Question

While measuring the diameter of a wire using a screw gauge, the following readings were noted: Main scale reading is \( 1 \, \text{mm} \) and circular scale reading is equal to 42 divisions. Pitch of the screw gauge is \( 1 \, \text{mm} \), and it has 100 divisions on the circular scale. The diameter of the wire is \( \frac{x}{50} \, \text{mm} \). The value of \( x \) is:

• A. 142
• B. 71
• C. 42
• D. 21

Answer 1

The Correct Option is B

Given:

MSR (Main Scale Reading) = 1 mm, CSD (Circular Scale Reading) = 42

Least Count (LC) is calculated as:

\[ \text{Least Count (LC)} = \frac{\text{Pitch}}{\text{Number of CSD}} = \frac{1}{100} \, \text{mm} = 0.01 \, \text{mm}. \]

The diameter is calculated as:

\[ \text{Diameter} = \text{MSR} + \text{LC} \times \text{CSD}. \]

Substitute the values:

\[ \text{Diameter} = 1 + (0.01) \times 42 \, \text{mm} = 1.42 \, \text{mm}. \]

Since the diameter is given as \( \frac{x}{50} \), equate:

\[ \frac{x}{50} = 1.42 \implies x = 71. \]

The value of \( x \) is: [71]