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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 $23^{rd}$ 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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7.23 cm

7.023 cm

7.073 cm

7.73 cm

answer is B.

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Detailed Solution

1MSD = 0.05 cm = 0.5 mm

50 VSD = 49 MSD

$1 VSD = \frac{49}{50} \times 0 .5 mm = 0 .49 mm$

L.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 + 0.023
$T R = 7.023 c m$

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