Question

The pitch of the screw gauge is $1 mm$ and there are $100$ divisions on the circular scale. When nothing is put in between the jaws, the zero of the circular scale lies $ 8$ divisions below the reference line. When a wire is placed between the jaws, the first linear scale division is clearly visible while $72^{\text {nd }}$ division on circular scale coincides with the reference line. The radius of the wire is

• A. $1.64\, mm$
• B. $0.82\, mm$
• C. $1.80\, mm$
• D. $0.90 \,mm$

Answer 1

The Correct Option is B

To solve this problem, we need to understand the working of a screw gauge and calculate the correct measurement of the wire's radius using given parameters.

1. Given that the pitch of the screw gauge is \(1 \, \text{mm}\). This means that one complete turn of the circular scale displaces the screw by \(1 \, \text{mm}\).
2. The number of divisions on the circular scale is \(100\). Therefore, the least count of the screw gauge is calculated as:

\[\text{Least Count} = \frac{\text{Pitch}}{\text{Number of Divisions on Circular Scale}} = \frac{1 \, \text{mm}}{100} = 0.01 \, \text{mm}\]

1. The zero error is mentioned as \(8\) divisions below the reference line. This translates to a value of: \(\text{Zero Error} = 8 \times \text{Least Count} = 8 \times 0.01 \, \text{mm} = 0.08 \, \text{mm}\).
2. When the wire is placed between the jaws, the first linear scale division is visible, and the circular scale shows the \(72^{\text{nd}}\) division aligned with the reference line. This indicates a reading from the circular scale of: \(72 \times \text{Least Count} = 72 \times 0.01 \, \text{mm} = 0.72 \, \text{mm}\).
3. Therefore, the total linear reading is: \(\text{Total Linear Reading} = 1 \, \text{mm} + 0.72 \, \text{mm} = 1.72 \, \text{mm}\) (since the first linear scale division is visible, we add \(1 \, \text{mm}\)).
4. To find the corrected diameter of the wire, we subtract the zero error from the total linear reading: \(\text{Corrected Diameter} = 1.72 \, \text{mm} - 0.08 \, \text{mm} = 1.64 \, \text{mm}\).
5. The radius of the wire is half of the corrected diameter, therefore: \(\text{Radius} = \frac{\text{Corrected Diameter}}{2} = \frac{1.64 \, \text{mm}}{2} = 0.82 \, \text{mm}\).

Hence, the radius of the wire is 0.82 mm, which matches the correct answer option.