Question

A student reports the density of an article to be 1.92 g/mL. The accepted value of density is 1.89 g/mL. Calculate the student's percent error.

Answer 1

Given :

Actual value observed ,
`v_a=1.92\ g/ml`.

Expected value ,
`v_e=1.89\ g/ml`.

To Find :

The student's percent error.

Solution :

Percentage error is given by :

`P=|(v_a-v_e)/(v_e)|* 100\ \%`

Putting given values in above equation, we get :

`P=|(1.92-1.89)/(1.89)|* 100\ \%\\\\P=1.587\ \%`

Therefore, percentage error is 1.587 %.

Hence, this is the required solution.