Question

A student in the laboratory measures thickness of a wire using screw gauge. The readings are 1.22 mm, 1.23 mm, 1.19 mm and 1.20 mm. The percentage error is
$\frac{x}{121}$%
. The value of x is _________.

Answer 1

The correct answer is 150
$I_{mean}=\frac{1.22+1.23+1.19+1.20}{4}=1.21$

$ΔI_{mean}=\frac{0.01+0.02+0.02+0.01}{4}=0.015$
So
%$I=\frac{ΔI_{mean}}{I_{mean}} ×100=\frac{0.015}{1.21}×100$
$=\frac{150}{121}$%
x = 150