Question

The values of kinetic energy K and potential energy U are measured as follows:
$
  K = 100.0 \pm 2.0J \\
  U = 200.0 \pm 1.0J \\
 $
Then the percentage error in the measurement of mechanical energy is
\[
  A.\;\;\;\;\;2.5\% \\
  B.\;\;\;\;\;1\% \\
  C.\;\;\;\;\;0.5\% \\
  D.\;\;\;\;\;1.5\% \\
 \]

Answer 1

Hint Mechanical energy is the sum of kinetic and potential energy. First find the percentage error of kinetic and potential energy separately using the given formula and then do the sum to get the total error in measurement of mechanical energy.

Complete step-by-step solution
The measuring process is essentially a process of comparison. In spite of it the measured values of a quantity is little away from the actual value, or true value. This difference in the true value and measured value of a quantity is called error of measurement.
Percentage Error in sum of the quantities for $A = (a + \Delta a)$ and $B = (b + \Delta b)$
The percentage error is given by the formula
$\% error = \dfrac{{\Delta a}}{a} \times 100$
The total percentage error in \[A{\text{ }} + {\text{ }}B{\text{ }} = {\text{ }}A{\text{ }}\% {\text{ }} + {\text{ }}B{\text{ }}\% \]
Using the same format,
Here,
 $
  K = 100.0 \pm 2.0J \\
  U = 200.0 \pm 1.0J \\
 $
Percentage error of kinetic energy K
$
  K = \dfrac{2}{{100}} \times 100 \\
  K = 2\% \\
 $
Percentage error of potential energy U
$
  U = \dfrac{1}{{200}} \times 100 \\
  U = 0.5\% \\
 $
The Percentage error in mechanical energy M is given by
$
  M = K + U \\
  M = 2 + 0.5 \\
  M = 2.5\% \\
 $

Hence the total percentage error in the measurement of mechanical energy is 2.5% and the correct option is A

Note Percentage error can also be calculated using this method
$
  M = K + U \\
  \dfrac{{dM}}{M} = \dfrac{{dK}}{K} + \dfrac{{dU}}{U} \\
   = \left( {\dfrac{2}{{100}} + \dfrac{1}{{200}}} \right) \times 100 \\
   = 2.5\% \\
 $