The zero error of a vernier is $+ 2$ . Its zero correction is (take L.C $= 0.01 c m$ )

0.2 c m

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- 0.2 c m

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- 0.02 c m

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+ 0.02 c m

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Solution

The correct option is C $- 0.02 c m$
When the zero error is positive, it will be subtract from the total reading.
Thus, zero correction $= - (2 \times L . C) = - 2 \times 0.01 = - 0.02 c m$

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The zero error of a vernier is +2. Its zero correction is (take L.C =0.01cm)

The correct option is C −0.02cmWhen the zero error is positive, it will be subtract from the total reading. Thus, zero correction =−(2×L.C)=−2×0.01=−0.02cm

The zero error of a vernier is $+ 2$ . Its zero correction is (take L.C $= 0.01 c m$ )

The correct option is C $- 0.02 c m$
When the zero error is positive, it will be subtract from the total reading.
Thus, zero correction $= - (2 \times L . C) = - 2 \times 0.01 = - 0.02 c m$