Question

In a vernier Calliper 1 MSD = 0.1 cm and the Least Count is 0.05mm. If the reading on Vernier calliper while taking measurement is 12.8mm , then the Vernier coinciding division (VCD) is ?

Answer 1

Given,

1 Main Scale division (1 MSD) = 0.1cm

Least count(L.C) = 0.05mm

Reading on Vernier Calliper = 12.8mm

Now, we know =>

Vernier Scale reading= Main Scale division reading + (L.C)×(V.S.D)

where, VSR = Vernier Scale Division and

LC = Least Count.

So we get=>

12.8 =( Main Scale Reading)×1mm + 0.05×VSD

Now, here main scale reading = 12

So we get->

=>12.8 = 12 + 0.05×VSD

=>0.8 = 0.05×VSD

=> $\dfrac{0.8}{0.05}$= VSD

=>VSD = 16

Hence the Vernier Scale decision or Vernier coniciding divisions is 16.

Answer 2

ANSWER:------------

count(L.C)

= 0.05mm

Reading on Vernier Calliper = 12.8mm

we know =>Vernier Scale reading=

division reading + (L.C)×(V.S.D) VSR = Vernier Scale Division andLC = Least Count

we get=>12.8 =

( Main Reading)×1mm +

0.05×VSR

12So we get-

=>12.8 = 12 + 0.05×VSD=>0.8 =

0.05×VSD

=> 0.80.05

{0.8}{0.05}

0.050.8 =

VSD=>VSD =

16

Vernier coniciding divisions is 16.

Hence proved:)

hope it helps:-

T!—!ANkS!!!