Question

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

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

Answer 1

Answer: (D)

$5\%$

Answer 2

Here, $T=2\pi\sqrt{\frac{l}{g}}$ $\therefore$ Relative error in $g$ is $\frac{\Delta g}{g}=\frac{\Delta l}{l}+\frac{2\Delta T}{T}$. 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 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 $\frac{\Delta g}{g}\times 100=\frac{\Delta l}{l}\times 100+2\left(\frac{\Delta T}{T}\right)\times 100$ $=\left[\frac{\Delta l}{l}+2\left(\frac{\Delta T}{T}\right)\right]\times 100$ $=0.05\times 100=5\%$