Question

The period of oscillation of a simple pendulum is $T=2\pi \sqrt{\frac{L}{g}}$. Measured value of $L$ is $10cm$ known to $1mm$ accuracy and time for $100$ oscillations of the pendulum is found to be $50s$ using a wristwatch of $1s$ resolution. What is the accuracy in the determination of $g$?

• A. 2 %
• B. 3 %
• C. 4 %
• D. 5 %

Answer 1

The correct option is D 5 %

Here, $T=2\pi \sqrt{\frac{L}{g}}$
Squaring both sides, we get $T^2=\frac{4{\pi }^{2}L}{g}org=\frac{4{\pi }^{2}L}{T^2}$
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.
$\therefore \frac{\Delta g}{g}=\frac{0.1}{10}+2\left(\frac{1}{50}\right)=0.01+0.04=0.05$
The percentage error in g is
$\frac{\Delta g}{g}\times 100=5$%