$A$ simple pendulum is being used to determine the value of gravitational acceleration $g$ at a certain place. The length of the pendulum is $25.0 \; cm$ and a stopwatch with $1 \; s$ resolution measures the time taken for $40$ oscillations to be $50 \; s$. The accuracy in $g$ is ....... $\%$ (in $.40$)

The potential difference across the ends of a conductor is $(50 \pm 3) \text{ V}$ and the current through it is $(5 \pm 0.1) \text{ A}$. The percentage error in the measurement of resistance of the conductor is (in $\%$)

What is the difference between a mistake and an error in the context of physical measurements?

The period of oscillation of a simple pendulum is given by $T = 2\pi\sqrt{\frac{l}{g}}$ where $l$ is about $80\text{ cm}$ and is known to have $0.1\text{ cm}$ accuracy. The period is about $1.5\text{ s}$. The time of $50$ oscillations is measured by a stop watch of least count $0.1\text{ s}$. The percentage error in $g$ is nearly

The time period of a simple harmonic oscillator is $T = 2 \pi \sqrt{\frac{m}{k}}$. The measured value of mass $(m)$ of the object is $10 \ g$ with an accuracy of $10 \ mg$, and the time for $50$ oscillations of the spring is found to be $60 \ s$ using a watch of $2 \ s$ resolution. The percentage error in the determination of the spring constant $(k)$ is . . . . . . %.

In Ohm's experiment,the values of an unknown resistance were found to be $4.12 \; \Omega, 4.08 \; \Omega, 4.22 \; \Omega$,and $4.14 \; \Omega$. Calculate the mean absolute error and relative error in these measurements.