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In a screw gauge, there are 100 divisions on the circular scale and the main scale moves by 0.5 mm on a complete rotation of the circular scale. The zero of circular scale lies 6 divisions below the line of graduation when two studs are brough in contact with each other. When a wire is placed between the studs, 4 linear scale divisions are clearly visible and 46^th division the circular scale coincide with the reference line. The diameter of the wire is $\times 10^{- 2} m m$ .

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2.2

answer is C.

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Detailed Solution

$L e a s t c o u n t = \frac{P i t c h}{N o . o f c i r c u l a r d i v i s i o n s}$

$= \frac{0.5 m m}{100}$

Least count $= 5 \times 10^{- 3} m m$

Positive error = MSR + CSR (LC)

$= O + 6 (5 \times 10^{- 3})$

$= 30 \times 10^{- 3} m m$

Diameter = $(4 \times 0.5) + (46 \times 5 \times 10^{- 3}) -$ $(6 \times 5 \times 10^{- 3})$

$= 2.2 m m$

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