Answer to Question #152734 in Classical Mechanics for GuardianVex

Question

In an experiment with a common balance the mass of a ring found to be 2.52g, 2.5g, 2,51g, 2.49g and 2.54g in successive measurements. Calculate a) The mean value of the mass of the ring b) The absolute error in each measurement c) Mean absolute error d) Relative error e) Percentage error 

Answer 1

Answer

a) mean mass is given

$m_a=\frac{2.52+2.5+2.51+2.49+2.54}{5}$

$=2.512g$

b) absolutele error is as below

$m_1=2.512-2.52=-0.008g$

$m_2=2.512-2.5=0.012g$

$m_3=2.512-2.51=0.002g$

$m_4=2.512-2.49=0.022g$

$m_5=2.512-2.54=-0.028g$

c)mean absolute error

$<m_a>=\frac{m_1+m_2+m_3+m_4+m_5}{5}$

=0

d) percentage error 0%