5. A screw gauge gives the following reading 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 of the circular scale,
the diameter of wire from the above data is:
a. 0.052 cm
b. 0.026 cm
c. 0.005 cm
d. 0.52 cm

The diameter of the wire measures 0.052 cm (option a).​

Screw Gauge Basics

Screw gauges measure thin wire diameters precisely. The least count (LC), smallest readable value, calculates as pitch divided by circular scale divisions. Here, 1 mm pitch equals 100 circular divisions, so LC=1100=0.01LC=1001=0.01 mm or 0.001 cm.​

Total reading uses $\text{Diameter}=\text{MSR}+(\text{CSR}\times \text{LC})$, where MSR is main scale reading (0 mm) and CSR is circular scale reading (52 divisions). Substituting gives $0+(52\times 0.01)=0.52$ mm, or $0.52÷10=0.052$ cm since options are in cm. No zero error applies.​

Option Analysis

• a. 0.052 cm: Correct. Matches $52\times 0.001$ cm directly or 0.52 mm converted.​

• b. 0.026 cm: Wrong. Halves correct value, possibly assuming radius instead of diameter.​

• c. 0.005 cm: Wrong. Uses 5 divisions or miscalculates LC as 0.0001 cm.​

• d. 0.52 cm: Wrong. Forgets mm-to-cm conversion (0.52 mm = 0.052 cm).​

This MCQ tests units, LC, and formula application, common in CSIR NET Life Sciences (instrumentation) and physics sections.​