Question

In a simple pendulum experiment for determination of acceleration due to gravity $\left(g\right)$, time taken for $20$ oscillations is measured by using a watch of $1$ second least count. The mean value of time taken comes out to be $30\text{sec}$. The length of pendulum is measured by using a meter scale of least count $1\text{mm}$ and the value obtained is $55.0\text{cm}$. The percentage error in the determination of $g$ is close to,

• A. $0.2\%$
• B. $3.5\%$
• C. $6.8\%$
• D. $0.7\%$

Answer 1

The correct option is C $6.8\%$

Here, $T=\frac{30}{20}=1.5\text{sec.}$

$\Delta T=\frac{1}{20}\text{sec.}$

$l=55\text{cm},\Delta l=1\text{mm}=0.1\text{cm}$

The time period of a simple pendulum is,

$T=2\pi \sqrt{\frac{l}{g}}$

$\Rightarrow g=4{\pi }^2\frac{l}{T^2}$

The percentage error in the value of $g$ is,

$\frac{\Delta g}{g}\times 100=\frac{\Delta l}{l}\times 100+2\frac{\Delta T}{T}\times 100$

$=\frac{0.1}{55}\times 100+2\left(\frac{1}{30}\right)\times 100\approx 6.8\%$

Hence, $\left(D\right)$ is the correct answer.