# The value of Kinetic energy $K$ and potential energy $V$ are measured as follows $K=100.0 \pm 2.0\; I \quad V=200.0 \pm 1.0\; I$. What is the percentage error in measurement of mechanical energy ?

$(a)\;2.5 \% \quad (b)\;1 \%\quad (c)\;0.5 \%\quad (d)\;1.5 \%$

$ME=K+V=E+\Delta E$
$ME=K+V=300 \pm 3$
$\%$ error in $ME=\large\frac{\Delta E}{E}$$\times 100 =\large\frac{3}{300}$$\times 100$
$=1 \%$
Hence b is the correct answer.

edited Mar 14, 2014