If error in measurement of radius of a sphere is 1%, what will be the error in measurement of volume?

Options

A. a) 0.01
B. b) `(1)/(3)%`
C. c) 0.03
D. d) None of these

Explanation

To solve the problem of determining the error in the measurement of the volume of a sphere when the error in the measurement of its radius is given, 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:
\[ V = \frac{4}{3} \pi r^3 \]

### Step 2: Differentiate the volume with respect to the radius
To find the error in volume, we need to differentiate the volume \( V \) with respect to the radius \( r \):
\[ \frac{dV}{dr} = 4 \pi r^2 \]

### Step 3: Relate the relative errors
The relative error in volume \( \frac{\Delta V}{V} \) can be related to the relative error in radius \( \frac{\Delta R}{R} \) using the formula:
\[ \frac{\Delta V}{V} = 3 \frac{\Delta R}{R} \]
This relationship arises because the volume is proportional to the cube of the radius.

### Step 4: Substitute the given error in radius
We are given that the error in the measurement of the radius \( \frac{\Delta R}{R} \) is 1%. We can express this as:
\[ \frac{\Delta R}{R} = 0.01 \]

### Step 5: Calculate the error in volume
Now, substituting this value into the equation from Step 3:
\[ \frac{\Delta V}{V} = 3 \times 0.01 = 0.03 \]

### Step 6: Convert to percentage
To express the error in volume as a percentage:
\[ \frac{\Delta V}{V} = 0.03 \times 100\% = 3\% \]

### Final Answer
Thus, the error in the measurement of the volume of the sphere is **3%**.

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