Question

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

• A. $2\%$
• B. $3\%$
• C. $4\%$
• D. $5\%$

Answer 1

Answer: (D)

Here, $T=2\pi \sqrt{\frac{L}{g}}$
Squaring both sides, we get $T^2=4{\pi }^2\frac{L}{g}$ or $g=4{\pi }^2\frac{L}{T^2}\dots (i)$
Take log and differentiate both sides of equation (i), we get
$\frac{\Delta g}{g}=\frac{\Delta L}{L}-2\left(\frac{\Delta T}{T}\right)$
For maximum relative error, the individual errors should be added.
$\therefore \frac{\Delta g}{g}=\frac{\Delta L}{L}+2\left(\frac{\Delta T}{T}\right)$
Here, $T=\frac{t}{n},\Delta T=\frac{\Delta t}{n}\therefore \frac{\Delta T}{T}=\frac{\Delta t}{t}$
As 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 $\frac{\Delta g}{g}\times 100=0.05\times 100=5\%$