Question

The length of a simple pendulum is about $100cm$, known to have an accuracy of $1mm$ . Its period of oscillation is $2s$ , determined by measuring the time for $100$ oscillations using a clock of $0.1s$ resolution. What is the accuracy of the determined value of $g$ ?

• A. $0.2\mathrm{\%}$
• B. $0.5\mathrm{\%}$
• C. $0.1\mathrm{\%}$
• D. $2\%$

Answer 1

The correct option is A

$0.2\mathrm{\%}$

Step 1. Given Data,
$l$ is the length of the pendulum,$l=$$100cm$

Change in length of the pendulum is $\mathrm{\Delta }l$,

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

$T$ is the time period of oscillation of the pendulum,$T=2s$

Step 2. Formula used,
$T=2\pi \sqrt{\frac{l}{g}}$
$T$ is the time period of oscillation of the pendulum,
$l$ is the length of the pendulum,
$g$ is the acceleration due to gravity.
$\Rightarrow {\displaystyle \frac{\mathrm{\Delta }g}{g}}={\displaystyle \frac{\mathrm{\Delta }l}{l}}+2{\displaystyle \frac{\mathrm{\Delta }T}{T}}$

Change in acceleration due to gravity is $\mathrm{\Delta }g$,

Change in length of the pendulum is $\mathrm{\Delta }l$,

Change in time period is $\mathrm{\Delta }T$
$\mathrm{\Delta }T={\displaystyle \frac{\text{Resolution of the clock}}{\text{Number of oscillations}}}$ .
Step 3. Calculating the accuracy of g,
Using the formula of the time period of oscillation, we get
$\Rightarrow T=2\pi \sqrt{{\displaystyle \frac{l}{g}}}$
The equation in terms of $g$,
$\therefore g={\left({\displaystyle \frac{2\pi }{T}}\right)}^2\times l$
Calculating the accuracy of $g$
$\Rightarrow {\displaystyle \frac{\mathrm{\Delta }g}{g}}\times 100\mathrm{\%}$
So, for $\mathrm{\%}$ accuracy, smallest change is needed in the variables $T$ and $l$ .
$\therefore {\displaystyle \frac{\mathrm{\Delta }g}{g}}={\displaystyle \frac{\mathrm{\Delta }l}{l}}+2{\displaystyle \frac{\mathrm{\Delta }T}{T}}$
The term $T$ is squared in the formula of $g$ so we have multiplied $2$ by ${\displaystyle \frac{\mathrm{\Delta }T}{T}}$
Accuracy percentage,

$\therefore {\displaystyle \frac{\mathrm{\Delta }g}{g}}\times 100\mathrm{\%}={\displaystyle \frac{\mathrm{\Delta }l}{l}}\times 100\mathrm{\%}+2{\displaystyle \frac{\mathrm{\Delta }T}{T}}\times 100\mathrm{\%}$
We know,
$\Rightarrow \mathrm{\Delta }l=1mm=0.1cm$
$\Rightarrow l=100cm$
$\mathrm{\Delta }T={\displaystyle \frac{\text{Resolution of the clock}}{\text{Number of oscillations}}}={\displaystyle \frac{0.1}{100}}=0.001s$
After substituting all values in the formula of g Accuracy percentage
$\Rightarrow {\displaystyle \frac{\mathrm{\Delta }g}{g}}\times 100\mathrm{\%}={\displaystyle \frac{0.1}{100}}\times 100\mathrm{\%}+2{\displaystyle \frac{0.001}{2}}\times 100\mathrm{\%}$
$\Rightarrow {\displaystyle \frac{\mathrm{\Delta }g}{g}}\times 100\mathrm{\%}=0.1\mathrm{\%}+0.1\mathrm{\%}=0.2\mathrm{\%}$
Thus the accuracy of $g$ is $0.2\mathrm{\%}$ .

Hence, the correct option is (A).