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NEET PHYSICSMedium

When the circular scale of a screw gauge completes 2 rotations, it covers 1 mm over the pitch scale. The total number of circular scale divisions is 50. The least count of the screw gauge in metres is:

A

10−410^{-4}

B

10−510^{-5}

C

10−210^{-2}

D

10−310^{-3}

Step-by-Step Solution

To find the least count of the screw gauge, we first need to calculate its pitch. Pitch is the distance moved by the spindle per revolution. Pitch=Distance covered on the pitch scaleNumber of rotations\text{Pitch} = \frac{\text{Distance covered on the pitch scale}}{\text{Number of rotations}} Pitch=1 mm2=0.5 mm\text{Pitch} = \frac{1 \text{ mm}}{2} = 0.5 \text{ mm}

The least count (LC) is given by the ratio of the pitch to the total number of divisions on the circular scale: Least Count=PitchTotal number of circular scale divisions\text{Least Count} = \frac{\text{Pitch}}{\text{Total number of circular scale divisions}} LC=0.5 mm50=0.01 mm\text{LC} = \frac{0.5 \text{ mm}}{50} = 0.01 \text{ mm}

Now, convert the least count into metres: LC=0.01×10−3 m=10−5 m\text{LC} = 0.01 \times 10^{-3} \text{ m} = 10^{-5} \text{ m}

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