The acceleration due to gravity is measured on the surface of the Earth by using a simple pendulum. If $\alpha$ and $\beta$ are relative errors in the measurement of length and time period respectively,then the percentage error in the measurement of acceleration due to gravity is:

The least count of a stopwatch is $\frac{1}{5} \ s$. 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 the error in measuring the diameter of a circle is $4\%$,the error in the radius of the circle would be (in $\%$)

Students $I$,$II$,and $III$ perform an experiment for measuring the acceleration due to gravity $(g)$ using a simple pendulum. They use different lengths of the pendulum and/or record time for different numbers of oscillations. The observations are shown in the table. Least count for length $= 0.1 \text{ cm}$. Least count for time $= 0.1 \text{ s}$.

Student	Length $(cm)$	Oscillations $(n)$	Total Time $(s)$	Time Period $(s)$
$I$	$64.0$	$8$	$128.0$	$16.0$
$II$	$64.0$	$4$	$64.0$	$16.0$
$III$	$20.0$	$4$	$36.0$	$9.0$

If $E_{I}$,$E_{II}$,and $E_{III}$ are the percentage errors in $g$,i.e.,$(\frac{\Delta g}{g} \times 100)$ for students $I$,$II$,and $III$ respectively,which of the following is correct?

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

If there is a positive error of $50\%$ in the measurement of velocity of a body,then the error in the measurement of kinetic energy is .............. $\%$