Question

In an experiment of simple pendulum, the errors in the measurement of length of the pendulum $(L)$ and time period $(T)$ are $3\%$ and $2\%$ respectively. The maximum percentage error in the value of $\frac{L}{T^2}$ is

• A. $5\%$
• B. $7\%$
• C. $8\%$
• D. $1\%$

Answer 1

Answer: (B)

The time period of simple pendulum is $T=2\pi \sqrt{\frac{L}{g}}$
Squaring both sides, we get
$T^2=4{\pi }^2\frac{L}{g}$ or $g=4{\pi }^2\frac{L}{T^2}\dots$ (i)
The maximum percentage error in $g$ is
$\frac{\Delta g}{g}\times 100=\frac{\Delta L}{L}\times 100+2\left(\frac{\Delta T}{T}\right)\times 100$
$=3\%+2\times 2\%=7\%$
From (i), we get $\frac{L}{T^2}=\frac{g}{4{\pi }^2}$
The maximum percentage error in
$\frac{L}{T^2}$ is $\frac{\Delta \left(\frac{L}{T^2}\right)}{\frac{L}{T^2}}\times 100=\frac{\Delta g}{g}\times 100=7\%$