Question

A spherical metal of radius 10 cm is molten and made into 1000 smaller spheres of equal sizes. In this process the surface area of the metal is increased by:

• A. 1000 times
• B. 100 times
• C. 10 times
• D. No change
• E. None of the above

Hint:

- In such problems, always use the principle of conservation of volume.
- Surface area does not scale linearly with volume; smaller subdivisions usually increase the total surface area.

Answer 1

Answer: (E)

Step 1: Volume conservation
The volume of the original sphere (radius \(R = 10\) cm) is: \[ V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (10^3) = \frac{4000}{3}\pi. \] This is redistributed into 1000 smaller spheres, each of radius \(r\). Step 2: Volume of one smaller sphere
Let the radius of each smaller sphere be \(r\). Then volume of one smaller sphere is: \[ v = \frac{4}{3}\pi r^3. \] Since there are 1000 spheres: \[ 1000 \times \frac{4}{3}\pi r^3 = \frac{4000}{3}\pi. \] Step 3: Solve for \(r\)
\[ 1000r^3 = 1000 \quad \Rightarrow \quad r^3 = 1 \quad \Rightarrow \quad r = 1 \, \text{cm}. \] Step 4: Surface area comparison
Original surface area: \[ A_{\text{big}} = 4\pi R^2 = 4\pi (10^2) = 400\pi. \] Surface area of one small sphere: \[ A_{\text{small}} = 4\pi r^2 = 4\pi (1^2) = 4\pi. \] For 1000 such spheres: \[ A_{\text{total}} = 1000 \times 4\pi = 4000\pi. \] Step 5: Ratio of increase
\[ \frac{A_{\text{total}}}{A_{\text{big}}} = \frac{3600\pi}{400\pi} = 9. \]