Experimental Physics

JEE Main level, single correct, medium

42 of 70

Question 42 of 70

The pitch of the screw gauge is 1mm1mm and there are 100100 divisions on the circular scale. When nothing is put in between the jaws, the zero of the circular scale lies 88 divisions below the reference line. When a wire is placed between the jaws, the first linear scale division is clearly visible while 72nd{72}^{\text{nd}} division on circular scale coincides with the reference line. The radius of the wire is

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Answer: (B) 0.82mm0.82mm

Least count =1mm100=0.01mm=\dfrac{1mm}{100}=0.01mm
zero error =+8×LC=+0.08mm=+8\times LC=+0.08mm
True reading (Diameter)
=(1mm+72×LC)−=\left(1mm+72\times LC\right)- (Zero error)
=(1mm+72×0.01mm)−0.08mm=\left(1mm+72\times 0.01mm\right)-0.08mm
=1.72mm−0.08mm=1.72mm-0.08mm
=1.64mm=1.64mm
therefore, radius =1.642=0.82mm.=\dfrac{1.64}{2}=0.82mm.

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