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In an experiment to find out the diameter of wire using screw gauge, the following observation were noted :

(a) Screw moves 0.5 mm on main scale in one complete rotation
(b) Total divisions on circular scale = 50
(c) Main scale reading is 2.5 mm
(d) 45 division of circular scale is in the pitch line
(e) Instrument has 0.03 mm negative error
Then the diameter of wire is :

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2.54 mm

3.45 mm

2.92 mm

2.98 mm

answer is C.

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Detailed Solution

$\begin{matrix} MSR = 2 .5 mm \\ CSR = 45 \times \frac{0 .5}{50} mm \end{matrix}$
Diameter reading = MSR + CSR – zero error
= 2.5 + 0.45 – (-0.03)
= 2.98 mm

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