Question

Three students $S_1,S_2$ and $S_3$ perform an experiment for determining the acceleration due to gravity $\left(g\right)$ using a simple pendulum. They use different lengths of pendulum and record time for different number of oscillations. The observations are as shown in the table.

$\begin{array}{ccccc}\text{Student No.}& \text{Length of}& \text{No. of}& \text{Total time for}& \text{Time}\\ & \text{pendulum (cm)}& \text{oscillations (n)}& \text{n oscillations}& \text{period (s)}\\ 1.& 64.0& 8& 128.0& 16.0\\ 2.& 64.0& 4& 64.0& 16.0\\ 3.& 20.0& 4& 36.0& 9.0\end{array}$
(Least count of length $=0.1\text{m}$
least count for time $=0.1\text{s}$
If $E_1,E_2$ and $E_3$ are the percentage errors in ${}^{\prime }{g}^{\prime }$ for students $1,2$ and $3$ respectively, then the minimum percentage error is obtained by student no.

• A. $2$
• B. Insufficient data.
• C. $1$
• D.
  $3$

Answer 1

The correct option is C $1$

We know that, time period

$T=2\pi \sqrt{\frac{l}{g}}\Rightarrow g=\frac{4{\pi }^{2}l}{T^2}$

$\Rightarrow \frac{\Delta g}{g}=\frac{\Delta l}{l}+\frac{2\Delta T}{T}$

$\Delta T=\frac{\text{least count of time}\left(\Delta T_0\right)}{\text{number of oscillations(n)}}$

$\Rightarrow \frac{\Delta g}{g}=\frac{\Delta l}{l}+\frac{2\Delta T_0}{nT}$

As $\Delta l$ and $\Delta T_0$ are same for all observations.

So, $\frac{\Delta g}{g}$ is minimum for highest value of $l,n$ and $T$.

$\Rightarrow$ Minimum percentage error in $g$ is for student number-$1$.

Correct answer: $1$