Hint - Q13

y = mx + c, where m = 2.2 ± 0.1 m^-3, x = 4.70 ± 0.05 g and c = 0.8 ± 0.2 g m^-3

SOLUTION PROCEDURE:

Define the following:
      m = 2.2 m^-3      m_max = 2.2 + 0.1 = 2.3 m^-3      m_min = 2.2 - 0.1 = 2.1 m^-3
      x = 4.70 g      x_max = 4.70 + 0.05 = 4.75 g      x_min = 4.70 - 0.05 = 4.65 g
      c = 0.8 g m^-3      c_max = 0.8 + 0.2 = 1.0 g m^-3      c_min = 0.8 - 0.2 = 0.6 g m^-3

Determine the value of y, using the equation y = mx + c

Find the maximum and minimum values of y using the following formulae:
      y_max = m_maxx_max + c_max            y_min = m_minx_min + c_min

Once values of y, y_max and y_min have been determined, calculate the difference between y and the limiting values, i.e. (y_max - y) and (y - y_min). The largest of these two differences should be used as the error estimate.

Question 13

The equation of a straight line is y = mx + c.

For a set of linear data, m has been determined as 2.2 ± 0.1 m^-3 and c has been determined as 0.8 ± 0.2 g m^-3.

When x = 4.70 ± 0.05 g, the value of y with its associated error (found from worst case scenarios) will be -

y = 11.1 ± 0.2 g m^-3

y = 11.1 ± 0.6 g m^-3

y = 11.1 ± 0.8 g m^-3

y = 11 ± 1 g m^-3