The image presents the following problem:

"While measuring $g$ using the simple pendulum method, the percentage error in the measurement of length is ±2%, and in the measurement of time is ±3%. Calculate the relative error in $g$."

Solution:

The formula for $g$ using the simple pendulum method is:

$g = \frac{4 \pi^2 L}{T^2}$

Where:

• $L$ is the length of the pendulum,
• $T$ is the time period.

To find the relative error in $g$, we use the rules for propagation of errors in multiplication and powers.

Error Propagation Rule:

1. When quantities are multiplied or divided, the relative errors add.
2. When a quantity is raised to a power, the relative error is multiplied by the absolute value of the exponent.

The relative error in $g$, $\frac{\Delta g}{g}$, is given by the following formula:

$\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2 \times \frac{\Delta T}{T}$

Given:

• The percentage error in length $\frac{\Delta L}{L} = 2\%$,
• The percentage error in time $\frac{\Delta T}{T} = 3\%$.

Substituting these values:

$\frac{\Delta g}{g} = 2\% + 2 \times 3\%$

$\frac{\Delta g}{g} = 2\% + 6\% = 8\%$

Thus, the relative error in $g$ is 8%.

Would you like more details, or do you have any further questions?

Here are some related questions you might consider:

1. What is the significance of error propagation in experimental physics?
2. How does the formula for the period of a pendulum relate to gravity?
3. How does increasing the length of the pendulum affect the percentage error in $g$?
4. Can you calculate the absolute error in $g$ if its value was given as $9.8 \, \text{m/s}^2$?
5. How does the square of the time period affect the overall calculation of the relative error?

Tip: Remember, when dealing with powers in error propagation, multiply the error by the power's exponent.