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A screw gauge with a pitch of 0.5mm a circuler scale with 50 divisions is used to measure thickness of a sheet. Before starting the measurement, it is found that when two jaws of the screw gauge are brought in contact, the $45^{t h}$ division coincides main scale line and zero of main scale is barely visible . what is the thickness of sheet if the mains reading is 0.5mm and $25^{t h}$ division coincides with mains scale line ?

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0.50 m

0.75mm

0.80 mm

0.70 mm

answer is C.

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Detailed Solution

$LC = \frac{pitch}{circular scale divisions} = \frac{0.5}{50} mm = 0.01 mm$

$N e g a t i v e z e r o e r r o r = - 5 \times (0.01) mm = - 0.05 mm$

$M . V . = P . S . R + [(H . S . R) \times L . C]$

$M e a s u r e d v a l u e = 0.5 mm + 25 \times 0.01 - (- 0.05) mm = 0.8 mm$

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