In an experiment of simple pendulum, the errors in the measurement of length of the pendulum `(L)` and time period `(T)` are `3%` and `2%` respectively. The maximum percentage error in the value of `L//T^(2)` is

Options

A. `5%`
B. `7%`
C. `8%`
D. `1%`

Answer

To solve the problem of finding the maximum percentage error in the value of \( \frac{L}{T^2} \), we will follow these steps:

### Step 1: Understand the given errors
We are given:
- The percentage error in the measurement of length \( L \) is \( 3\% \).
- The percentage error in the measurement of time period \( T \) is \( 2\% \).

### Step 2: Identify the formula for maximum percentage error
The formula for the maximum percentage error in a quantity that is a function of multiple variables can be derived from the general rule of error propagation. For a function \( x = \frac{L}{T^2} \), the maximum percentage error can be calculated as:

\[ \frac{\Delta x}{x} \times 100 = \frac{\Delta L}{L} \times 100 + 2 \times \frac{\Delta T}{T} \times 100 \]

### Step 3: Substitute the given values
Now, we substitute the given percentage errors into the formula:

1. The error in length \( \Delta L \) is \( 3\% \), so \( \frac{\Delta L}{L} \times 100 = 3 \).
2. The error in time period \( \Delta T \) is \( 2\% \), so \( \frac{\Delta T}{T} \times 100 = 2 \).

### Step 4: Calculate the maximum percentage error
Using the formula:

\[ \frac{\Delta x}{x} \times 100 = 3 + 2 \times 2 \]

Calculating this gives:

\[ \frac{\Delta x}{x} \times 100 = 3 + 4 = 7\% \]

### Step 5: Conclusion
Thus, the maximum percentage error in the value of \( \frac{L}{T^2} \) is \( 7\% \).

---