The mean time period of a second's pendulum is $2.00 \, s$ and the mean absolute error in the time period is $0.05 \, s$. To express the maximum estimate of error,the time period should be written as:

The measured value of the length of a simple pendulum is $20 \ cm$ with $2 \ mm$ accuracy. The time for $50$ oscillations was measured to be $40 \ s$ with $1 \ s$ resolution. From these measurements,the accuracy in the measurement of acceleration due to gravity is $N \%$. The value of $N$ is:

$A$ physical quantity is given by $X = M^a L^b T^c$. The percentage errors in the measurement of $M, L,$ and $T$ are $\alpha, \beta,$ and $\gamma$ respectively. The maximum percentage error in the quantity $X$ is:

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?

Two resistors of resistances $R_{1} = 100 \pm 3 \ \Omega$ and $R_{2} = 200 \pm 4 \ \Omega$ are connected $(a)$ in series,$(b)$ in parallel. Find the equivalent resistance of the $(a)$ series combination,$(b)$ parallel combination. Use for $(a)$ the relation $R = R_{1} + R_{2}$ and for $(b)$ $\frac{1}{R^{\prime}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$ and $\frac{\Delta R^{\prime}}{R^{\prime 2}} = \frac{\Delta R_{1}}{R_{1}^{2}} + \frac{\Delta R_{2}}{R_{2}^{2}}$.

The errors in the measurement which arise due to unpredictable fluctuations in temperature and voltage supply are: