In the expression for time period $T$ of a simple pendulum $T = 2 \pi \sqrt{\frac{l}{g}}$,if the percentage error in time period $T$ and length $l$ are $2 \%$ and $2 \%$ respectively,then the percentage error in acceleration due to gravity $g$ is equal to ......... $\%$

Following observations were taken with a vernier callipers while measuring the length of a cylinder:
$3.29 \, cm, 3.28 \, cm, 3.29 \, cm, 3.31 \, cm, 3.28 \, cm, 3.27 \, cm, 3.29 \, cm, 3.30 \, cm$
Find the absolute error in the fourth and eighth observations.

$A$ wire has mass $(0.3 \pm 0.003) \text{ g}$, radius $(0.5 \pm 0.005) \text{ cm}$, and length $(6 \pm 0.06) \text{ cm}$. The maximum percentage error in the measurement of density is (in $\%$)

If the error in measuring the diameter of a circle is $4\%$,the error in the circumference of the circle would be (in $\%$)

In a simple pendulum experiment,the maximum percentage error in the measurement of length is $2\%$ and that in acceleration due to gravity $g$ is $4\%$. Then the maximum percentage error in determination of the time-period is

$A$ cuboid has volume $V = l \times l \times \sqrt{l} = l^{2.5}$, where $l$ is the length of one side. When the length $l$ is measured by a meter scale of least count $1 \text{ mm}$, the relative percentage error in the measurement of its volume is $6.25\%$. Then the relative percentage error in measurement of its length and the value of the length $l$ is: