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Chapter 29: Problem 7

Six rotations are given to a screw to turn it through a distance of \(3 \mathrm{~mm}\) and there are 50 divisions on the circular scale. What is the least count of the system? (a) \(0.01 \mathrm{~cm}\) (b) \(0.02 \mathrm{~mm}\) (c) \(0.001 \mathrm{~cm}\) (d) \(0.001 \mathrm{~mm}\)

Short Answer

Expert verified
The least count of the system is 0.01 mm, matching option (c) \(0.001 \text{ cm}\).

Step by step solution

01

Understand the Problem

We need to find the least count of a screw gauge system given the distance moved per rotation and the number of divisions on the circular scale.
02

Calculate Pitch

First, determine the pitch, which is the distance moved by the screw per full rotation. It is given that the screw moves a total of 3 mm in 6 rotations. Therefore, the pitch can be calculated as follows:\[\text{Pitch} = \frac{\text{Total Distance}}{\text{Number of Rotations}} = \frac{3 \text{ mm}}{6} = 0.5 \text{ mm}\]
03

Calculate Least Count

The least count is found using the formula:\[ \text{Least Count} = \frac{\text{Pitch}}{\text{Number of Divisions on the Circular Scale}} \]Plug in the values we have:\[ \text{Least Count} = \frac{0.5 \text{ mm}}{50} = 0.01 \text{ mm} \]
04

Compare with Options

Now, let's compare the calculated least count of 0.01 mm with the provided options. Convert units if necessary to match the choices: - Option (a) 0.01 cm is equivalent to 0.1 mm. - Option (b) 0.02 mm. - Option (c) 0.001 cm is equivalent to 0.01 mm. - Option (d) 0.001 mm. The correct match is option (c) as it is equivalent to 0.01 mm.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Screw Gauge
A screw gauge is a precision instrument used to measure dimensions with a high degree of accuracy. It is often used in laboratories and engineering fields to measure the thickness of small objects. The fundamental parts of a screw gauge include the main scale, circular scale, screw, and spindle. These components work together to enable precise measurements.

In operation, the screw gauge uses a finely threaded screw to traverse a known distance across each rotation. This distance is critical when understanding how accurate the screw gauge can be in its measurements. By knowing how much the screw moves with each turn, we can calculate something known as the pitch. Accurate measurements depend on proper calibration and understanding of these intervals.
Pitch Calculation
Pitch calculation in a screw gauge is essential. It provides us the incremental distance that the screw moves per a complete rotation. This value is pivotal for determining other parameters such as the least count.

To calculate the pitch:
  • Take the total distance moved by the screw.
  • Divide it by the total number of rotations made by the screw.
For instance, if a screw gauge moves 3 mm over 6 rotations, the pitch is calculated as follows: \[\text{Pitch} = \frac{3 \text{ mm}}{6} = 0.5 \text{ mm}\]The pitch tells us how far the screw travels during each complete turn. Knowing this helps us understand how the screw translates rotational motion to linear motion. This conversion is key when using the screw gauge for precise measurements.
Circular Scale Divisions
Circular scale divisions are important when measuring with a screw gauge because they dictate the instrument's precision. These divisions act as increments on the circular scale, allowing us to measure not just in whole units but in precise fractions.

Each mark on the circular scale indicates a fraction of the pitch. For example, if there are 50 divisions, each division represents a fraction of the screw's pitch, as calculated by:
  • Total divisions on the circular scale: 50
  • Each division equals: \( \frac{\text{Pitch}}{\text{Divisions}} \)
This equation provides us the least count, or the smallest measurement increment possible with that gauge:\[\text{Least Count} = \frac{0.5 \text{ mm}}{50} = 0.01 \text{ mm}\]This value indicates how precisely the screw gauge can measure. Understanding these divisions is crucial because they enable users to achieve highly accurate measurements, down to fractions of a millimeter.

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Most popular questions from this chapter

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