Question

The circular scale of a screw gauge has
200 divisions. When it is given 4
complete rotations, it moves through 2
mm. The least count of the screw gauge
is​

Answer 1

Answer:

the last answer is 0.0005cm

Answer 2

The least count of the screw gauge is 0.00025cm.

Given:-

Circular scale division (CSD) = 200

No. of rotations (n) = 4

Distance covered (d) = 2mm

To Find:-

The least count of the screw gauge.

Solution:-

We can easily find the least count of the screw gauge by following the given steps.

As

Circular scale division (CSD) = 200

No. of rotations (n) = 4

Pitch (d) = 2mm

Least Count (LC) =?

According to the formula,

$Least Count = \frac{Pitch}{Circular scale division (CSD) \times No. of rotations }$

$LC = \frac{d}{CSD \times n}$

$LC = \frac{2}{200 \times 4}$

$LC = \frac{1}{400}$

$LC = 0.0025mm$

$LC = 0.00025cm$

Hence, The least count of the screw gauge is 0.00025cm.

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