Question

The period of oscillation of a simple pendulum is $T=2\pi \sqrt{\frac{L}{g}}$. Measured value of $Lis1.0m$ from meter-scale having a minimum division of $1mm$ and time of one complete oscillation is $1.95s$ measured from the stopwatch of $0.01s$ resolution. The percentage error in the determination of $‘g’$ will be:

• A. $1.33\%$
• B. $1.30\%$
• C. $1.13\%$
• D. $1.03\%$

Answer 1

The correct option is C

$1.13\%$

Step 1: Given Data:

Length $\left(L\right)=1m$

Minimum division $\left(∆L\right)=1mm=1\times {10}^{-3}m$

Time to complete one oscillation $\left(T\right)=1.95sec$

Stopwatch having resolution $=0.01s$

Step 2: Formula Used:

The period of oscillation of a simple pendulum is

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

Where $g=$the acceleration due to gravity

Following formula can be used to compute percentage error:

$Percentageerror=\frac{∆x}{x}\times 100$

Where $x=$actual amount, $∆x=$change in amount

Step 3: Calculating the percentage error in the determination of acceleration due to gravity

The given formula-

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

On squaring both sides,

$\begin{array}{rcl}{T}^2& =& 4{\pi }^2\frac{L}{g}\end{array}$

$\begin{array}{rcl}g& =& 4{\pi }^2\frac{L}{T^2}······\left(1\right)\end{array}$

On applying the formula of percentage error in equation $\left(1\right)$ we get,

$\begin{array}{rcl}\frac{∆g}{g}& =& \frac{∆L}{L}+\frac{2∆T}{T}\\ \frac{∆g}{g}\times 100& =& \frac{∆L}{L}\times 100+\frac{2∆T}{T}\times 100·····\left(2\right)\end{array}$

Thus we can say that percentage error in the value of acceleration due to gravity is equal to the sum of percentage error in measuring the length and twice of the percentage error in measuring the time period.

Substitute the known values in equation $\left(2\right)$,

$\begin{array}{rcl}\frac{∆g}{g}& =& \left(\frac{1\times {10}^{-3}}{1}+2\frac{0.01}{1.95}\right)\times 100\\ & =& 1.13\%\end{array}$

Hence, option C is the correct answer.