Question

Students I, II and III perform an experiment for measuring the acceleration due to gravity (g) using a simple pendulum.They use different lenghts of the pendulum and record time for different number of oscillations.The observations are shown in the table.Least count for length = 0.1cm, Least count for time = 0.1s

$\begin{array}{ccccc}\text{Student}& \text{Length of the pendulum(cm)}& \text{Number of oscillations(n)}& \text{Total time for (n) oscillation(s)}& \text{Time period(s)}\\ \text{I}& 64.0& 8& 128.0& 16.0\\ \text{II}& 64.0& 4& 64.0& 16.0\\ \text{III}& 20.0& 4& 36.0& 9.0\end{array}$

If $E_I$, $E_{II}$ and $E_{III}$ are the percentage errors in g, i.e. $(\frac{△g}{g}\times 100)$ for student I, II and III, respectively, t = $2\pi \sqrt{\frac{l}{g}}$ , then

• A. $E_l$ = 0
• B. $E_l$ is minimum
• C. $E_l$ and $E_{ll}$ = 0
• D. $E_{ll}$ is maximum

Answer 1

The correct option is B

$E_l$ is minimum

Given $△l$ = 0.1cm and $△t$ = 0.1s.

$T=2\pi \sqrt{\frac{L}{g}}$ $\Rightarrow g=4{\pi }^2\frac{1}{T^2}$ $\Rightarrow \frac{△g}{g}=\frac{△l}{l}+\frac{2△T}{T}$

(As relative error gets added when quantities are multiplied or divided)

Comparing $E_1$ and $E_2$

$△l$ and l will be the same but,

$△T_1<△T_2$ as no. of observations of student 1 is higher.
So

$△E_1<△E_2$

Comparing $E_2$ and $E_3$

$L_2>l_3$, $T_2>T_3$, $△l_2=△l_3$, as least count is same , also $△T_2=△T_3$ as number of observations is same.

$\therefore E_2<E_3$

$\therefore E_1<E_2<E_3$

$\therefore E_1$ is minimum.