Experimental Physics

JEE Main level, single correct, medium

57 of 70

Question 57 of 70

In a screw gauge, 55 complete rotations of the screw cause it to move a linear distance of 0.25cm0.25cm. There are100100 circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of 44 main scale divisions and 3030 circular scale divisions. Assuming negligible error, the thickness of the wire is

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Answer: (D) 0.2150cm0.2150cm

The Least count=0.255×100cm=5×10−4cm=\dfrac{0.25}{5\times 100}cm=5\times {10}^{-4}cm.
Reading =4×0.05cm+30×5×10−4cm=4\times 0.05cm+30\times 5\times {10}^{-4}cm
=(0.2+0.0150)cm=0.2150cm=\left(0.2+0.0150\right)cm=0.2150cm

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