Screw Gauge

Practice Questions:

Question 1: easy

A screw gauge gives the following readings when used to measure the diameter of a wire
Main scale reading: \(0 \text{mm}\)
Circular scale reading: \(52 \text{divisions}\)
Given that \(1 \text{mm}\) on main scale corresponds to \(100 \text{divisions}\) on the circular scale. The diameter of the wire from the above data is:

[NEET 2020]

1. \(0.026 text{cm}\)

2. \(0.26 text{cm}\)

3. \(0.052 text{cm}\)

4. \(0.52 text{cm}\)

View Answer

Pitch = \(1 \text{mm}\), Number of divisions = \(100\). Least Count (LC) = Pitch / Number of divisions = \(1 \text{mm} / 100 = 0.01 \text{mm}\). Total reading = MSR + (CSR \(\times\) LC) = \(0 \text{mm} + (52 \times 0.01 \text{mm}) = 0.52 \text{mm}\). Convert to cm: \(0.52 \text{mm} = 0.052 \text{cm}\).

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Question 2: easy

A screw gauge has least count of \(0.01 \text{mm}\) and there are \(50 \text{divisions}\) in its circular scale. The pitch of the screw gauge is:

[NEET 2020]

1. \(0.25 \text{mm}\)

2. \(0.5 \text{mm}\)

3. \(1.0 \text{mm}\)

4. \(0.01 \text{mm}\)

View Answer

Least Count (LC) = \(0.01 \text{mm}\), Number of divisions on circular scale = \(50\). The formula for LC is: LC = Pitch / Number of divisions. Therefore, Pitch = LC \(\times\) Number of divisions = \(0.01 \text{mm} \times 50 = 0.5 \text{mm}\).

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Question 3: easy

A student measured the diameter of a small steel ball using a screw gauge of least count \(0.001 \text{cm}\). The main scale reading is \(5 \text{mm}\) and zero of circular scale division coincides with \(25 \text{divisions}\) above the reference level. If screw gauge has a zero error of \(-0.004 \text{cm}\), the correct diameter of the ball is :

[NEET 2018]

1. \(0.053 \text{cm}\)

2. \(0.525 \text{cm}\)

3. \(0.521 \text{cm}\)

4. \(0.529 \text{cm}\)

View Answer

Least Count (LC) = \(0.001 \text{cm}\). Main Scale Reading (MSR) = \(5 \text{mm} = 0.5 \text{cm}\). Circular Scale Reading (CSR) = \(25 \text{divisions}\). Observed Reading = MSR + (CSR \(\times\) LC) = \(0.5 \text{cm} + (25 \times 0.001 \text{cm}) = 0.525 \text{cm}\). Correct Reading = Observed Reading - Zero Error = \(0.525 \text{cm} - (-0.004 \text{cm}) = 0.525 \text{cm} + 0.004 \text{cm} = 0.529 \text{cm}\).

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