Question

The circular head of the screw gauge is divided into 20 divisions and the screw moves 1 mm ahead in two rotations of the head of the screw. Find its least count.​

Answer 1

$\huge \rm {Answer:-}$

__________________________________

$\sf \red {Given:-}$

▸No of divisions,the circular head of the screw gauge is divided = 20

▸Distance moved =1mm

▸Number of rotations=2

__________________________________

$\sf \blue {To\: Find:-}$

▸Least count of the screw gauge=?

__________________________________

$\sf \pink {We\: Know:-}$

$\bf \mapsto {Least\: Count=\frac{pitch}{No.\: of\: divisions}}$

▸Inorder to calculate the least count,pitch is to be determined.

▸To determine pitch we have a formula,

$\bf \mapsto {Pitch=\frac{Distance\: moved}{No.\: of\: rotations}}$

$\bf \mapsto {Pitch=\frac{1mm}{2}}$

$\bf \implies \green {Pitch=0.5mm}$

__________________________________

$\sf \orange {Now,}$

$\bf \mapsto {Least\: Count=\frac{pitch}{No.\: of\: divisions}}$

$\bf \mapsto {Least\: Count=\frac{0.5mm}{20}}$

$\bf \mapsto {Least\: Count=\frac{5mm}{20\times10}}$

$\bf \mapsto {Least\: Count=\frac{5mm}{200}}$

$\bf \mapsto {Least\: Count=\frac{\cancel{5}^{1}mm}{\cancel{200}_{40}}}$

$\bf \mapsto {Least\: Count=\frac{1mm}{40}}$

$\bf \implies \green {\fbox{Least\: Count=0.025mm}}$

__________________________________

$\sf \purple {Thence,}$

▸The least count of the screw gauge is 0.025mm.

Answer 2

Explanation:

hope it helps you ok mark me brainlist