Question

What is the percentage error in the measurement of time period of a pendulum if maximum errors in measurement of $l$ and $g$ are $2$% and $4$% respectively

• A. $6$%
• B. $4$%
• C. $3$%
• D. $5$%

Answer 1

Answer: (C)

$3$%

For a simple pendulum time period is given by:

$T=2\pi \sqrt{\frac{l}{g}}$

Take log and differentiate,

$\frac{△T}{T}=\frac{1}{2}(\frac{△l}{l}-\frac{△g}{g})$

To maximize error consider $+$ sign

$\frac{△T}{T}=\frac{1}{2}(\frac{△l}{l}+\frac{△g}{g})$

For percentage error multiply by $100$ on both the sides,

$\frac{△T}{T}\times 100=\frac{1}{2}(\frac{△l}{l}+\frac{△g}{g})\times 100$

$\frac{△T}{T}\times 100=\frac{1}{2}(2\%+4\%)=3\%$

Therefore, the percentage error in the measurement of the time period of the pendulum is $3$