Question

In an experiment to find acceleration due to gravity $\left(g\right)$ using simple pendulum, time period of $0.5\text{s}$ is measured from time of $100$ oscillations with a watchof $1\text{s}$ resolution. If measured value of length is $10\text{cm}$ known to $1\text{mm}$ accuracy, the accuracy in the determination of $g$ is found to be $x$%. The value of $x$ is _________

• A. $5$
• B. 5.0
• C. 5.00

Answer 1

Answer: (A), (B), (C)

Time period is given as
$T=2\pi \sqrt{\frac{l}{g}}$

$\frac{dg}{g}\times 100=\frac{2dT}{T}\times 100+\frac{dl}{l}\times 100$

$=2\times \frac{1}{50}\times 100+\frac{1}{100}\times 100=5$ %