In an experiment of determine acceleration due to gravity by simple pendulum, a student commits `1%` positive error in the measurement of length and `3%` negative error in the time then `%` error in `g` will be

Options

A. (a)`3%`
B. (b)`4%`
C. (c)`7%`
D. (d)`10%`

Answer

To determine the percentage error in the acceleration due to gravity (g) based on the errors in the measurement of length (l) and time (t) in a simple pendulum experiment, we can follow these steps:

### Step 1: Understand the relationship between g, l, and t
The time period (T) of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{l}{g}} \]
Squaring both sides gives:
\[ T^2 = \frac{4\pi^2 l}{g} \]
From this, we can rearrange to find g:
\[ g = \frac{4\pi^2 l}{T^2} \]

### Step 2: Identify the errors in measurements
- The student commits a **1% positive error** in the measurement of length (l).
- The student commits a **3% negative error** in the measurement of time (T).

### Step 3: Calculate the percentage error in g
The formula for the percentage error in g based on the errors in l and T is:
\[ \frac{\Delta g}{g} \times 100 = \frac{\Delta l}{l} \times 100 + 2 \times \frac{\Delta T}{T} \times 100 \]
Where:
- \(\Delta l\) is the error in length,
- \(\Delta T\) is the error in time.

### Step 4: Substitute the values of errors
- The percentage error in length (\(\frac{\Delta l}{l} \times 100\)) is **1%**.
- The percentage error in time (\(\frac{\Delta T}{T} \times 100\)) is **3%**. Since this is a negative error, we take its absolute value for the calculation.

Now substituting these values into the formula:
\[ \frac{\Delta g}{g} \times 100 = 1 + 2 \times 3 \]
\[ \frac{\Delta g}{g} \times 100 = 1 + 6 \]
\[ \frac{\Delta g}{g} \times 100 = 7\% \]

### Step 5: Conclusion
The percentage error in the acceleration due to gravity (g) is **7%**.

### Final Answer
The percentage error in g is **7%**.

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