The pitch of screw gauge is 1 mm and its circular scale is divided into 50 divisions. When nothing is put between the studs $44^{t h}$ division of circular scale coincides with the reference line and circular scale lies between $3^{r d}$ and $4^{t h}$ division of main scale. When a glass plate is placed between the studs, the circular scale lies between $13^{t h}$ and $14^{t h}$ division of main scale and circular scale reads 26 divisions. The thickness of the plate is 2.41 K (in mm). Find the value of K.

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answer is 4.

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Detailed Solution

$E r r o r = + 3.88 l = 13.52 L_{c o r r u} = 13.52 - 3.88 = 9.64$

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