Question

A solid metallic sphere of 16 cm diameter is melted and cast into smaller solid spheres of diameter 4 cm. The number of smaller spheres formed is:

• A. \(32 \)
• B. \(48 \)
• C. \(64 \)
• D. \(84 \)

Hint:

When solids are melted and recast, always use volume ratio. For spheres, number of new spheres is the cube of the ratio of radii.

Answer 1

The Correct Option is C

Concept: When a solid object is melted and recast into smaller identical solids of the same shape, the total volume remains conserved. Therefore, the number of smaller solids formed is equal to the ratio of the volume of the original solid to the volume of one smaller solid.

Step 1: Find the radius of the original and smaller spheres.
The diameter of the original sphere is \(16\) cm, so its radius is: \[ R = \frac{16}{2} = 8 \, \text{cm} \] The diameter of the smaller sphere is \(4\) cm, so its radius is: \[ r = \frac{4}{2} = 2 \, \text{cm} \]

Step 2: Write the formula for volume of a sphere.
The volume of a sphere is given by: \[ V = \frac{4}{3}\pi r^3 \] Since \(\frac{4}{3}\pi\) is common for both spheres, it cancels when taking ratio.

Step 3: Find the ratio of volumes.
Number of smaller spheres formed: \[ N = \frac{\frac{4}{3}\pi R^3}{\frac{4}{3}\pi r^3} = \frac{R^3}{r^3} \]

Step 4: Substitute the values carefully.
\[ N = \frac{8^3}{2^3} \] Now compute powers: \[ 8^3 = 512, \quad 2^3 = 8 \] So, \[ N = \frac{512}{8} \]

Step 5: Perform final division.
\[ N = 64 \] Final Answer: \(64\)