Question

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