The pitch of screw gauge is 0.75 mm and there are 50 divisions on the circular scale. When nothing is put between the studs it shows the following where zero of mains scale is not visible. Now thickness of a metal plate is measured and when the plate is put between the studs it shows the following. The thickness of metal plate is (in mm) :

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answer is 4.2.

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Detailed Solution

$L C = \frac{p i t c h}{n o . o f c i r c u l a r d i v i s i o n s} = \frac{0.75}{50} = 0.015 m m$
Negative zero error.
20 divisions $\times$ LC
$\Rightarrow 20 \times 0.015 = (- 0.3)$
Which metal is put
Measured value = 3.75 mm + 10 divisions LC
$= 3.75 m m + 10 \times 0.015 = 3.90 m m$
$T V = M V - (z e r o e r r o r)$
$= 3.9 - (- 0.3) = 4.20 m m$

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