Question

In an experiment to find out the diameter of the wire using a screw gauge, the following observations 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^th division of circular scale is in the pitch line
(E) Instrument has 0.03 mm negative error
Then the diameter of wire is :

• A. 2.92 mm
• B. 2.54 mm
• C. 2.98 mm
• D. 3.45 mm

Answer 1

The Correct Option is C

\(\text{L.C.} = \frac{0.5}{50} \, \text{mm} = 0.01 \, \text{mm}\)
\(d = (2.5 + 45 × 0.01 + 0.03)\) mm
\(= 2.98\)
So, the correct option is (C): 2.98 mm