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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

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2.66 cm

2.5 cm

3 cm

3.5 cm

answer is B.

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Detailed Solution

The combined volumes of the new three balls will equal the initial spherical ball's volume.
A sphere's volume with radius r can be given as,

V = \frac{4}{3} π r^{3}

Here, diameter = 3 cm
So, radius

= \frac{3}{2} = 1.5 cm

As, the diameter of two balls is 1.5 cm and 2 cm
Then, radius of other two balls = .75 cm and 1 cm.
Suppose, radius of the third new ball = R cm.
Here,
Volume of the three new spheres = Volume of old sphere

\frac{4}{3} π \times 0 \cdot 75^{3} + \frac{4}{3} π \cdot 1^{3} + \frac{4}{3} π \cdot R^{3} = \frac{4}{3} π {1.5}^{3}

0 {. 75}^{3} + 1^{3} + R^{3} = {1.5}^{3}

0.421875 + 1 + R^{3} = 3.375

1.421875 + R^{3} = 3.375

R^{3} = 1.953125

R = {1.953125}^{\frac{1}{3}}

R = 1.25

So, radius of third new ball = 1.25 cm.
Therefore, its diameter D=

1.25 \times 2

2.5 cm

Correct option is 2.

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