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

The problem involves finding the diameter of a wire using the measurements obtained from a screw gauge. Let us solve this step-by-step using the observations provided: 

1. Pitch of the Screw Gauge:
   • The pitch is defined as the distance moved by the screw on the main scale in one complete rotation of the circular scale.
   • Given that the screw moves 0.5 mm per complete rotation, the pitch is \(0.5\ mm\).
2. Least Count of the Screw Gauge:
   • The least count can be calculated using the formula: \(\text{Least Count} = \frac{\text{Pitch}}{\text{Total Divisions on Circular Scale}}\).
   • Substitute the values to get: \(\text{Least Count} = \frac{0.5\ mm}{50} = 0.01\ mm\).
3. Main Scale Reading (MSR):
   • The main scale reading given is 2.5 mm.
4. Circular Scale Reading (CSR):
   • The 45^th division of the circular scale coincides with the pitch line, so CSR = 45.
5. Total Reading:
   • Total Reading (TR) is calculated as: \(\text{TR} = \text{MSR} + (\text{CSR} \times \text{Least Count})\).
   • Substitute the known values: \(\text{TR} = 2.5\ mm + (45 \times 0.01\ mm) = 2.5\ mm + 0.45\ mm = 2.95\ mm\).
6. Correction for the Negative Error:
   • Negative error provided is 0.03 mm. This error needs to be added to the total reading to get the corrected reading.
   • Corrected Diameter = \(2.95\ mm + 0.03\ mm = 2.98\ mm\)

Therefore, the diameter of the wire is 2.98 mm, which matches the given correct answer.

Answer 2

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