Problem 7
Question
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) \(0.001 \mathrm{~mm}\)
Step-by-Step Solution
Verified Answer
The least count of the system is 0.01 mm, matching option (c) \(0.001 \text{ cm}\).
1Step 1: 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.
2Step 2: 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:\[\text{Pitch} = \frac{\text{Total Distance}}{\text{Number of Rotations}} = \frac{3 \text{ mm}}{6} = 0.5 \text{ mm}\]
3Step 3: 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:\[ \text{Least Count} = \frac{0.5 \text{ mm}}{50} = 0.01 \text{ mm} \]
4Step 4: 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.
Key Concepts
Screw GaugePitch CalculationCircular Scale Divisions
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.
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:
To calculate the pitch:
- Take the total distance moved by the screw.
- Divide it by the total number of rotations made by the screw.
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:
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: \( \frac{\text{Pitch}}{\text{Divisions}} \)
Other exercises in this chapter
Problem 5
On measuring diameter of a wire with help of screw gauge, main scale reading is \(1 \mathrm{~mm}\) and 6 th division of circular scale lying over reference line
View solution Problem 6
Two screw gauges \(A\) and \(B\) have equal number of divisions on circular scale. \(A\) has pitch \(1 \mathrm{~mm}\) and \(B\) has pitch \(0.5 \mathrm{~mm}\).
View solution Problem 10
If \(m\) and \(M\) are the masses of two bodies that are tied at two ends of a meter scale that is balanced on a sharp edge of a heavy broad wedge. If \(M=20 \m
View solution Problem 11
If two masses \(M\) and \(m\) are tied to two ends of a meter scale. If a balanced point is obtained at point \(P\) and if \(M>m\), then (a) \(P A=P B\) (b) \(P
View solution