Question

If the force acting on a body of mass $(2kg\pm 20g)$ is $(3N\pm 2\%)$, then maximum percentage error in acceleration is?

• A. $6\%$
• B. $12\%$
• C. $2\%$
• D. $3\%$

Answer 1

The correct option is C $2\%$

$\begin{array}{cc}\text{Given}-& \text{Mass}=2kg\pm 20g=2kg\pm 0.02kg\\ & \text{Force}=3N\pm 2\%=3N\pm 0.06N\end{array}$

$\begin{array}{c}\text{We know}\\ F=M\cdot a\\ \Rightarrow a=\frac{F}{M}\end{array}$

$\begin{array}{c}\text{By error analysis method -}\\ \text{Taking log Both sides}\\ \log a=\log F-\log M\\ \text{Differentiate both sides -}\\ \frac{da}{a}=\frac{dF}{F}-\frac{dM}{M}\\ \text{For maximum possible error -}\\ \begin{array}{cc}\frac{da}{a}& =\frac{dF}{F}+\frac{dM}{M}\\ \Rightarrow & \frac{da}{a}=\frac{0.02}{20}+\frac{0.06}{3}=0.001+0.02\\ & =0.021=2.1\%\approx 2\%\end{array}\end{array}$