The period of oscillation of a simple pendulum is $T = 2\pi \sqrt{\frac{l}{g}}$. The measured value of $l$ is $20.0 \text{ cm}$ known to $1 \text{ mm}$ accuracy,and the time for $100$ oscillations of the pendulum is found to be $90 \text{ s}$ using a wristwatch of $1 \text{ s}$ resolution. The accuracy in the determination of $g$ is ........ $\%$

Accuracy of measurement is determined by

In the measurement of a physical quantity $X = \frac{A^2 B}{C^{1/3} D^3}$, the percentage errors introduced in the measurements of the quantities $A, B, C$, and $D$ are $2 \%, 2 \%, 4 \%$, and $5 \%$, respectively. Then, the minimum amount of percentage error in the measurement of $X$ is contributed by:

An experiment is performed to obtain the value of acceleration due to gravity $g$ by using a simple pendulum of length $L$. In this experiment,the time for $100$ oscillations is measured by using a watch of $1$ second least count,and the value is $90.0$ seconds. The length $L$ is measured by using a meter scale of least count $1$ mm,and the value is $20.0$ cm. The error in the determination of $g$ would be ........... $\%$.

The resistance $R = V / I$ where $V = (100 \pm 5) \; V$ and $I = (10 \pm 0.2) \; A$. Find the percentage error in $R$. (in $\%$)

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: