Question

The percentage errors in the measurement of length and time period of a simple pendulum are $1\%and2\%$ respectively. Then, the maximum error in the measurement of the acceleration due to gravity is

• A. $8\%$
• B. $3\%$
• C. $4\%$
• D. $6\%$
• E. $5\%$

Answer 1

The correct option is E

$5\%$

Step 1: Given Data:

The percentage errors in the measurement of length$=1\%$

The percentage errors in the measurement of the time period$=2\%$

Step 2: Formula Used:

$\begin{array}{l}Timeperiod\left(T\right)=2\pi \sqrt{\frac{l}{g}}\end{array}$, where $g$ is the acceleration due to gravity

Step 3: Calculating the maximum error in the measurement of the acceleration due to gravity:

The time period of a simple pendulum is given by-

$$\begin{gathered} \begin{array}{l}Timeperiod\left(T\right)=2\pi \sqrt{\frac{l}{g}}\end{array} \\ \begin{array}{l}\frac{\mathrm{\Delta }T}{T}=\frac{1}{2}(\frac{\mathrm{\Delta }l}{l}-\frac{\mathrm{\Delta }g}{g})\end{array} \\ \begin{array}{l}\frac{\mathrm{\Delta }g}{g}=\frac{\mathrm{\Delta }l}{l}-\frac{2\mathrm{\Delta }T}{T}\end{array} \end{gathered}$$

Maximum percentage error in the measurement of the acceleration due to gravity is-

$$\begin{gathered} \begin{array}{l}\frac{\mathrm{\Delta }g}{g}\times 100=\frac{\mathrm{\Delta }l}{l}\times 100+\frac{2\mathrm{\Delta }T}{T}\times 100\end{array} \\ =1\times 100+2\times 2\times 100 \\ =5\times 100=5\% \end{gathered}$$

Hence option E is the correct answer.