Question

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 number of oscillations. The observations are shown in the table.
Least count for length = $0.1\, cm$, Least count for time = $0.1\, s$

Student	Length of the pendulum (cm)	Number of oscillations (n)	Total time for(n) oscillations (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_1,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

• A. $E_I=0 $
• B. $E_I $ is minimum
• C. $E_I=E_{II} $
• D. $E_{II} $ is maximum

Answer 1

Answer: (B)

$T=2 \pi \sqrt {\frac {1}{g}} $
or $\frac {t}{n}= 2 \pi \sqrt {\frac {1}{g}} $
$\therefore g=\frac {(4 \pi^2)(n^2)l}{t^2} $
$\%$ error in $g=\frac {\Delta g}{g} \times 100=(\frac {\Delta l}{l}+ \frac {2 \Delta t}{t}) \times 100 $
$E_I=( \frac {0.1}{64}+ \frac {2 \times 0.1}{128}) \times 100=0.3125 \%$
$E_{II}=( \frac {0.1}{64}+ \frac {2 \times 0.1}{64}) \times 100=0.46875 \%$
$E_{III}=( \frac {0.1}{20}+ \frac {2 \times 0.1}{36}) \times 100=1.055 \%$
Hence $E_I $ is minimum.