A spherical balloon was deflated till its radius was halved. If its volume becomes n times its original volume, then value of n is

Options

A. `1/4`
B. `1/16`
C. `1/8`
D. `1/2`

Answer

To solve the problem, we need to find the value of \( n \) when the radius of a spherical balloon is halved, and we want to determine how the volume changes.

### Step-by-Step Solution:

1. **Define the Original Radius and Volume**:
Let the original radius of the balloon be \( R \).
The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi R^3 \]

2. **Define the New Radius**:
When the radius is halved, the new radius \( R' \) becomes:
\[ R' = \frac{R}{2} \]

3. **Calculate the New Volume**:
The new volume \( V' \) with the new radius \( R' \) is:
\[ V' = \frac{4}{3} \pi (R')^3 = \frac{4}{3} \pi \left(\frac{R}{2}\right)^3 \]
Simplifying this gives:
\[ V' = \frac{4}{3} \pi \left(\frac{R^3}{8}\right) = \frac{4}{3} \pi \frac{R^3}{8} = \frac{1}{6} \pi R^3 \]

4. **Relate the New Volume to the Original Volume**:
We know from the problem that the new volume \( V' \) is \( n \) times the original volume \( V \):
\[ V' = n \cdot V \]
Substituting the expressions for \( V' \) and \( V \):
\[ \frac{1}{6} \pi R^3 = n \cdot \frac{4}{3} \pi R^3 \]

5. **Cancel Out Common Terms**:
We can cancel \( \pi R^3 \) from both sides (assuming \( R \neq 0 \)):
\[ \frac{1}{6} = n \cdot \frac{4}{3} \]

6. **Solve for \( n \)**:
Rearranging the equation to isolate \( n \):
\[ n = \frac{1}{6} \cdot \frac{3}{4} = \frac{1}{8} \]

### Final Answer:
Thus, the value of \( n \) is:
\[ n = \frac{1}{8} \]