JEE Challenger
More from Units and Measurements
PhyMedAdv 2026 • Paper 2

Determine Unknown Wire Diameter Using Screw Gauge Readings

Two thin wires, Wire-1 of diameter 0.650 mm0.650 \text{ mm} and Wire-2 of unknown diameter dd are given. To obtain the value of dd, the diameters of the two wires are measured with a screw gauge. The screw gauge has a pitch of 0.5 mm0.5 \text{ mm} and there are 100100 divisions on the circular scale (CS). The smallest division on the linear scale (LS) is 0.5 mm0.5 \text{ mm}. The table shows the readings of LS and CS for the measurements. The value of dd (in μm\mu\text{m}) is:

Question Diagram 1
Official Numerical Answer1915

Topics & Concepts

Units and MeasurementsMeasuring Instruments

Step-by-Step Solution

To find the unknown diameter dd of Wire-2 in micrometers (μm\mu\text{m}), we follow these steps:

1. Calculate the Least Count (LC) of the Screw Gauge

The least count (LC\text{LC}) of a screw gauge is given by the formula: LC=PitchTotal number of divisions on circular scale (CS)\text{LC} = \frac{\text{Pitch}}{\text{Total number of divisions on circular scale (CS)}}

Given:

  • Pitch=0.5 mm\text{Pitch} = 0.5\text{ mm}
  • Number of CS divisions=100\text{Number of CS divisions} = 100

LC=0.5 mm100=0.005 mm\text{LC} = \frac{0.5\text{ mm}}{100} = 0.005\text{ mm}

2. Determine the Zero Error (ZZ) Using Wire-1

The measured (uncorrected) diameter of Wire-1, d1′d_1', is: d1′=LS1+(CS1×LC)d_1' = \text{LS}_1 + (\text{CS}_1 \times \text{LC})

Given for Wire-1:

  • LS1=0.5 mm\text{LS}_1 = 0.5\text{ mm}
  • CS1=42\text{CS}_1 = 42

d1′=0.5 mm+(42×0.005 mm)=0.5 mm+0.210 mm=0.710 mmd_1' = 0.5\text{ mm} + (42 \times 0.005\text{ mm}) = 0.5\text{ mm} + 0.210\text{ mm} = 0.710\text{ mm}

The actual diameter of Wire-1 is d1=0.650 mmd_1 = 0.650\text{ mm}.

The zero error ZZ is defined by: Actual Diameter=Measured Value−Z\text{Actual Diameter} = \text{Measured Value} - Z d1=d1′−Z  ⟹  Z=d1′−d1d_1 = d_1' - Z \implies Z = d_1' - d_1 Z=0.710 mm−0.650 mm=+0.060 mmZ = 0.710\text{ mm} - 0.650\text{ mm} = +0.060\text{ mm}

3. Calculate the Diameter of Wire-2 (dd)

The measured (uncorrected) diameter of Wire-2, d2′d_2', is: d2′=LS2+(CS2×LC)d_2' = \text{LS}_2 + (\text{CS}_2 \times \text{LC})

Given for Wire-2:

  • LS2=1.5 mm\text{LS}_2 = 1.5\text{ mm}
  • CS2=95\text{CS}_2 = 95

d2′=1.5 mm+(95×0.005 mm)=1.5 mm+0.475 mm=1.975 mmd_2' = 1.5\text{ mm} + (95 \times 0.005\text{ mm}) = 1.5\text{ mm} + 0.475\text{ mm} = 1.975\text{ mm}

Applying the zero error correction: d=d2′−Z=1.975 mm−0.060 mm=1.915 mmd = d_2' - Z = 1.975\text{ mm} - 0.060\text{ mm} = 1.915\text{ mm}

4. Convert the Value to μm\mu\text{m}

d=1.915 mm×1000 μmmm=1915 μmd = 1.915\text{ mm} \times 1000\ \frac{\mu\text{m}}{\text{mm}} = 1915\ \mu\text{m}

Final Answer: The value of dd is 1915.

Determine Unknown Wire Diameter Using Screw Gauge Readings | Physics PYQ Solution - JEE Challenger