What is the fractional error in g calculated from `T = 2 pi sqrt((l)/(g))` ? Given that fractional error in `T` and `l` are `pm x` and `pm y` respectively.

Options

A. x + y
B. x - y
C. 2x +y
D. 2x - y

Explanation

To find the fractional error in \( g \) calculated from the equation \( T = 2\pi \sqrt{\frac{l}{g}} \), we can follow these steps:

### Step 1: Rearranging the Equation
We start with the equation for the time period \( T \):
\[ T = 2\pi \sqrt{\frac{l}{g}} \]
To eliminate the square root, we can square both sides:
\[ T^2 = 4\pi^2 \frac{l}{g} \]
Rearranging gives:
\[ g = \frac{4\pi^2 l}{T^2} \]

### Step 2: Finding the Fractional Error in \( g \)
The fractional error in a quantity can be found using the formula:
\[ \frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta T}{T} \]
Where:
- \( \Delta g \) is the absolute error in \( g \)
- \( g \) is the calculated value of \( g \)
- \( \Delta l \) is the absolute error in \( l \)
- \( l \) is the measured value of \( l \)
- \( \Delta T \) is the absolute error in \( T \)
- \( T \) is the measured value of \( T \)

### Step 3: Substituting the Given Errors
From the problem, we know:
- The fractional error in \( T \) is \( \pm x \), so \( \frac{\Delta T}{T} = x \)
- The fractional error in \( l \) is \( \pm y \), so \( \frac{\Delta l}{l} = y \)

Substituting these into the fractional error equation gives:
\[ \frac{\Delta g}{g} = y + 2x \]

### Step 4: Conclusion
Thus, the fractional error in \( g \) is:
\[ \frac{\Delta g}{g} = 2x + y \]

### Final Answer
The fractional error in \( g \) calculated from the given equation is \( 2x + y \).

---