There are $100$ divisions on the circular scale of a screw gauge of pitch $1 \mathrm{~mm}$. With no measuring quantity in between the jaws, the zero of the circular scale lies $5$ divisions below the reference line. The diameter of a wire is then measured using this screw gauge. It is found the $4$ linear scale divisions are clearly visible while $60$ divisions on circular scale coincide with the reference line. The diameter of the wire is :
A$4.65 \mathrm{~mm}$
B$4.55 \mathrm{~mm}$
C$4.60 \mathrm{~mm}$
D$3.35 \mathrm{~mm}$
JEE MAIN 2024, Diffcult
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B$4.55 \mathrm{~mm}$
b $\text { Least count }=\frac{1}{100} \mathrm{~mm}=0.01 \mathrm{~mm}$
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