Subject: General Aptitude · Chapter: Volume and Surface Area · Exam: 2009 · Marks: · Difficulty:
A spherical ball of lead, 3 cm in diameter is melted and recast into three spherical balls. The diameter of two of these are 1.5 cm and 2 cm respectively. The diameter of the third ball is
(a)2.5 cm
(b)2.66 cm
(c)3 cm
(d)3.5 cm
Answer
Answer (as printed): A
Explanation
Let the radius of the third ball be $R$ cm. Then, $$\begin{aligned} & \frac{4}{3} \pi \times\left(\frac{3}{4}\right)^{3}+\frac{4}{3} \pi \times(1)^{3}+\frac{4}{3} \pi \times R^{3}=\frac{4}{3} \pi \times\left(\frac{3}{2}\right)^{3} \\ & \Rightarrow \frac{27}{64}+1+R^{3}=\frac{27}{8} \Rightarrow R^{3}=\frac{125}{64}=\left(\frac{5}{4}\right)^{3} \Rightarrow R=\frac{5}{4} \\ & \therefore \text { Diameter of the third ball }=2 R=\frac{5}{2} \mathrm{cm}=2.5 \mathrm{cm} \end{aligned}$$