The main scale reading of a vernier calliper is obtained as 3.2 cm for an object placed in it. If the 4th division of the vernier scale coincides with a division on the main scale, and the magnitude of the positive zero error is 0.03 cm, what is the true length of the object in cm? Take L.C. = 0.01 cm.
3.21

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Solution

The correct option is A 3.21
Given:
Main scale reading (MSR) = $3.2 c m$
Vernier scale reading that coincides with a division on the main scale, $p = 4$
VSR = $p \times L.C.$
$= 4 \times 0.01 = 0.04 c m$
Observed length of the object, $X$ = MSR + VSR $= 3.2 + 0.04 = 3.24 c m$
Magnitude of positive zero error, PZE = $0.03 c m$
Positive zero error is subtracted from the observed length to get the true length.
$⟹$ True length = X - PZE = $3.24 - 0.03 = 3.21 c m$ .

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