Question

A screw gauge with a pitch of $0.5\, mm$ and a circular scale with $50$ divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the $45^{th}$ division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is $0.5\, mm$ and the $25^{th}$ division coincides with the main scale line ?

• A. 0.75 mm
• B. 0.80 mm
• C. 0.70 mm
• D. 0.50 mm

Answer 1

The Correct Option is B

To determine the thickness of the aluminium sheet using a screw gauge, we need to consider the provided readings and implement the screw gauge formula for measurement:

1. Understanding the Device Parameters:
   • Pitch of screw gauge: 0.5\, \text{mm}.
   • Circular scale divisions: 50.
   • Least count (LC) of the screw gauge is calculated as: \text{LC} = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}} = \frac{0.5\, \text{mm}}{50} = 0.01\, \text{mm}
2. Determining Zero Error:
   • Zero error occurs when the screw gauge shows a reading, even if there is no object to measure (i.e., jaws are in contact).
   • In this case, the 45^{th} division coincides with the main scale line when jaws are touching.
   • Zero error: E = 45 \times \text{Least Count} = 45 \times 0.01\, \text{mm} = 0.45\, \text{mm} (Positive zero error)
3. Measurement Reading:
   • Main scale reading (MSR): 0.5\, \text{mm}.
   • Coinciding circular division: 25^{th}
   • Circular scale reading (CSR): = 25 \times \text{Least Count} = 25 \times 0.01\, \text{mm} = 0.25\, \text{mm}
4. Calculating the Thickness:
   • True reading (corrected for zero error): \text{Reading} = \text{MSR} + \text{CSR} - \text{Zero Error}
   • Substituting the values: = 0.5\, \text{mm} + 0.25\, \text{mm} - 0.45\, \text{mm} = 0.8\, \text{mm}

Thus, the thickness of the aluminium sheet is 0.80\, \text{mm}, which corresponds to the given correct answer.