Question

The time period of oscillation of simple pendulum given by $T=2\pi \sqrt{\frac{L}{g}}$ where $L=(200\pm 0.1)cm$. The time period, T =4s and the time of 100 oscillations is measured using a stopwatch of least count 0.1 s. The percentage error in g is

• A. $0.1$%
• B. $1.5$%
• C. $2$%
• D. $4$%

Answer 1

The correct option is A $0.1$%

$\begin{array}{c}\text{Given:}\\ T=2\pi \sqrt{\frac{l}{g}}\end{array}$

$\begin{array}{c}L=(200\pm 0.1)cm\\ T=4s\\ \text{L.}C\text{. of stopwitch}=0.1s\end{array}$

$\begin{array}{c}\text{Taking log in eq (i)}\\ \log T=\log \left(2\pi \sqrt{\frac{1}{g}}\right)\end{array}$

$\begin{array}{cc}\log T& =\log 2\pi +\log \sqrt{t}\log \sqrt{g}\\ \log T& =\log 2\pi +\frac{1}{2}\log l-\frac{1}{2}\log g\end{array}$

$\begin{array}{c}\text{differentiating both side}\\ \begin{array}{cc}\delta T& =\frac{1}{2}\frac{\delta l}{l}-\frac{1}{2}\frac{\delta g}{g}\\ 2\frac{\delta T}{T}& =\frac{\delta l}{l}\frac{\delta g}{g}\end{array}\end{array}$

$T=\frac{t}{n}$

t= time period for 100 osullation

by stap wates.

T= Time perial of I oscillation

n=no.of oscillation

i.e. 100

$\begin{array}{c}\text{Taking log both srole}\\ \log T=\log \left(\frac{t}{n}\right)\\ \text{Differentiating both side}\\ \frac{\delta T}{T}=\frac{\delta t}{t}\end{array}$

$\begin{array}{c}\text{putting (ii) in (i)}\\ \frac{2\delta t}{t}=\frac{\delta l}{d}+\frac{\delta g}{g}\end{array}$

$\begin{array}{c}\frac{2\delta t}{nT}=\frac{\delta l}{l}+\frac{\delta g}{g}\\ \left(2\right)\frac{0.1}{4\times 100}=\frac{0.1}{200}+\frac{\delta g}{g}\end{array}$

$\text{\%. error}$=$0.1\%=\frac{\delta g}{g}\times 100\%\text{=Ans}$