Two spheres of the same metal weigh 1 kg and 7 kg. The radius of the smaller sphere is 3 cm. The two spheres are melted to form a single big sphere. Find the diameter of the big sphere.

Radius of the smaller sphere = r = 3 cm.

Let R be the radius of a larger new sphere.

Mass of small sphere (m₁) = 1 kg.

Mass of bigger sphere (m₂) = 7 kg.

The spheres are melted to form a new sphere.

So, the mass of the new sphere (M) = 1 + 7 = 8 kg.

Density of smaller sphere = Density of new sphere.

Let v be volume of small sphere and V be volume of new bigger sphere.

$\dfrac{m_1}{v} = \dfrac{M}{V} \\[1em] \dfrac{1}{v} = \dfrac{8}{V} \\[1em] \dfrac{v}{V} = \dfrac{1}{8} ….(i)$

Given, radius of smaller sphere, r = 3 cm.

Volume of smaller sphere = v = $\dfrac{4}{3}$πr³

= $\dfrac{4}{3}$π(3)³

= 36π cm³.

Volume of new sphere = V = $\dfrac{4}{3}$πR³

Putting these values in (i),

$\dfrac{36π}{\dfrac{4}{3}πR^3} = \dfrac{1}{8} \\[1em] \dfrac{36 \times 3}{4 \times R^3} = \dfrac{1}{8} \\[1em] \dfrac{108}{4R^3} = \dfrac{1}{8} \\[1em] R^3 = \dfrac{108 \times 8}{4} \\[1em] R^3 = 216 \\[1em] R^3 = 6^3 \\[1em] R = 6 \text{ cm}.$

Diameter = 2 × 6 = 12 cm.

Hence, the diameter of the new bigger sphere = 12 cm.