Question

Find the least count for a given screw gauge, If pitch is 0.1mm and there are 100 divisions on the circular scale

Answer 1

Given:

Pitch = 0.1 mm

Number of divisions on circular scale = 100

To Find:

Least count of screw gauge (L.C.)

Answer:

Least count of screw gauge is given by:

$\bf \leadsto L.C. = \dfrac{Pitch/Main \ scale \ division \ value}{Number \ of \ divisions \ on \ circular \ scale} \\ \\ \rm \leadsto L.C. = \dfrac{0.1}{100} \\ \\ \rm \leadsto L.C. = 0.001 \: mm$

$\therefore$ Least count of screw gauge (L.C.) = 0.001 mm

Answer 2

Answer:

• The least count for a given screw gauge is 10 $\mu$ m.

Explanation:

Given that,

• Pitch (P) = 0.1 mm
• Number of divisions on circular scale (N) = 100.

As we know that,

$\purple\bigstar$ L.C = P/N

[ Putting values ]

↪ L.C = 0.1mm/100

↪ L.C = 0.001mm

↪ L.C = 10 $\mu$ m. $\red \bigstar$