General Physics Ques 2
- 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 $E_1, E_{11}$ 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
(a) $E_1=0$
(b) $E_1$ is minimum
(c) $E_1=E_{\mathrm{II}}$
(d) $E_{\mathrm{II}}$ is maximum
Show Answer
Answer:
Correct Answer: 2.( b )
Solution:
$ \begin{aligned} & T=2 \pi \sqrt{\frac{l}{g}} \text { or } \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} \\ & % \text { error in } g=\frac{\Delta g}{g} \times 100=\left(\frac{\Delta l}{l}+\frac{2 \Delta t}{t}\right) \times 100 \\ E_1=\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 % \\ & E_{\mathrm{III}}=\left(\frac{0.1}{20}+\frac{2 \times 0.1}{36}\right) \times 100=1.055 % \\ & \end{aligned} $
Hence $E_1$ is minimum. $\therefore$ Correct option is (b).
BETA
