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 = 4π²(l/T) will be

The correct option is 7%.

Step-by-step Explanation:

The acceleration due to gravity $g$ using a simple pendulum is given by the standard formula:
$g=4{\pi }^2\frac{l}{T^2}$
where:
• $l$ is the length of the pendulum
• $T$ is the time period of oscillation

To find the maximum percentage error in the measurement of $g$, we take the relative error on both sides of the equation. Since constants like $4{\pi }^2$ have no error, we ignore them. For a quotient or product, the relative errors of the individual terms are added to estimate the maximum possible error, and power factors are multiplied by their respective fractional errors:
$\frac{\Delta g}{g}=\frac{\Delta l}{l}+2\frac{\Delta T}{T}$

Note that errors are always added up to find the maximum possible uncertainty (worst-case error analysis), so we use the absolute value of the error for $T$ even though the error in the time period is negative.

Multiplying by 100 to convert to percentage errors:
$\left(\frac{\Delta g}{g}\times 100\right)\%=\left(\frac{\Delta l}{l}\times 100\right)\%+2\left(\frac{\Delta T}{T}\times 100\right)\%$

Given in the problem:
• Percentage error in length $\left(\frac{\Delta l}{l}\times 100\right)\%=1\%$
• Percentage error in time period $\left(\frac{\Delta T}{T}\times 100\right)\%=3\%$ (absolute value)

Substituting these values into the error equation:
$\text{Percentage error in}g=1\%+2\times 3\%$
$\text{Percentage error in}g=1\%+6\%=7\%$

Therefore, the percentage error in the measurement of $g$ is 7%.