The period of oscillation of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$,where $l$ is about $100 \, cm$ and is known to have $1 \, mm$ accuracy. The period is about $2 \, s$. The time of $100$ oscillations is measured by a stopwatch of least count $0.1 \, s$. The percentage error in $g$ is ......... $\%$

In an experiment to determine the acceleration due to gravity $g$,the formula used for the time period of a periodic motion is $T=2 \pi \sqrt{\frac{7(R-r)}{5 g}}$. The values of $R$ and $r$ are measured to be $(60 \pm 1) \text{ mm}$ and $(10 \pm 1) \text{ mm}$,respectively. In five successive measurements,the time period is found to be $0.52 \text{ s}, 0.56 \text{ s}, 0.57 \text{ s}, 0.54 \text{ s}$ and $0.59 \text{ s}$. The least count of the watch used for the measurement of time period is $0.01 \text{ s}$. Which of the following statement$(s)$ is(are) true?
$(A)$ The error in the measurement of $r$ is $10 \%$
$(B)$ The error in the measurement of $T$ is $3.57 \%$
$(C)$ The error in the measurement of $T$ is $2 \%$
$(D)$ The error in the determined value of $g$ is $11 \%$

Four persons measure the length of a rod as $20.00 \ cm$, $19.75 \ cm$, $17.01 \ cm$, and $18.25 \ cm$. The relative error in the measurement of the average length of the rod is:

The least count of a stopwatch is $0.2\, s$. The time for $20\, oscillations$ of a pendulum is measured to be $25\, s$. The percentage error in the measurement of time will be ........ $\%$

If two resistors have resistances $R_1 = (350 \pm 3) \text{ } \Omega$ and $R_2 = (140 \pm 4) \text{ } \Omega$, then the percentage error for the sum and difference of $R_1$ and $R_2$ are respectively:

The percentage error in the measurement of mass and velocity are $3 \%$ and $4 \%$ respectively. The percentage error in the measurement of kinetic energy is (in $\%$)