In four complete rotations, the distance moved by the screw on the linear scale is $2 m m$ . Its circular scale contains $50$ divisions. Find the least count of the screw gauge

0.05 m m

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0.01 m m

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0.1 m m

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0.02 m m

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Solution

The correct option is B $0.01 m m$
Circular scale has 50 divisions so $one rotation=50 divisions$
Thus $4 rotation=200 division$
and it is given that $4 rotation=2mm$ so $200 division=2mm$
So $l e a s t$ count = $1 D i v i s i o n = (2 / 200) m m = .01 m m$
Option B is correct.

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In four complete rotations, the distance moved by the screw on the linear scale is 2mm. Its circular scale contains 50 divisions. Find the least count of the screw gauge

The correct option is B 0.01mmCircular scale has 50 divisions so one rotation=50 divisions Thus 4 rotation=200 divisionand it is given that 4 rotation=2mm so 200 division=2mmSo least count = 1Division=(2/200)mm=.01mm Option B is correct.

In four complete rotations, the distance moved by the screw on the linear scale is $2 m m$ . Its circular scale contains $50$ divisions. Find the least count of the screw gauge

The correct option is B $0.01 m m$
Circular scale has 50 divisions so $one rotation=50 divisions$
Thus $4 rotation=200 division$
and it is given that $4 rotation=2mm$ so $200 division=2mm$
So $l e a s t$ count = $1 D i v i s i o n = (2 / 200) m m = .01 m m$
Option B is correct.