A student performs an experiment of measuring the thickness of a slab with a vernier calliper whose $$50$$ divisions of the vernier scale are equal to $$49$$ divisions of the main scale. He noted that zero of the vernier scale is between $$7.00$$ cm and $$7.05$$ cm mark of the main scale and $$23rd$$ division of the vernier scale exactly coincides with the main scale. The measured value of the thickness of the given slab using the calliper will be :

Question

A student performs an experiment of measuring the thickness of a slab with a vernier calliper whose $$50$$ divisions of the vernier scale are equal to $$49$$ divisions of the main scale. He noted that zero of the vernier scale is between $$7.00$$ cm and $$7.05$$ cm mark of the main scale and $$23rd$$ division of the vernier scale exactly coincides with the main scale. The measured value of the thickness of the given slab using the calliper will be :

Infinity Learn · Accepted Answer

$$1MSD$$ $$=$$ $$0.05$$ cm $$=$$ $$0.5$$ $$mm50$$ VSD $$=$$ $$49$$ MSD $$1$$ $$VSD=4950$$  ×  $$0.5$$  mm  $$=$$  $$0.49$$  mmL.C $$=$$ $$1MSD$$ – $$1VSD$$       $$=$$ $$0.5$$ mm – $$0.49$$ mm       $$=$$ $$0.01$$ mm $$=$$ $$0.001$$ cm T.R $$=$$ MSR $$+$$ V.C. x L.C.$$=$$ $$7.00$$ cm $$+$$ $$23$$ x $$0.001$$ $$cm=$$ $$7.00$$ $$+$$ $$0.023TR=7.023$$   cm

Detailed Solution

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