Question

A screw gauge has least count of 0.01 mm and
there are 50 divisions in its circular scale.​

Answer 1

Answer :

Least count of screw gauge = 0.01mm

No of divisons on circular scale = 50

We have to find pitch of screw gauge$.$

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◈ Least count of screw guage is given by

$\dag\:\boxed{\bf{LC=\dfrac{pitch(p)}{no.\:of\:divisons\:on\:circular\:scale}}}$

⇒ LC = p / n

⇒ p = LC × n

⇒ p = 0.01 × 50

⇒ p = 50/100

⇒ p = 0.5 mm

Learn More :

Least count of vernier scale is given by

• LC = MSD - VSD

MSD denotes main scale divisons

VSD denotes vernier scale divisons

Answer 2

Step-by-step explanation:

least count 0.01

number of divisions50

lc=P/N

P= 0.01×50

P=0.5NM