Question : A metallic solid spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of two of these balls are 2 cm and 1.5 cm. What is the surface area (in cm2) of the third ball?
Option 1: $50 \pi$
Option 2: $\frac{25}{4} \pi$
Option 3: $25 \pi$
Option 4: $\frac{25}{2} \pi$
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Correct Answer: $25 \pi$
Solution : Let the radius of the smaller balls be r 1 , r 2, and r 3. The radius of the larger sphere is r. So, $\frac{4}{3}\pi$r 3 = $\frac{4}{3}\pi$(r 1 3 + r 2 3 + r 3 3 ) ⇒ r 3 = r 1 3 + r 2 3 + r 3 3 ⇒ 3 3 = 2 3 + 1.5 3 + r 3 3 ⇒ 3 3 = 8 + 3.375 + r 3 3 ⇒ r 3 3 = 27 – 8 – 3.375 ⇒ r 3 3 = 15.625 ⇒ r 3 = 2.5 The surface area of the third ball = 4$\pi$r 3 2 = 4$\pi$(2.5) 2 = 25$\pi$ cm 2 Hence, the correct answer is $25\pi$.
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