On decreasing the radius of a sphere by `20%`. By what per cent its volume will be decreased?

Options

A. `20%`
B. `28.8%`
C. `57.6%`
D. `48.8%`

Answer

To solve the problem of how much the volume of a sphere decreases when its radius is decreased by 20%, we can follow these steps:

### Step 1: Understand the relationship between radius and volume
The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
where \( r \) is the radius of the sphere.

### Step 2: Determine the new radius after a 20% decrease
If the original radius is \( r \), a decrease of 20% means the new radius \( r' \) will be:
\[ r' = r - 0.2r = 0.8r \]

### Step 3: Calculate the new volume using the new radius
Substituting \( r' \) into the volume formula, we get:
\[ V' = \frac{4}{3} \pi (r')^3 = \frac{4}{3} \pi (0.8r)^3 \]
Calculating \( (0.8r)^3 \):
\[ (0.8r)^3 = 0.512r^3 \]
Thus, the new volume \( V' \) can be expressed as:
\[ V' = \frac{4}{3} \pi (0.512r^3) = 0.512 \left(\frac{4}{3} \pi r^3\right) = 0.512 V \]

### Step 4: Calculate the decrease in volume
The decrease in volume \( \Delta V \) is given by:
\[ \Delta V = V - V' = V - 0.512 V = (1 - 0.512)V = 0.488 V \]

### Step 5: Calculate the percentage decrease in volume
To find the percentage decrease in volume, we use the formula:
\[ \text{Percentage Decrease} = \left(\frac{\Delta V}{V}\right) \times 100 = \left(\frac{0.488 V}{V}\right) \times 100 = 48.8\% \]

### Conclusion
The volume of the sphere decreases by **48.8%** when the radius is decreased by 20%.