A cuboid has volume `V=lxx2lxx3l`, where l is the length of one side. If the relative percentage error in the measurment of l is 1%, then the relative percentage error in measurement of V is

Options

A. 0.18
B. 0.06
C. 0.03
D. 0.01

Explanation

To solve the problem, we need to find the relative percentage error in the volume \( V \) of a cuboid given the relative percentage error in the measurement of its length \( l \).

### Step-by-Step Solution:

1. **Understand the Volume Formula**:
The volume \( V \) of the cuboid is given by:
\[ V = l \times 2l \times 3l \]
This simplifies to:
\[ V = 6l^3 \]

2. **Identify the Relative Error in Length**:
The problem states that the relative percentage error in the measurement of \( l \) is 1%. This can be expressed mathematically as:
\[ \frac{\Delta l}{l} \times 100 = 1\% \]
Thus,
\[ \frac{\Delta l}{l} = 0.01 \]

3. **Calculate the Relative Error in Volume**:
To find the relative error in volume, we use the formula for the relative error in a product:
\[ \frac{\Delta V}{V} = \frac{\Delta (6l^3)}{6l^3} \]
Since \( 6 \) is a constant, it can be neglected in the differentiation. Therefore:
\[ \frac{\Delta V}{V} = 3 \cdot \frac{\Delta l}{l} \]

4. **Substitute the Relative Error in Length**:
Now, substituting \( \frac{\Delta l}{l} = 0.01 \) into the equation:
\[ \frac{\Delta V}{V} = 3 \cdot 0.01 = 0.03 \]

5. **Convert to Percentage**:
To express this as a percentage, we multiply by 100:
\[ \frac{\Delta V}{V} \times 100 = 0.03 \times 100 = 3\% \]

### Conclusion:
The relative percentage error in the measurement of the volume \( V \) is **3%**.