If the error in the measurement of radius of a sphere in `2%` then the error in the determination of volume of the spahere will be

Options

A. `4%`
B. `6%`
C. `8%`
D. `2%`

Explanation

To solve the problem of determining the error in the volume of a sphere based on the error in the radius measurement, we can follow these steps:

### Step 1: Understand the formula for the volume of a sphere
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 error in the radius
The problem states that the error in the measurement of the radius \( R \) is \( 2\% \). This can be expressed as:
\[ \frac{\Delta R}{R} = 0.02 \]
where \( \Delta R \) is the absolute error in the radius.

### Step 3: Differentiate the volume formula
To find the error in volume, we can use the concept of differentiation. Taking the logarithm of both sides of the volume formula, we have:
\[ \log V = \log\left(\frac{4}{3} \pi\right) + 3 \log R \]
Differentiating both sides with respect to \( R \) gives:
\[ \frac{dV}{V} = 3 \frac{dR}{R} \]
This implies:
\[ \frac{\Delta V}{V} = 3 \frac{\Delta R}{R} \]

### Step 4: Substitute the error in radius into the equation
Now, substituting the known error in the radius into the equation:
\[ \frac{\Delta V}{V} = 3 \times 0.02 = 0.06 \]

### Step 5: Convert the error into percentage
To express this error as a percentage, we multiply by 100:
\[ \Delta V \text{ (percentage)} = 0.06 \times 100 = 6\% \]

### Conclusion
Thus, the error in the determination of the volume of the sphere is \( 6\% \).

### Final Answer
The error in the determination of the volume of the sphere will be **6%**.

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