While measuring the acceleration due to gravity by a simple pendulum , a student makes a positive error of `1%` in the length of the pendulum and a negative error of `3%` in the value of time period . His percentage error in the measurement of `g` by the relation ` g = 4 pi^(2) ( l // T^(2))` will be

Options

A. `2 %`
B. `4 %`
C. ` 7%`
D. `10%`

Explanation

To find the percentage error in the measurement of acceleration due to gravity \( g \) using the formula \( g = 4 \pi^2 \frac{l}{T^2} \), we will follow these steps:

### Step 1: Identify the given errors
- Positive error in length \( l \): \( +1\% \)
- Negative error in time period \( T \): \( -3\% \)

### Step 2: Write the formula for percentage error in \( g \)
The formula for \( g \) is given as:
\[ g = 4 \pi^2 \frac{l}{T^2} \]
To find the percentage error in \( g \), we use the formula for the propagation of errors:
\[ \frac{\Delta g}{g} \times 100 = \frac{\Delta l}{l} \times 100 + \left(-2 \times \frac{\Delta T}{T} \times 100\right) \]
Where:
- \(\Delta l\) is the error in length
- \(\Delta T\) is the error in time period

### Step 3: Substitute the known values
From the problem, we have:
- \(\frac{\Delta l}{l} \times 100 = +1\%\)
- \(\frac{\Delta T}{T} \times 100 = -3\%\)

Substituting these values into the error formula:
\[ \frac{\Delta g}{g} \times 100 = 1\% + \left(-2 \times (-3\%)\right) \]

### Step 4: Calculate the percentage error
Now, calculate the second term:
\[ -2 \times (-3\%) = +6\% \]
Thus, the total percentage error in \( g \) becomes:
\[ \frac{\Delta g}{g} \times 100 = 1\% + 6\% = 7\% \]

### Conclusion
The percentage error in the measurement of \( g \) is \( 7\% \).

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