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 will follow these steps:

### Step 1: Rearranging the Equation
We start with the formula for the time period:
\[ T = 2\pi \sqrt{\frac{l}{g}} \]
To express \( g \) in terms of \( T \) and \( l \), we rearrange the equation:
\[ g = \frac{4\pi^2 l}{T^2} \]

### Step 2: Understanding the Errors
We are given the fractional errors in \( T \) and \( l \):
- The fractional error in \( T \) is \( \pm x \).
- The fractional error in \( l \) is \( \pm y \).

### Step 3: Applying the Error Propagation Formula
To find the fractional error in \( g \), we use the formula for error propagation. For a function \( g = \frac{A}{B} \), the fractional error is given by:
\[ \frac{\Delta g}{g} = \frac{\Delta A}{A} + \frac{\Delta B}{B} \]
In our case, \( g \) is dependent on \( l \) (which appears in the numerator) and \( T^2 \) (which appears in the denominator).

### Step 4: Finding the Errors
From our expression for \( g \):
- The term \( l \) contributes a fractional error of \( \frac{\Delta l}{l} \), which is \( y \).
- The term \( T^2 \) contributes a fractional error of \( \frac{\Delta T^2}{T^2} \). Since \( T^2 \) is the square of \( T \), the fractional error in \( T^2 \) is \( 2 \times \frac{\Delta T}{T} \), which is \( 2x \).

### Step 5: Combining the Errors
Now we can combine these contributions to find the total fractional error in \( g \):
\[ \frac{\Delta g}{g} = \frac{\Delta l}{l} + \frac{\Delta T^2}{T^2} = y + 2x \]

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