Question

A students performs an experiment to determine the acceleration due to gravity (g) at a place using a simple pendulum. The length of the pendulum is 60 cm and the total time for 30 oscillations is 100s. What is maximum percentage error for the measurement g ? Given, least count for time $=0.1s$ and least count for length $=0.1cm$.

• A. $0.26\%$
• B. $0.3\%$
• C. $0.36\%$
• D. $3.6\%$

Answer 1

Answer: (C)

$0.36\%$

For a pendulum, the time period is: $T=2\pi \sqrt{\frac{l}{g}}$
$\Rightarrow g=\frac{4{\pi }^{2}l}{T^2}=\frac{4{\pi }^{2}l{n}^2}{t^2}$

Take $ln$ and then differentiate:
$\frac{\Delta g}{g}=\frac{\Delta l}{l}+2\frac{\Delta t}{t}$ (as n is constant so its derivative will be zero)

The % error in g $=\frac{\Delta g}{g}\times 100=\frac{\Delta l}{l}\times 100+2\frac{\Delta t}{t}\times 100=\frac{0.1}{60}\times 100+2\frac{0.1}{100}\times 100=0.36\%$