Question

What is the approximate percentage error in the measurement of time period of a simple pendulum if maximum errors in the measurement of length $l$ and gravitational acceleration $g$ are $3$% and $7$% respectively?

• A. $5$%
• B. $10$%
• C. $2$%
• D. $3$%

Answer 1

The correct option is A $5$%

Time-period of a simple pendulum,
$T=2\pi \sqrt{\frac{l}{g}}\Rightarrow \frac{△T}{T}=\frac{△l}{2l}+\frac{△g}{2g}$
So, % change in time-period $=(\frac{3}{2}+\frac{7}{2})$% $=\frac{10}{2}$% $=5$%