The least count of the main scale of the screw gauge is 1mm. The minimum number of divisions on its circular scale required to, measure 5μm diameter of the wire is(A) 50(B) 100(C) 200(D) 500

The correct option is (C) 200
In a screw gauge, the least count is the smallest length that can be measured with the instrument. The least count is determined by the ratio of the pitch of the screw (the distance the spindle moves in one complete rotation) and the number of divisions on the circular scale.
Given:
• Least count of the main scale: 1 mm
• Required measurement: 5 μm (which is 0.005 mm)
• We need to find the number of divisions on the circular scale to measure 5 μm accurately.
Step-by-Step Calculation:
1. Formula for Least Count:
Least count=Pitch of the screwNumber of divisions on the circular scale\text{Least count} = \frac{\text{Pitch of the screw}}{\text{Number of divisions on the circular scale}}
2. Pitch of the screw:
Since the least count of the main scale is given as 1 mm, we assume the pitch of the screw to be 1 mm (this is a typical value for screw gauges).
3. Let the number of divisions on the circular scale be NN. The least count is then:
Least count=1N mm\text{Least count} = \frac{1}{N} \, \text{mm}
4. To measure 5 μm (0.005 mm): To measure the required length with the required precision (5 μm), we need to make sure that the least count is small enough to measure this value. We set:
1N=0.005\frac{1}{N} = 0.005
Solving for NN:
N=10.005=200N = \frac{1}{0.005} = 200
Thus, the minimum number of divisions on the circular scale required to measure 5 μm diameter of the wire is 200.
The correct answer is (C) 200.