The diameter of a rod was found to be 23.1 mm using a standrad vernier calliper.The MSR and VCD on the another vernier calliper in which 1 MSD =0.6mm and least count is 0.06mm are______ and _________.

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HERE'S THE ANS

ok try to find out VSD devisions which you can find out very simply.... least count = MS count/ total VSD so you get total vernier scale division as 10 now its given that for new caliper main scale division is 0.6 mm . Standrd caliper has value 1 mm so if you divided 23.1 by 0.6 you will get 38.5 38 comes from main scale division  so 0.5 has to come from VSD which point on VCD is nearer to any MSD? so divide .5 by 0.06 that will be VCD

Solution:

( 23.1) / ( 0.6 )  =  (38 * 0.6 ) + 0.3

i.e. MSD = 38

Now VSD =  0.3 / 0.06 = 5

Thus MSD = 38      &      VSD   = 5