The least count of a screw gauge is given by: 

1. \(\dfrac{\text {Pitch}}{\text { Number of divisions on circular scale }}\)

2. \(\dfrac{\text{One circular scale division}}{\text{Number of divisions on main scale}}\)

3. \(\dfrac{\text{Number of divisions on circular scale }}{\text{One main scale division}}\)

4. \(\dfrac{\text{One circular scale division}}{\text{Pitch}}\)
Subtopic:  Measurement & Measuring Devices |
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The diameter of a thin wire is measured by:
1. Screw gauge 2. Spherometer
3. Spectrometer 4. Venturimeter
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What is the reading displayed on the micrometer screw gauge shown in the figure, given that its least count is \(0.01~ \text{mm} \text{?}\)
1. \(6.41 ~\text{mm}\) 2. \(6.38~\text{mm}\)
3. \(6.26 ~\text{mm}\) 4. \(6.48~\text{mm}\)
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A vernier calliper is designed such that \(10\) main scale divisions coincide with \(11\) vernier scale divisions. If one main scale division corresponds to \(5~\text{mm},\) the least count of the instrument is:
1. \(\dfrac{1}{2}~\text{mm}\) 2. \(\dfrac{5}{12}~\text{mm}\)
3. \(\dfrac{5}{11}~\text{mm}\) 4. \(0.3~\text{mm}\)
Subtopic:  Measurement & Measuring Devices |
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A student measured the diameter of a small steel ball using a screw gauge of least count \(0.001 \mathrm{~cm} \text {. }\)The main scale reading is \(5 \mathrm{~mm} \) and zero of circular scale division coincides with 25 divisions above the reference level. If screw gauge has a zero error of \(-0.004 \mathrm{~cm}\), the correct diameter of ball is: 
1. \(0.521 \mathrm{~cm}\)
2. \(0.525 \mathrm{~cm}\)
3. \(0.529 \mathrm{~cm}\)
4. \(0.053 \mathrm{~cm}\)
 
Subtopic:  Measurement & Measuring Devices |
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The side of a cube is measured by vernier calipers (\(20\) divisions of the vernier scale coincide with \(19\) divisions of the main scale, where \(1\) division of the main scale is \(1~\text{mm}\)). The main scale reads \(10~\text{mm}\) and the first division of the vernier scale coincides with the main scale. The side length of a cube is:
1. \(10.02\) mm
2. \(10.05\) mm
3. \(10.04\) mm
4. \(10.06\) mm
Subtopic:  Measurement & Measuring Devices |
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A screw gauge has least count of \(10 ~\mu \mathrm{m}\) and its circular scale is divided into \(100\) equal divisions. One main scale division is: 
1. \(0.5 \mathrm{~mm}\)
2. \(1 \mathrm{~mm}\)
3. \(0.5 \mathrm{~cm}\)
4. \(1 \mathrm{~cm}\)
Subtopic:  Measurement & Measuring Devices |
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One picometre (pm) is equal to:
1. \(10^{-9} \) m
2. \(10^{-10} \) m
3. \(10^{-11} \) m
4. \(10^{-12} \) m
Subtopic:  Measurement & Measuring Devices |
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Given below are two statements:
Statement I: A positive zero error is added to the observed reading to obtain the correct measurement.
Statement II: Measuring instruments may have defects due to imperfections during the manufacturing process.
 
1. Statement I is correct and Statement II is incorrect.
2. Statement I is incorrect and Statement II is correct.
3. Both Statement I and Statement II are incorrect.
4. Both Statement I and Statement II are correct.
Subtopic:  Measurement & Measuring Devices |
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\(n-\)divisions on the main scale of a vernier calipers coincide with \(n+1 \) divisions on the vernier scale. If each division on the main scale is of \(x\) units, determine the least count of the instrument:
1. \(\dfrac{x}{n}\)
2. \(\dfrac{x}{n+1}\)
3. \(\dfrac{x}{n-1}\)
4. \(\dfrac{2 x}{n+1}\)
Subtopic:  Measurement & Measuring Devices |
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