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:

The percentage error in the measurement of a physical quantity $m$ given by $m = \pi \tan \theta$ is minimum when $\theta = \dots \dots \dots \dots \dots ^\circ$ (Assume that the error in $\theta$ remains constant).

$A$ physical quantity $P$ is related to four observables $a, b, c$ and $d$ as follows: $P = \frac{a^3 b^2}{\sqrt{c} d}$. The percentage errors of measurement in $a, b, c$ and $d$ are $1 \%, 3 \%, 4 \%$ and $2 \%$ respectively. What is the percentage error in the quantity $P$? If the value of $P$ calculated using the above relation turns out to be $3.763$,to what value should you round off the result?

In an experiment of a simple pendulum,the time period measured was $50 \, s$ for $25$ vibrations when the length of the simple pendulum was taken as $100 \, cm$. If the least count of the stopwatch is $0.1 \, s$ and that of the meter scale is $0.1 \, cm$,then the maximum possible error in the value of $g$ is .......... $\%$

$A$ force $F$ is applied on a square plate of length $L$. If the percentage error in the determination of $L$ is $3 \%$ and in $F$ is $4 \%$, then the permissible error in the calculation of pressure is (in $\%$)

$A$ balance is made using a uniform metre scale of mass $100 \ g$ and two plates each of mass $200 \ g$ fixed at the two ends of the scale. The balance is pivoted at the $45 \ cm$ mark. If a $300 \ g$ weight is placed in the plate at $0 \ cm$ to weigh vegetables placed in the plate at $100 \ cm$, what is the error in the measurement (in $g$)?