Why is the error in a measurement expressed with both positive and negative signs?

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 ........ $\%$

The unit of percentage error is

The density of a cube is measured by measuring its mass and the length of its sides. If the maximum error in the measurement of mass and length are $3\%$ and $2\%$ respectively,then find the maximum error in the measurement of the density of the cube.

In an experiment of a simple pendulum,the errors in the measurement of the length of the pendulum $(L)$ and time period $(T)$ are $3 \%$ and $2 \%$,respectively. The maximum percentage error in the value of $\frac{L}{T^2}$ is (in $\%$)

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: