In a screw gauge,there are 100 divisions on t | vedclass

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(D) The pitch of the screw gauge is $0.5\,mm$ and the number of circular divisions is $100$.$	ext{Least Count (LC)} = \frac{	ext{Pitch}}{	ext{Total

Vedclass · Accepted Answer

$22$

Vedclass · Answer

(D) The pitch of the screw gauge is $0.5\,mm$ and the number of circular divisions is $100$.$	ext{Least Count (LC)} = \frac{	ext{Pitch}}{	ext{Total circular divisions}} = \frac{0.5\,mm}{100} = 5 	imes 10^{-3}\,mm$.Since the zero of the circular scale is $6$ divisions below the reference line,the zero error is positive.$	ext{Zero Error} = +6 	imes 	ext{LC} = 6 	imes 5 	imes 10^{-3}\,mm = 30 	imes 10^{-3}\,mm$.The observed reading is $MSR + (CSR 	imes LC) = 4 	imes 0.5\,mm + 46 	imes 5 	imes 10^{-3}\,mm = 2.0\,mm + 0.23\,mm = 2.23\,mm$.The corrected diameter is $	ext{Observed Reading} - 	ext{Zero Error} = 2.23\,mm - 0.03\,mm = 2.20\,mm$.Converting to the required format: $2.20\,mm = 220 	imes 10^{-2}\,mm$. Wait,checking the calculation: $2.20\,mm = 220 	imes 10^{-2}\,mm$. Given the options,the value is $220 	imes 10^{-2}\,mm$. The question asks for the value in the blank,which is $220$.

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