Q. Student 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 cm. Least count for time = 0.1 s. [table] Student | Length of the pendulum (cm) | Number of oscillations (n) | Total time for ($\mathbf{n}$) oscillations (s) | Time period (s) I | 64.0 | 8 | 128.0 | 16.0 II | 64.0 | 4 | 64.0 | 16.0 III | 20.0 | 4 | 36.0 | 9.0 [/table] If $E_{1},\ E_{II}\ and\ E_{III}$ are the percentage errors in g, $ie,\ \left( \frac{\mathrm{\Delta}g}{g} \times 100 \right),$ for students I, II and III respectively.
A) $E_{I} = 0$
B) $E_{I}$ is minimum
C) $E_{I} = E_{II}$
D) $E_{II}$ is maximum
Explanation
Time period $T = 2\pi\sqrt{\frac{l}{g}}$ Or $\ \ \ \ \frac{t}{n} = 2\pi\sqrt{\frac{l}{g}}$ $\therefore\ \ g = \frac{\left( 4\pi^{2} \right)\left( n^{2} \right)l}{t^{2}}$ $\% error\ in\ g = \frac{\mathrm{\Delta}g}{g} \times 100 = \left( \frac{\mathrm{\Delta}l}{l} + \frac{2\mathrm{\Delta}t}{t} \right) \times 100$ $\ \ E_{I} = \left( \frac{0.1}{64} + \frac{2 \times 0.1}{128} \right) \times 100 = 0.3125\%$ $\ \ E_{II} = \left( \frac{0.1}{64} + \frac{2 \times 0.1}{64} \right) \times 100 = 0.46875\%$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E_{III} = \left( \frac{0.1}{20} + \frac{2 \times 0.1}{36} \right) \times 100 = 1.055\%$ Hence, $E_{I}$ is minimum.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.