The diameter of a cylinder is measured using a Vernier calipers with no zero error. If is found that the zero of the Vernier scale lies between $$5.10cm$$ and $$5.15cm$$ of the main scale. The Vernier scale has $$50$$ divisions equivalent to $$2.45cm$$. The $$24division$$ of the Vernier scale exactly coincides with one of the main scale divisions. The diameter of cylinder is

Question

Infinity Learn · Accepted Answer

value of $$1$$ Vernier Scale $$Division=$$ $$1$$ $$VSD=2.45cm50=0.049cm$$ value of $$1$$ Main Scale $$Division=1MSD$$ $$=5.15$$−$$5.10=0.05cm$$ Least $$Count=L$$.C  $$=$$ value of $$1MSD$$ −value $$of1VSD$$ $$=0.05$$−$$0.049$$                                         $$=0.001cm$$ given Vernier $$Coincidence=$$ $$VC=24$$ Diameter of wire $$=MSD+VC$$×L.C                                $$=5.10+24$$×$$0.001$$                                $$=5.10+0.024$$ Diameter of wire $$=5.124cm$$

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