In an experiment, on the measurement of g, using a simple pendulum, the time period was measured with an accuracy of 0.2% while the length was measured with an accuracy of 0.5%. The percentage accuracy in the value of g thus obtained is

The correct option is 0.9%.

Step-by-Step Explanation:

The relationship between the time period ($T$), the length of the pendulum ($L$), and the acceleration due to gravity ($g$) is given by the formula for a simple pendulum:

$T=2\pi \sqrt{\frac{L}{g}}$

To find the expression for $g$, we square both sides of the equation:

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

Rearranging the equation to solve for $g$:

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

In error analysis, the constant term ($4{\pi }^2$) has no error. Therefore, the relative maximum error in $g$ is the sum of the relative errors of the individual quantities, multiplied by their respective exponents:

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

To express this as percentage accuracy (or percentage error), we multiply both sides by 100:

$\left(\frac{\Delta g}{g}\times 100\right)=\left(\frac{\Delta L}{L}\times 100\right)+2\left(\frac{\Delta T}{T}\times 100\right)$

We are given the following values from the problem statement:
1. Percentage accuracy in length: $\frac{\Delta L}{L}\times 100=0.5\%$
2. Percentage accuracy in time period: $\frac{\Delta T}{T}\times 100=0.2\%$

Substituting these values into our error equation:

$\text{Percentage accuracy in}g=0.5\%+2(0.2\%)$

$\text{Percentage accuracy in}g=0.5\%+0.4\%=0.9\%$

Thus, the percentage accuracy in the value of $g$ obtained is 0.9%.