Question

Consider a Vernier callipers in which each \(1 \mathrm{~cm}\) on the main scale is divided into 8 equal divisions and a screw gauge with 100 divisions on its circular scale. In the Vernier callipers, 5 divisions of the Vernier scale coincide with 4 divisions on the main scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then: (A) If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is \(0.01 \mathrm{~mm}\). (B) If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is \(0.005 \mathrm{~mm}\). (C) If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is \(0.01 \mathrm{~mm}\). (D) If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is \(0.005 \mathrm{~mm}\).

Short answer

None of the given options match the calculated least count. The correct least count of the screw gauge should be 0.02 mm based on the calculations.

Step-by-step solution

Find the least count of the Vernier calipers

The least count of the Vernier calipers is calculated by the formula: Least Count = (Value of one main scale division) / (Number of divisions on Vernier scale). So we have 1 cm divided by 10 on the main scale, which is 0.1 cm or 1 mm, and 5 divisions of the Vernier scale coincide with 4 on the main scale, which means the Vernier scale must be subdivided more finely. Therefore, we have 1 mm / (5 - 4) = 1 mm / 1 = 1 mm as the least count of the Vernier calipers.

Determine the least count of the screw gauge

For the screw gauge, one complete rotation moves the scale by two divisions of the linear scale. Since there are 100 divisions on the circular scale, each division on the circular scale represents a movement of (2 divisions on linear scale)/(100 divisions on circular scale). If the pitch of the screw gauge is twice the least count of the Vernier calipers, then the pitch (distance moved by one complete rotation) is twice of 1 mm, i.e., 2 mm. Thus, the least count of the screw gauge is 2 mm / 100 = 0.02 mm.

Analyze the options given in the exercise

Only option (C) correctly matches the calculated least count of the screw gauge with 0.01 mm if the pitch is considered as 1 mm (which is twice the least count of the Vernier calipers). However, since we calculated the pitch to be 2 mm to match the least count of the Vernier calipers and found the least count of the screw gauge to be 0.02 mm, none of the options provided match our calculation. There seems to be an error either in the options or in the understanding of the 'pitch' in relation to the 'least count' provided by the question. The least count should be the pitch divided by the number of divisions on the circular scale, which gave us a value of 0.02 mm rather than the options given which suggest a least count of 0.01 mm or 0.005 mm.

Key concepts

Least Count Calculation

Understanding the least count of any measuring instrument is crucial for precision and accuracy in measurements. The least count refers to the smallest measurement that can be taken accurately with the instrument. For vernier calipers, the least count is determined by the difference in the size of divisions between the main scale and the vernier scale. In the exercise provided, the main scale division is 1 cm which is divided into 8 equal parts so each division is 1.25 mm (since 1 cm = 10 mm, then 10 mm / 8 = 1.25 mm). However, the 5 divisions on the Vernier scale are equal to 4 divisions on the main scale.

This would mean that each vernier scale division is less than 1.25 mm. To calculate the least count, you subtract the main scale division size from the vernier scale division size and divide by the number of vernier divisions. The correct formula for calculating the least count of a vernier caliper is: \
Least Count = \frac{Value \text{ of one main scale division (MSD)}}{Number \text{ of divisions on Vernier scale}} - \frac{Value \text{ of one Vernier scale division (VSD)}}{Number \text{ of divisions on Vernier scale}}\.\

In this scenario, the least count of the vernier calipers would actually be \( \frac{1.25 \text{ mm}}{5} \), since each division on the main scale is 1.25 mm and there are 5 divisions on the vernier scale that match 4 main scale divisions. This discrepancy in understanding the main scale division size led to an incorrect earlier calculation.

Vernier Scale Reading

When it comes to reading a vernier caliper, the process involves two scales: the main scale and the vernier scale. The main scale provides the coarse measurement and the vernier scale offers a refined measurement, allowing you to read fractions of the main scale's smallest division.

The way to read the vernier scale is to look for a line on the vernier scale that aligns exactly with a line on the main scale. The number of this vernier line gives you the number of least counts to add to the main scale reading. It's essential to read both scales starting from the zero end of the vernier scale, not from any other arbitrary point. Combining the value from the vernier scale with that of the main scale gives you the precise measurement of the object you are measuring.

For visual learners, it's often helpful to demonstrate the technique with a real vernier caliper. Practice is key to proficiency in reading these measurements accurately. Precision is achieved by careful observation and steady hands to ensure that parallax errors are minimized when aligning the scales.

Screw Gauge Measurement

The screw gauge, also known as a micrometer, is another instrument for taking precise measurements, capable of measuring even smaller dimensions than the vernier caliper. The two key components are the spindle which moves linearly when twisted and the thimble that houses a circular scale.

The pitch of the screw gauge is the distance the spindle moves for one complete rotation of the thimble. If the pitch is known, the least count can be found by dividing the pitch by the number of divisions on the thimble's circular scale. In the initial solution presented, the misunderstanding of the pitch and the incorrect value used for the least count calculation led to an incorrect answer.

To accurately measure the pitch and consequently find the least count, one should ensure that the screw gauge is properly zeroed out and that readings are taken without any undue force applied to the instrument, which may cause errors. Keeping the instrument clean and free of dust will also enhance its accuracy. Measurements with a screw gauge are highly reliable due to its design, which minimizes human error and allows for repeated consistency in taking measurements.