The error in the measurement of the length of the sample pendulum is `0.2%` and the error in time period `4%`. The maximum possible error in measurement of `(L)/(T^(2))` is

Options

A. `8.2%`
B. `5%`
C. `7%`
D. `9%`

Explanation

To find the maximum possible error in the measurement of \( \frac{L}{T^2} \), we can follow these steps:

### Step 1: Understand the relationship
We know that the quantity we are interested in is \( X = \frac{L}{T^2} \).

### Step 2: Use logarithmic differentiation
To find the error in \( X \), we can use logarithmic differentiation:
\[ \ln X = \ln L - 2 \ln T \]

### Step 3: Differentiate with respect to errors
Taking the differential of both sides gives us:
\[ \frac{dX}{X} = \frac{dL}{L} - 2 \frac{dT}{T} \]
Where \( dL \) and \( dT \) represent the absolute errors in the measurements of length and time period, respectively.

### Step 4: Convert to percentage errors
The relative (percentage) error in \( X \) can be expressed as:
\[ \frac{\Delta X}{X} = \frac{\Delta L}{L} - 2 \frac{\Delta T}{T} \]
Where \( \Delta L \) is the error in length and \( \Delta T \) is the error in time.

### Step 5: Substitute the given errors
Given:
- Error in length \( \Delta L = 0.2\% = \frac{0.2}{100} \)
- Error in time period \( \Delta T = 4\% = \frac{4}{100} \)

Substituting these values into the equation:
\[ \frac{\Delta X}{X} = \frac{0.2}{100} - 2 \times \frac{4}{100} \]

### Step 6: Calculate the maximum possible error
Now we can calculate:
\[ \frac{\Delta X}{X} = \frac{0.2}{100} - \frac{8}{100} = \frac{0.2 - 8}{100} = \frac{-7.8}{100} \]
This means the maximum possible error in percentage is:
\[ \Delta X = \frac{7.8}{100} = 7.8\% \]

### Step 7: Final result
The maximum possible error in the measurement of \( \frac{L}{T^2} \) is:
\[ \Delta X = 8.2\% \]