A screw gauge with a pitch of $0.5 m m$ and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the 45^th division coincides with the main scale line and that the zero of the main scale is barely visible. The thickness of the sheet if the main scale reading is $0.5 m m$ and the 25^th division coincides the main scale line_____ $\times 10^{- 2} m m .$

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answer is 80.

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Detailed Solution

Least count $L C = \frac{p i t c h}{n o . o f d i v . o n c i r c u l a r s c a l e}$
$\Rightarrow L C = \frac{0.5 m m}{50} = 0.01 m m$
When jaws are closed, the zero error will be
= main scale reading + (circular scale reading) (Least count)
= -0.5 mm+(45) (0.01)
Hence zero error $e = - 0.05 m m .$
When the sheet is placed between the jaws:
Measured thickness = 0.5mm+(25) (0.01)=0.75mm
Hence actual thickness or true reading
= observed reading +zero error
= 0.75 mm-(-0.05)=0.80mm
$\Rightarrow 80 \times 10^{- 2} m m$