Physics
Vernier scale readings are as shown in the figures given below. Give the actual reading provided :
(a) the instrument has a zero error of +0.02 cm.

(b) the instrument has a zero error of -0.03 cm.

Measurements
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Answer
(a) From the figure, the zero of the vernier scale lies between 1.0 cm and 1.1 cm on the main scale.
So, the main scale reading (MSR) = 1.0 cm.
As we know,
There are 10 divisions on 1 cm on the main scale
Total number of divisions on the vernier scale (n) = 10
Thus, the least count = 0.01 cm.
Since 6th division of vernier scale coincides with the main scale then,
Vernier scale reading (VSR) = 6 × 0.01 = 0.06 cm
Observed reading = MSR + VSR = 1.0 + 0.06 = 1.06 cm
Given
Zero error = +0.02 cm,
So the zero correction = −0.02 cm.
Then,
Actual reading = 1.06 − 0.02 = 1.04 cm
Hence, the actual reading is 1.04 cm.
(b) From the figure, the zero of the vernier scale lies between 1.0 cm and 1.1 cm on the main scale.
So, the main scale reading (MSR) = 1.0 cm.
As we know,
There are 10 divisions on 1 cm on the main scale
Total number of divisions on the vernier scale (n) = 10
Thus, the least count = 0.01 cm.
Since 4th division of vernier scale coincides with the main scale then,
Vernier scale reading (VSR) = 4 × 0.01 = 0.04 cm
Observed reading = MSR + VSR = 1.0 + 0.04 = 1.04 cm
Given
Zero error = -0.03 cm,
So the zero correction = +0.03 cm.
Then,
Actual reading = 1.04 + 0.03 = 1.07 cm
Hence, the actual reading is 1.07 cm.
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