Students I, II and III perform an experiment for measuring the...

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}

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

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.

Options

A $E_{$ \mathrm{I} $}=0$
B $E_{$ \mathrm{I} $}$ is minimum
C $E_{$ \mathrm{I} $}=E_{$ \mathrm{II} $}$
D $E_{$ \mathrm{II} $}$ is maximum

Correct answer

B. $E_{$ \mathrm{I} $}$ is minimum

Step-by-step 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) \times100=0.3125 \% \\ & E_{\mathrm{II}}=\left(\frac{0.1}{64}+\frac{2 \times 0.1}{64}\right) \times100=0.46875 \% \quad E_{\mathrm{III}}=\left(\frac{0.1}{20}+\frac{2 \times 0.1}{36}\right) \times100=1.055 \% \end{aligned}

Hence,

E_{

\mathrm{I}

}

is minimum.

\therefore

correct option is (b).

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