Experimental Physics

JEE Main level, single correct, medium

58 of 70

Question 58 of 70

In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of 0.250.25 cm\mathrm{cm}. There are 100 circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of 4 main scale divisions and 30 circular scale divisions. Assuming negligible zero error, the thickness of the wire is:

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Answer: (D) 0.2150 cm0.2150 \mathrm{~cm}

Least count
= Value of 1 part on main scale  Number of parts on vernier scale =\frac{\text { Value of } 1 \text { part on main scale }}{\text { Number of parts on vernier scale }}
=0.255×100 cm=5×10−4 cm=\frac{0.25}{5 \times 100} \mathrm{~cm}=5 \times 10^{-4} \mathrm{~cm}
Reading =4×0.05 cm+30×5×10−4 cm=4 \times 0.05 \mathrm{~cm}+30 \times 5 \times 10^{-4} \mathrm{~cm} =(0.2+0.0150)cm=0.2150 cm=(0.2+0.0150) \mathrm{cm}=0.2150 \mathrm{~cm} (Thickness of wire)

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