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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 the balls are $1.5 cm$ and $2 cm$ respectively. The diameter of the third balls is

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2.4 cm

2.5 cm

2.7 cm

2.8 cm

answer is B.

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Detailed Solution

It is given that the diameter of the big spherical ball is 3cm .
So, the radius is 1.5 cm .
The diameter of two of the balls are 1.5 cm, and 2 cm respectively.
So, the radii are

1 .5 2 cm,

, and 1 cm respectively.
Volume of the sphere is,

V = 43 π r 3

, where

r

is the radius of the sphere.
The volume of the big spherical ball will be the sum of the volumes of the new three balls.
So,

V = V 1 + V 2 + V 3 ⇒ 43 π r 3 = 43 π r 1 3 + 43 π r 2 3 + 43 π r) 3 3 ⇒ r 3 = r 1 3 + r 2 3 + r 3 3 ⇒ 1.5 3 = 0.75)

Solving further,

r 3 3 = 0.75 3 + 13 − 1.5 3 ⇒ r 3 = 0.75 3 + 13 − 1.5 3 3 ⇒ r 3 = 1.953125 3 ⇒ r 3 = 1.25

Thus, the diameter is, 2.5 cm.
Therefore, option (2) is correct.

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