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 following table. Least count for length = $0.1cm$, Least count for time = $0.1s$.

Student	Length of 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 $E_I,E_{II},E_{III}$ are the percentage errors in g, i.e., $(\frac{△g}{g}\times 100)$ for students I, II and III, respectively, then

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

Answer 1

The correct option is B $E_I$ is minimum

The period of oscillation $\left(T\right)$ of a simple pendulum of length $l$ is given by $T=2\pi \sqrt{\frac{l}{g}}$

Therefore $g=\frac{{4\pi }^{2}L}{T^2}$ so the fractional error is $g$ is given by

$\frac{\Delta g}{g}=\frac{\Delta L}{L}+2\frac{\Delta T}{T}$

[The above expression is obtained by taking logarithm of both sides and then differentiating]

Here $\Delta L=0.1$ and $\Delta T=0.1s$

The percentage error is $100$ times the fractional error

$E_1=\frac{\Delta g}{g}=[\left(\frac{0.1}{64}\right)+2\left(\frac{0.1}{128}\right)]\times 100$

$=\frac{5}{16}$%

$E_2=\frac{\Delta g}{g}=[\left(\frac{0.1}{64}\right)+2\left(\frac{0.1}{64}\right)]\times 100=\frac{15}{32}$%

$E_3=\frac{\Delta g}{g}=[\left(\frac{0.1}{20}\right)+2\left(\frac{0.1}{36}\right)]\times 100=\frac{19}{18}$%

$E_1$ is minimum correct option is B.