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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