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NEET PHYSICSUNITS AND MEASUREMENTSMedium

Question

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}

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from UNITS AND MEASUREMENTS. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSUNITS AND MEASUREMENTScircularcompletesrotationscoversnumber

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