A student performs an experiment for determination of $g \left(=\frac{4 \pi^{2} l }{ T ^{2}}\right), \ell =1 m$ and he commits an error of $\Delta \ell$. For $T$ he takes the time of $n$ oscillations with the stop watch of least count $\Delta T$ and he commits a human error of $0.1 s$ For which of the following data, the measurement of $g$ will be most accurate?

The current voltage relation of diode is given by $I=(e^{1000V/T} -1)\;mA$, where the applied voltage $V$ is in volts and the temperature $T$ is in degree Kelvin. If a student makes an error measuring $ \mp 0.01\;V$ while measuring the current of $5\; mA$ at $300\; K$, what will be the error in the value of current in $mA$ ?

The percentage errors in the measurement of mass and speed are $2\%$ and $3\%$ respectively. How much will be the maximum error in the estimation of the kinetic energy obtained by measuring mass and speed  ……… $\%$

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

In the expression for time period $T$ of simple pendulum $T=2 \pi \sqrt{\frac{l}{g}}$, if the percentage error in time period $T$ and length $l$ are $2 \%$ and $2 \%$ respectively then percentage error in acceleration due to gravity $g$ is equal to ……… $\%$