The percentage error in measuring M, L and T are `1%, 1.5 %` and `3 %` respectively. Then the percentage error in measuring the physical quantity with dimensions `ML^(-1)T^(-1)` is :

Options

A. `1 %`
B. `3.5 %`
C. `3 %`
D. `5.5 %`

Explanation

To solve the problem of finding the percentage error in measuring the physical quantity with dimensions \( ML^{-1}T^{-1} \), we can follow these steps:

### Step 1: Understand the dimensions and their respective errors
The physical quantity has dimensions \( M^1 L^{-1} T^{-1} \). The percentage errors in measuring \( M \), \( L \), and \( T \) are given as:
- Percentage error in \( M \) (mass) = \( 1\% \)
- Percentage error in \( L \) (length) = \( 1.5\% \)
- Percentage error in \( T \) (time) = \( 3\% \)

### Step 2: Identify the powers of each dimension
From the dimensions \( ML^{-1}T^{-1} \):
- The power of \( M \) is \( 1 \)
- The power of \( L \) is \( -1 \)
- The power of \( T \) is \( -1 \)

### Step 3: Use the formula for percentage error
The formula for the percentage error in a quantity based on its dimensions is given by:
\[ \text{Percentage Error} = |a| \cdot \frac{\Delta M}{M} + |b| \cdot \frac{\Delta L}{L} + |c| \cdot \frac{\Delta T}{T} \]
where \( a \), \( b \), and \( c \) are the powers of \( M \), \( L \), and \( T \) respectively.

### Step 4: Substitute the values into the formula
Here, we substitute:
- \( a = 1 \), \( b = -1 \), \( c = -1 \)
- \( \Delta M/M = 1\% \) (or \( 0.01 \))
- \( \Delta L/L = 1.5\% \) (or \( 0.015 \))
- \( \Delta T/T = 3\% \) (or \( 0.03 \))

Thus, the percentage error becomes:
\[ \text{Percentage Error} = |1| \cdot 1\% + |-1| \cdot 1.5\% + |-1| \cdot 3\% \]
This simplifies to:
\[ \text{Percentage Error} = 1\% + 1.5\% + 3\% \]

### Step 5: Calculate the total percentage error
Now, we add these values together:
\[ \text{Percentage Error} = 1 + 1.5 + 3 = 5.5\% \]

### Final Answer
The percentage error in measuring the physical quantity with dimensions \( ML^{-1}T^{-1} \) is \( 5.5\% \).

---