Question

The time period of a simple pendulum is given by $T=2\pi \sqrt{\frac{l}{g}}$ The measured value of the length of the pendulum is $10cm$ known to a $1mm$ accuracy. The time for $200$ oscillations of the pendulum is found to be $100$ seconds using a clock of $1s$ resolution. The percentage accuracy in the determination of $‘g’$ using this pendulum is $‘x’$. The value of $‘x’$ to the nearest integer is.

• A. $5\%$
• B. $4\%$
• C. $3\%$
• D. $2\%$

Answer 1

The correct option is C

$3\%$

Step 1: Given Data:

The measured value of the length of the pendulum $\left(l\right)=10cm=10\times {10}^{-2}m$

Accuracy in length$\left(∆l\right)=1mm={10}^{-3}m$

The time for $200$ oscillations of the pendulum $\left(t\right)=100$ seconds

Time period $T=\frac{Timetakenbypendulum}{Noofoscillation}=\frac{100}{200}=0.5$

Resolution of clock$\left(∆t\right)=1s$

Accuracy in the time period $∆T=\frac{1}{200}=5\times {10}^{-3}$

The percentage accuracy in the determination of $‘g’$is $‘x’$

Step 2: Formula Used:

The given formula

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

Where $\left(l\right)=$length, $\left(T\right)=$time, and $\left(g\right)=$gravity

Step 3: Calculating “g”

The given formula

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

On squaring both the sides

$T^2=4{\pi }^2\left(\frac{l}{g}\right)$

Rearranging the formula to calculate the value of $g$

$\begin{array}{l}g=4{\pi }^2\frac{l}{T^2}\end{array}\dots \dots \dots \dots \dots \dots \dots \left(1\right)$

Step 3: Calculating the percentage accuracy

The maximum accuracy in the determination of $g$, from equation (1)

$\begin{array}{l}g=4{\pi }^2\frac{l}{T^2}\end{array}$

$$\begin{gathered} \begin{array}{l}\frac{\mathrm{\Delta }g}{g}=\frac{\mathrm{\Delta }l}{l}+\frac{2\mathrm{\Delta }T}{T}\end{array} \\ \begin{array}{c}\frac{\mathrm{\Delta }g}{g}=\frac{{10}^{-3}}{10\times {10}^{-2}}+\frac{2\times \left(5\times {10}^{-3}\right)}{0.5}\end{array} \\ \begin{array}{c}\frac{\mathrm{\Delta }g}{g}=0.01+0.02\\ \frac{\mathrm{\Delta }g}{g}=0.03\end{array} \end{gathered}$$

The percentage accuracy

$\begin{array}{rcl}\frac{∆g}{g}\times 100& =& 0.03\times 100\\ x& =& 3\%\end{array}$

Hence option C is the correct answer.