Question

The period of oscillation of a simple pendulum is $T=2\pi \sqrt{\frac{L}{g}}$. Measured value of $L$ is $20.0cm$ known to $1mm$ accuracy and time for $100$ oscillations of the pendulum is found to be $90s$ using a wrist watch of $1s$ resolution. The accuracy in the determination of $g$ is

• A. 3 %
• B. 2 %
• C. 1 %
• D. 5 %

Answer 1

Answer: (D)

Here, $T=2\pi \sqrt{\frac{L}{g}}$
Squaring both sides, we get, $T^2=\frac{4{\pi }^{2}L}{g}$ or $g=\frac{4{\pi }^{2}L}{T^2}$
The relative error in $g$ is, $\frac{\Delta g}{g}=\frac{\Delta L}{L}+2\frac{\Delta T}{T}$
Here, $T=\frac{t}{n}$ and $\Delta T=\frac{\Delta t}{n}\therefore \frac{\Delta T}{T}=\frac{\Delta t}{t}$
The errors in both $L$ and $t$ are the least count errors.
$<br/>\therefore \frac{\Delta g}{g}=\frac{0.1}{10}+2\left(\frac{1}{50}\right)=0.01+0.04=0.05<br/>$
The percentage error in $g$ is
$<br/>\frac{\Delta g}{g}\times 100=\frac{\Delta L}{L}\times 100+2\left(\frac{\Delta T}{T}\right)\times 100=\left[\frac{\Delta L}{L}+2\left(\frac{\Delta T}{T}\right)\right]\times 100=0.05\times 100=<br/>$