Question

In an experiment to determine acceleration due to gravity by simple pendulum, a student commits 1 % positive error in the measurement of length and 3% negative error in the measurement of time period. The percentage error in the value of g will be :

• A. 7%
• B. 10%
• C. 4%
• D. 3%

Answer 1

The correct option is A 7%

We know that the acceleration due to gravity $g=4{\pi }^2\left(\frac{l}{t^2}\right)$.
Here, this student commits 1 % +ve error in the measurement of length and 3% +ve in the measurement of time period.
$\therefore \frac{\Delta g}{g}\times 100=(\frac{\Delta l}{l}\times 100+2\frac{\Delta t}{t}\times 100)$
Thus, the percentage error in the value of $g=\pm (1+2\times 3)100$ =7%.