The period of oscillation of a simple pendulum is $T = 2 \pi \sqrt{\frac{L}{g}}$. The measured value of $L$ is $1.0 \text{ m}$ using a meter scale with a minimum division of $1 \text{ mm}$,and the time for one complete oscillation is $1.95 \text{ s}$ measured using a stopwatch with a resolution of $0.01 \text{ s}$. The percentage error in the determination of $g$ will be ..... $\%$.

  • A
    $1.13$
  • B
    $1.03$
  • C
    $1.33$
  • D
    $1.30$

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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 $= 0.1 \text{ s}$.
StudentLength $(cm)$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?

The resistance $R = \frac{V}{I}$ where $V = (100 \pm 5) \text{ V}$ and $I = (10 \pm 0.2) \text{ A}$. The percentage error in $R$ is: (in $\%$)

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