What is the percentage error in the measurement of the time period of a pendulum if the maximum errors in the measurement of $l$ and $g$ are $2\%$ and $4\%$ respectively?

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 $11 \%$

Find the relative error in $Z$,if $Z = \frac{A^{4} B^{1/3}}{C D^{3/2}}$.

The maximum percentage error in the measurement of the density of a wire is: [Given: mass of wire $= (0.60 \pm 0.003) \ g$,radius of wire $= (0.50 \pm 0.01) \ cm$,length of wire $= (10.00 \pm 0.05) \ cm$]

The period of oscillation of a simple pendulum is $T = 2 \pi \sqrt{\frac{L}{g}}$. The measured value of $L$ is $1.0 \text{ m}$ using a meter scale with a minimum division of $1 \text{ mm}$,and the time for one complete oscillation is $1.95 \text{ s}$ measured using a stopwatch with a resolution of $0.01 \text{ s}$. The percentage error in the determination of $g$ will be ..... $\%$.

Explain the error of a sum or a difference.