In an experiment to find out the diameter of wire using screw gauge, the following observations were noted:

(A) Screw moves $0.5 mm$ on main scale in one complete rotation
(B) Total divisions on circular scale $= 50$
(C) Main scale reading is $2.5 mm$
(D) $45^{th}$ division of circular scale is in the pitch line
(E) Instrument has $0.03 mm$ negative error

Then the diameter of wire is:

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Solution

$Least count = \frac{pitch}{number of divisions on circular scale}$

$= \frac{0.5 mm}{50} = 0.01 mm$

Measured diameter = (main scale reading) $+$ ( circular scale reading ) ( least count )
$= 2.5 mm + (45) (0.01 mm) = 2.95 mm$
Actual diameter = measured diameter - zero error

$= 2.95 mm - (- 0.03 mm)$

$= 2.98 mm$

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Least count =pitchnumber of divisions on circular scale =0.5 mm50=0.01 mm Measured diameter = (main scale reading) + ( circular scale reading ) ( least count ) =2.5 mm+(45)(0.01 mm)=2.95 mm Actual diameter = measured diameter - zero error =2.95 mm−(−0.03 mm) =2.98 mm