Students I, II and III perform an experiment for measuring the acceleration due to gravity $(g)$ using a…

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 $=0.1 \mathrm{~cm}$ Least count for time $=0.1 \mathrm{~s}$
If $E_{\mathrm{I}}, E_{\mathrm{II}}$ and $E_{\mathrm{III}}$ are the percentage errors in $g$, i.e., $\left(\frac{\Delta g}{g} \times 100\right)$ for students I, II and III, respectively.
  1. $E_{\mathrm{I}}=0$
  2. $E_{\mathrm{I}}$ is minimum
  3. $E_{\mathrm{I}}=E_{\mathrm{II}}$
  4. $E_{\mathrm{II}}$ is maximum

Solution

$ \begin{aligned} & T & =2 \pi \sqrt{\frac{l}{g}} \quad \text { or } \quad \frac{t}{n}=2 \pi \sqrt{\frac{l}{g}} \\ \therefore \quad & g & =\frac{\left(4 \pi^2\right)\left(n^2\right) l}{t^2} \end{aligned} $ $\%$ error in $g=\frac{\Delta g}{g} \times 100=\left(\frac{\Delta l}{l}+\frac{2 \cdot \Delta t}{t}\right) \times 100$ $ \begin{aligned} & E_{\mathrm{I}}=\left(\frac{0.1}{64}+\frac{2 \times 0.1}{128}\right) \times 100=0.3125 \% \\ & E_{\mathrm{II}}=\left(\frac{0.1}{64}+\frac{2 \times 0.1}{64}\right) \times 100=0.46875 \% \quad E_{\mathrm{III}}=\left(\frac{0.1}{20}+\frac{2 \times 0.1}{36}\right) \times 100=1.055 \% \end{aligned} $ Hence, $E_{\mathrm{I}}$ is minimum. $\therefore$ correct option is (b)

Asked in: JEE Advanced 2008 (Paper 1)

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