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.

Question

asked Dec 24, 2021 228k views

1 Answer

Andre Kirpitch · Answer 1 · 2021-12-28T10:02:47+0000

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.