"It is advisable to use an instrument with the smallest possible least count." Explain this statement.

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 time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{\ell}{g}}$. The measured value of the length of the pendulum is $10 \ cm$ known to a $1 \ mm$ accuracy. The time for $200$ oscillations of the pendulum is found to be $100 \ s$ using a clock of $1 \ s$ resolution. The percentage accuracy in the determination of $g$ using this pendulum is $x$. The value of $x$ to the nearest integer is ...........$\%$

Three students $S_{1}, S_{2}$ and $S_{3}$ perform an experiment for determining the acceleration due to gravity $(g)$ using a simple pendulum. They use different lengths of pendulum and record time for different number of oscillations. The observations are as shown in the table.

Student No.	Length of pendulum $(cm)$	No. of oscillations $(n)$	Total time for oscillations $(s)$	Time period $(s)$
$1.$	$64.0$	$8$	$128.0$	$16.0$
$2.$	$64.0$	$4$	$64.0$	$16.0$
$3.$	$20.0$	$4$	$36.0$	$9.0$

(Least count of length $= 0.1 \, cm$,least count for time $= 0.1 \, s$)
If $E_{1}, E_{2}$ and $E_{3}$ are the percentage errors in $g$ for students $1, 2$ and $3$ respectively,then the minimum percentage error is obtained by student no. ....... .

In a simple pendulum experiment,the maximum percentage error in the measurement of length is $2\%$ and that in acceleration due to gravity $g$ is $4\%$. Then the maximum percentage error in determination of the time-period is

The voltage across a lamp is $(6.0 \pm 0.3) \text{ V}$ and the current passing through it is $(4.0 \pm 0.1) \text{ A}$. The power consumed in watt will be (using percentage error):