If potential $V = 100 \pm 0.5\,V$ and current $I = 10 \pm 0.2\,A$ are given,what will be the value of resistance?

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?

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:

Measurement of two quantities along with the error is: $A = 2.5 \ ms^{-1} \pm 0.5 \ ms^{-1}$,$B = 0.10 \ s \pm 0.01 \ s$. The value of $AB$ will be:

$A$ physical quantity $X$ is given by $X = \frac{2k^3l^2}{m\sqrt{n}}$. The percentage errors in the measurements of $k, l, m$ and $n$ are $1\%, 2\%, 3\%$ and $4\%$ respectively. The value of $X$ is uncertain by .......... $\%$

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 ..... $\%$.