the surface areas of the spheres, so formed?
Option 1: 2 : 1
Option 2: 2 : 3
Option 3: 1 : 2
Option 4: 1 : 4
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Correct Answer: 1 : 2
Solution :
Given: Radius of bigger sphere = 4 cm
The radius of a smaller sphere = 2 cm
Let the radius of the bigger sphere be $R$ and smaller be $r$.
According to the question,
Number of spheres formed = $\frac{\text{Volume of bigger sphere}}{\text{Volume of smaller spheres}}$
= $\frac{\frac{4}{3}\pi R^3}{\frac{4}{3}\pi r^3}$
= $\frac{R^3}{r^3}$
= $\frac{4^3}{2^3}$
= 8
Now, the surface area of the original sphere = $4\pi R^2$ = $4\pi (4)^2$ = $64\pi$
And sum of surface area of small spheres = $8 × 4\pi (2)^2$ = $128\pi$
The required ratio = $64\pi : 128\pi= 1 : 2$
Hence, the correct answer is 1 : 2.
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