Screw Gauge

Practice Questions:

Question 1: easy

A screw gauge gives the following reading while measuring diameter of a wire. Main Scale Reading = \(7\text{ mm}\), Circular Scale Reading = \(67\). Given that \(1\text{ mm}\) on main scale corresponds to \(100\text{ divisions}\) on circular scale. The diameter of the wire is:

1. \(76.7\text{ mm}\)

2. \(7.67\text{ mm}\)

3. \(0.767\text{ mm}\)

4. \(7.067\text{ mm}\)

View Answer

Least count is \(LC = \frac{1\text{ mm}}{100} = 0.01\text{ mm}\). Total Reading is \(MSR + CSR \times LC = 7\text{ mm} + 67 \times 0.01\text{ mm} = 7.67\text{ mm}\).

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Question 2: easy

A screw gauge gives the following readings when used to measure the diameter of a wire:
Main scale reading : 0 mm
Circular scale reading : 52 divisions
Given that 1 mm on main scale corresponds to 100 divisions on the circular scale.

The diameter of the wire from the above data is

1. 0.052 cm

2. 0.52 cm

3. 0.026 cm

4. 0.26 cm

View Answer

Least Count \(\frac{1\text{ mm}}{100} = 0.01\text{ mm} = 0.001\text{ cm}\. Diameter = MSR + (CSR times LC) = 0\text{ mm} + 52 \times 0.001\text{ cm} = 0.052\text{ cm}\.

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Question 3: easy

The pitch of a screw gauge is 1 mm and there are 100 divisions on circular scale. While measuring the thickness of a sheet, the main scale reads 1 mm and \(52^{\text{nd}}\) division on circular scale coincide with the reference line. The thickness of the sheet is

1. 1.52 cm

2. 0.152 mm

3. 0.152 cm

4. 1.052 mm

View Answer

Least count \(LC = \frac{\text{Pitch}}{\text{Number of circular divisions}} = \frac{1 \text{ mm}}{100} = 0.01 \text{ mm}\). Thickness \(= \text{MSR} + \text{CSR} \times LC = 1 \text{ mm} + 52 \times 0.01 \text{ mm} = 1.52 \text{ mm} = 0.152 \text{ cm}\).

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Question 4: easy

Dashrath measures the length of a wire using a meter scale with a least count of 1 mm and finds it to be L = 75.0 cm. He also measures diameter of thin wire using a screw gauge with a least count of 0.01 mm and finds it to be d = 0.500 cm. He uses these measurements to calculate the volume of wire. The maximum percentage error in volume of wire is nearly

1. 0.53%

2. 0.32%

3. 0.11%

4. 0.45%

View Answer

Volume of wire is \(V = \pi \left(\frac{d}{2}\right)^2 L ⇒ \frac{\Delta V}{V} = 2\frac{\Delta d}{d} + \frac{\Delta L}{L} = 2\left(\frac{0.001}{0.500}\right) + \frac{0.1}{75.0} \approx 0.53%\).

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Question 5: easy

Dashrath measures the length of a wire using a meter scale with a least count of \( 1\text{ mm} \) and finds it to be \( L = 75.0\text{ cm} \). He also measures diameter of thin wire using a screw gauge with a least count of \( 0.01\text{ mm} \) and finds it to be \( d = 0.500\text{ cm} \). He uses these measurements to calculate the volume of wire. The maximum percentage error in volume of wire is nearly

1. \( 0.53% \)

2. \( 0.32% \)

3. \( 0.11% \)

4. \( 0.45% \)

View Answer

Volume \( V = \frac{\pi d^2 L}{4} ⇒ \frac{\Delta V}{V} = 2 \frac{\Delta d}{d} + \frac{\Delta L}{L} \). Here, \( \Delta d = 0.001\text{ cm} \) and \( \Delta L = 0.1\text{ cm} \). Thus, \( \frac{\Delta V}{V} = 2\left(\frac{0.001}{0.500}\right) + \frac{0.1}{75} = 0.4% + 0.13% = 0.53% \).

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Question 6: 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 7: 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 8: 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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