Question

An experiment is performed to obtain the value of acceleration due to gravity g by using a simple pendulum of length L. In this experiment, time for $100$ oscillations is measured by using a watch of least count 1 second and the value is $90.0$ seconds. The length L is measured by using a meter scale of least count 1 mm and the value is $20.0$ cm. The error in the determination of g would be :

• A. $1.7$%
• B. $2.7$%
• C. $4.4$%
• D. $2.27$%

Answer 1

The correct option is B $2.7$%

Here, $T=2\pi \sqrt{\frac{L}{g}}$ or $T^2=4{\pi }^2\left(L/g\right)$

So, $g=\frac{4{\pi }^{2}L}{T^2}$

Thus, $\frac{\Delta g}{g}=\frac{\Delta L}{L}+2\frac{\Delta T}{T}$

% error in g $=\frac{\Delta g}{g}\times 100$

$=(\frac{\Delta L}{L}+2\frac{\Delta T}{T})\times 100$

$=(\frac{\left(1/10\right)}{20}+2\times \frac{1}{90})\times 100=2.72\%$