If the length and time period of an oscillating pendulum have errors of 1% and 2% respectively, what is the error in the estimate of g ?

The correct option is 5%.

To find the percentage error in the estimation of the acceleration due to gravity ($g$), we start with the formula for the time period ($T$) of a simple pendulum:
$T=2\pi \sqrt{\frac{l}{g}}$
where:
$T$ is the time period,
$l$ is the length of the pendulum, and
$g$ is the acceleration due to gravity.

Squaring both sides of the equation to isolate $g$ gives:
$T^2=4{\pi }^2\frac{l}{g}$

Rearranging the formula to solve for $g$:
$g=4{\pi }^2\frac{l}{T^2}$

To calculate the relative error, we take the natural logarithm on both sides:
$\ln \left(g\right)=\ln \left(4{\pi }^2\right)+\ln \left(l\right)-2\ln \left(T\right)$

Differentiating both sides to find the fractional errors (and considering the maximum possible error, where errors add up):
$\frac{\Delta g}{g}=\frac{\Delta l}{l}+2\frac{\Delta T}{T}$

Multiplying by 100 on both sides to express the relationship in terms of percentage errors:
$\frac{\Delta g}{g}\times 100=(\frac{\Delta l}{l}\times 100)+2(\frac{\Delta T}{T}\times 100)$

Given in the problem:
Percentage error in length, $\frac{\Delta l}{l}\times 100=1\%$
Percentage error in time period, $\frac{\Delta T}{T}\times 100=2\%$

Substituting these values into our error equation:
$\text{Percentage error in}g=1\%+2(2\%)$
$\text{Percentage error in}g=1\%+4\%=5\%$

Therefore, the estimated error in the measurement of $g$ is 5%.