Question

In a simple pendulum experiment for the 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 $30s$. The length of the pendulum is measured by using a meter scale of least count $1mm$ and the value obtained is $55.0cm$. The percentage error in the determination of $g$ is close to

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

Answer 1

The correct option is D

$6.8\%$

Step 1: Given

$△L=1mm$, $L=55cm=550mm$

$△T=1s,T=30s$

Step 2: Formula used

We know that the time of simple pendulum,

$T=2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$

$\therefore g=\frac{4{\mathrm{\pi }}^2\mathrm{L}}{T^2}$

Step 3: Calculate the percentage error

The percentage error of this equation can be given as,

$\left(\frac{△g}{g}\times 100\right)=\left(\frac{△L}{L}\times 100\right)+\left(2\frac{△T}{T}\times 100\right)$

Upon substituting the values we get,

$\left(\frac{△g}{g}\times 100\right)=\left(\frac{1}{550}\times 100\right)+\left(2\frac{1}{30}\times 100\right)$

$=\frac{10}{55}+\frac{20}{3}$

$=0.18+6.6$

$\approx 0.2+6.6$

$=6.8\%$

Hence, the correct answer is option (D).