Question

Twenty-seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S'. Find the 
(i) radius r' of the new sphere, 
(ii) ratio of S and S'

Answer 1

Step 1: Surface Area of a Sphere

The surface area \( S \) of a sphere is given by the formula:

\[ S = 4 \pi r^2 \]

Step 2: Volume of a Sphere

The volume \( V \) of a sphere is given by the formula:

\[ V = \frac{4}{3} \pi r^3 \]

Step 3: Total Volume of the 27 Spheres

Each of the 27 spheres has a volume of \( \frac{4}{3} \pi r^3 \), so the total volume of all 27 spheres is:

\[ V_{\text{total}} = 27 \times \frac{4}{3} \pi r^3 = 36 \pi r^3 \]

Step 4: Volume of the New Sphere

The total volume of the new sphere is equal to the total volume of the 27 smaller spheres. Let \( r' \) be the radius of the new sphere. The volume of the new sphere is:

\[ V' = \frac{4}{3} \pi r'^3 \] \] Since the volume is conserved, we equate the total volume of the new sphere to the total volume of the 27 smaller spheres: \[ \frac{4}{3} \pi r'^3 = 36 \pi r^3 \] Canceling \( \pi \) and \( \frac{4}{3} \) from both sides: \[ r'^3 = 27 r^3 \] Taking the cube root of both sides: \[ r' = 3r \]

Answer for (i):

The radius \( r' \) of the new sphere is \( 3r \).

Step 5: Surface Area of the New Sphere

The surface area \( S' \) of the new sphere with radius \( r' = 3r \) is given by:

\[ S' = 4 \pi (r')^2 = 4 \pi (3r)^2 = 36 \pi r^2 \]

Answer for (ii):

The ratio of the surface area of the original sphere \( S \) to the surface area of the new sphere \( S' \) is:

\[ \frac{S}{S'} = \frac{4 \pi r^2}{36 \pi r^2} = \frac{1}{9} \]

Conclusion:

1. The radius \( r' \) of the new sphere is \( 3r \).
2. The ratio of the surface areas is \( \frac{S}{S'} = \frac{1}{9} \).