Question

While measuring the acceleration due to gravity by a simple pendulum, a student makes a positive error of 1% in the length of the pendulum and a negative error of 3% in the value of time period. His per-centre error in the n=measurement of g by the relation $g={4\pi }^2\left(I/T^2\right)$ will be

• A. 2%
• B. 4%
• C. 7%
• D. 10%

Answer 1

Answer: (C)

7%

$\begin{array}{c}g=4{\pi }^2\left(\frac{l}{T^2}\right)\\ \text{Taking log}\end{array}$

$\begin{array}{c}\log g=\log \left(4x^2\right)+\log f\\ \text{Differentiating both}\\ \frac{dg}{g}=\frac{dl}{l}-\frac{2dT}{T}\end{array}$

$\begin{array}{c}\text{error is always added}\\ \begin{array}{cc}\text{100}\times \frac{dg}{g}=(\frac{dl}{l}+\frac{2dT}{T})\times 100& \\ \text{\%error in}g=& 1+2\times 3\%\\ & =7\%\end{array}\end{array}$