The dimensional formula for a physical quantity `x` is `[M^(-1) L^(3) T^(-2) ]` . The errors in measuring the quantities `M , L , and T,` respectively are `2%, 3% , and 4%`. The maximum percentage of error that occurs in measuring the quantity `x` is

Options

A. `9`
B. `10`
C. `14`
D. `19`

Explanation

To solve the problem, we need to find the maximum percentage error in measuring the physical quantity \( x \) given its dimensional formula and the errors in measuring \( M \), \( L \), and \( T \).

### Step-by-Step Solution:

1. **Identify the Dimensional Formula**:
The dimensional formula for the physical quantity \( x \) is given as:
\[ [M^{-1} L^{3} T^{-2}] \]
This indicates that \( x \) depends on \( M \), \( L \), and \( T \) with the respective powers of -1, 3, and -2.

2. **List the Errors in Measurement**:
The errors in measuring the quantities are:
- Error in \( M \) = 2%
- Error in \( L \) = 3%
- Error in \( T \) = 4%

3. **Apply the Formula for Maximum Percentage Error**:
The maximum percentage error in a quantity that is a product of several factors can be calculated using the formula:
\[ \text{Maximum Percentage Error} = \left| n_1 \cdot \text{Error in } M \right| + \left| n_2 \cdot \text{Error in } L \right| + \left| n_3 \cdot \text{Error in } T \right| \]
where \( n_1, n_2, n_3 \) are the powers of \( M, L, T \) in the dimensional formula.

4. **Substitute the Values**:
Here, we substitute the values:
- For \( M^{-1} \): \( n_1 = -1 \), Error = 2%
- For \( L^{3} \): \( n_2 = 3 \), Error = 3%
- For \( T^{-2} \): \( n_3 = -2 \), Error = 4%

Therefore, the maximum percentage error can be calculated as:
\[ \text{Maximum Percentage Error} = |-1 \cdot 2| + |3 \cdot 3| + |-2 \cdot 4| \]
\[ = 2 + 9 + 8 \]
\[ = 19\% \]

5. **Final Answer**:
The maximum percentage error that occurs in measuring the quantity \( x \) is:
\[ \boxed{19\%} \]