Question

A student measured the diameter of a small steel ball

Answer 1

O K A Y

But please provide the complete Q

Answer 2

LC=0.001cm =0.01mm

25 division

VSR=LC×DIVISION

=0.01×25

=0.25

MSR=5 mm

OR=MSR +VSR

=5+0.25

=5.25mm

true reading=OR-(ZERO ERROR with sign)

zero error= -0.004cm = -0.04mm

5.25 -(-0.04)

5.25+0.04

5.29 mm

You have not written the full question. but I know this questions and it's answer