The error in the measument of radius of a sphere is 0.4% . The relative error in its volume is

Options

A. 0.0016
B. 0.0012
C. 0.0024
D. 0.012

Answer

To find the relative error in the volume of a sphere given the relative error in the radius, 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: Identify the Given Information
We are given that the error in the measurement of the radius \( R \) is 0.4%. This can be expressed as:
\[ \frac{\Delta R}{R} = \frac{0.4}{100} = 0.004 \]
where \( \Delta R \) is the absolute error in the radius.

### Step 3: Relate the Error in Volume to the Error in Radius
To find the relative error in the volume, we can use the formula for the propagation of errors. For a function of the form \( V = k R^n \) (where \( k \) is a constant and \( n \) is the exponent), the relative error in volume is given by:
\[ \frac{\Delta V}{V} = n \cdot \frac{\Delta R}{R} \]
In our case, since \( V = \frac{4}{3} \pi R^3 \), we have \( n = 3 \).

### Step 4: Calculate the Relative Error in Volume
Substituting the values we have:
\[ \frac{\Delta V}{V} = 3 \cdot \frac{\Delta R}{R} = 3 \cdot 0.004 = 0.012 \]

### Step 5: Convert to Percentage
To express this as a percentage, we multiply by 100:
\[ \text{Relative error in volume} = 0.012 \times 100 = 1.2\% \]

### Final Answer
Thus, the relative error in the volume of the sphere is:
\[ 0.012 \text{ or } 1.2\% \]