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

Answer: (B)

71

Answer 2

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]