While measuring acceleration due to gravity by a simple pendulum , a student makes a positive error of `2%` in the length of the pendulum and a positive error of `1%` in the measurement of the value of `g` will be

Options

A. `3%`
B. `0%`
C. `4%`
D. `5%`

Explanation

To solve the problem of determining the percentage error in the measurement of acceleration due to gravity \( g \) when there are errors in the length of the pendulum and the time period, we can follow these steps:

### Step 1: Understand the relationship between the variables
The formula for the acceleration due to gravity \( g \) using a simple pendulum is given by:
\[ g = \frac{4\pi^2 L}{T^2} \]
where \( L \) is the length of the pendulum and \( T \) is the time period of the pendulum.

### Step 2: Identify the errors in measurements
The problem states:
- There is a positive error of \( 2\% \) in the length \( L \).
- There is a positive error of \( 1\% \) in the time period \( T \).

### Step 3: Express the errors in terms of differentials
The percentage error in \( g \) can be determined using the formula for propagation of errors. For a function of multiple variables, the total differential can be expressed as:
\[ \frac{dg}{g} = \frac{dL}{L} - 2\frac{dT}{T} \]
This means that the percentage error in \( g \) is related to the percentage errors in \( L \) and \( T \).

### Step 4: Substitute the known errors
From the problem:
- The error in length \( \frac{dL}{L} = 2\% = 0.02 \)
- The error in time \( \frac{dT}{T} = 1\% = 0.01 \)

Substituting these values into the equation:
\[ \frac{dg}{g} = 0.02 - 2(0.01) \]

### Step 5: Calculate the total error
Now, calculate the right-hand side:
\[ \frac{dg}{g} = 0.02 - 0.02 = 0 \]

### Step 6: Convert to percentage
To express this as a percentage error, multiply by 100:
\[ \text{Percentage error in } g = 0 \times 100 = 0\% \]

### Conclusion
The percentage error in the measurement of the value of \( g \) is \( 0\% \).

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