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In 4 complete rotations, the distance moved by the screw on the linear scale is 2mm. Its circular scale contain 50 divisions. Find least count?
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In 4 complete rotations, the distance moved by the screw on the linear...
**Introduction**
To find the least count of a screw gauge, we need to consider the number of divisions on the circular scale and the distance moved by the screw on the linear scale in a certain number of rotations.

**Given Data**
- Distance moved by the screw on the linear scale in 4 rotations = 2mm
- Number of divisions on the circular scale = 50

**Calculating the Distance Moved in One Rotation**
To find the distance moved by the screw in one rotation, we divide the total distance moved in 4 rotations by 4.

Distance moved in one rotation = (2mm) / (4 rotations) = 0.5mm

**Calculating the Least Count**
The least count is defined as the minimum measurement that can be made using the instrument. It is the smallest difference that can be observed between two readings.

To calculate the least count, we need to consider the number of divisions on the circular scale and the distance moved in one rotation.

The formula to calculate the least count is:
Least Count = (Distance moved in one rotation) / (Number of divisions on the circular scale)

Substituting the given values:
Least Count = (0.5mm) / (50 divisions) = 0.01mm

**Interpretation**
The least count of the screw gauge is 0.01mm. This means that the instrument can measure a minimum difference of 0.01mm between two readings on the linear scale. Any difference smaller than this cannot be observed or measured accurately using this screw gauge.

**Conclusion**
The least count of the screw gauge is an important factor to consider when using the instrument for precise measurements. By understanding the number of divisions on the circular scale and the distance moved by the screw, we can calculate the least count. In this case, the least count is found to be 0.01mm, indicating that the instrument can measure differences as small as 0.01mm accurately.
Community Answer
In 4 complete rotations, the distance moved by the screw on the linear...
0.01
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In 4 complete rotations, the distance moved by the screw on the linear scale is 2mm. Its circular scale contain 50 divisions. Find least count?
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