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 the given problem using a screw gauge, we need to determine the radius of the wire based on the provided measurements. 

1. The pitch of the screw gauge is given as \(1 \text{ mm}\). Pitch indicates the distance traveled by the screw in one complete rotation.
2. There are \(100\) divisions on the circular scale. This implies that each smallest division (least count) on the circular scale equals \(\frac{1 \text{ mm}}{100} = 0.01 \text{ mm}\).
3. The zero error is given as \(8\) divisions below the reference line. Since it's below, the zero error is negative: \(-8 \times 0.01 \text{ mm} = -0.08 \text{ mm}\).
4. When the wire is placed in the jaws, the first linear scale division is visible, meaning the main scale reading is \(1 \text{ mm}\).
5. The circular scale reads the \(72^{\text{nd}}\) division, which corresponds to \(72 \times 0.01 \text{ mm} = 0.72 \text{ mm}\).
6. Calculate the total measured length as \(1 \text{ mm} + 0.72 \text{ mm} = 1.72 \text{ mm}\).
7. Factor in the zero error (since it's a negative error, we add its absolute value): \(1.72 \text{ mm} + 0.08 \text{ mm} = 1.80 \text{ mm}\).
8. This is the diameter of the wire. To find the radius, divide by 2: 1\text{mm}
9. Circular scale reading = \(0.72\text{mm}\)
10. Total reading before correcting zero error = \(1.72\text{mm}\)
11. Correction factor from zero error = 1.72 \text{ mm} - 0.08 \text{ mm} = 1.64 \text{ mm}
12. Corrected diameter of the wire is actually \(1.64 \text{ mm}\). Thus, the radius of the wire:
13. Radius = 0.82 \text{ mm}. This matches with the correct option given.