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 :
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answer is 2.
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Detailed Solution
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