# Mensuration-61

### From Homeworkwiki

**A spherical ball of lead 3 cm in diameter is melted and recast into 3 spherical balls. The diameters**

**of the two of these are 1 cm and 1.5 cm respectively. What is the diameter of the third spherical ball?**

**Solution:** Radius of the sphere = 3/2 cm

Volume of the sphere = 4/3 Π r^{3} = 4/3 x 22/7 x (3/2)^{3} = 99/7 cm^{3}

Diameter of the 1st small sphere = 1 cm; Therefore, radius = 1/2 cm

Volume of the 1st small sphere = 4/3 Π r^{3} = 4/3 x 22/7 x (1/2)^{3} = 11/21 cm^{3}

Diameter of the 2nd small sphere = 1.5 cm; Therefore, radius = 1.5/2 = 3/4 cm

Volume of the 2nd small sphere = 4/3 Π r^{3} = 4/3 x 22/7 x (3/4)^{3} = 99/56 cm^{3}

Total volume of the two small spheres = 11/21 + 99/56 = (88 + 297) / 168 = 385 / 168 cm^{3}

Volume of the 3rd small sphere = (99 / 7 – 385 / 168) cm^{3} = (2376 – 385) / 168 cm^{3} 1991 / 168 cm^{3}

Let the radius of the 3rd small sphere be r cm. Then

4/3 Π r^{3} = 1991 / 168

Implies, r^{3} = 1991 / 168 x 3/4 x 7 / 22 = 181 / 64

Implies, r = (181)^{1/3} / 4 cm

Hence, diameter of the 3rd small spherical ball = 2r = 2 x 1/4 x (181)^{1/3} cm = 1/2(181)^{1/3} cm