Question

If the length and time period of an oscillating pendulum have errors of $1\%$ and $3\%$ respectively, then the error in measurement of acceleration due to gravity is

• A. $4\%$
• B. $5\%$
• C. $6\%$
• D. $7\%$

Answer 1

Answer: (D)

$7\%$

Answer 2

As $g=\frac{4\pi^{2} l}{T^{2}}$ $\therefore \frac{\Delta g}{g}\times100=\frac{\Delta l}{l}\times100+2\frac{\Delta T}{T}\times100$ $=1\% + 2\times 3\% = 7\%$