Experimental Physics

JEE Main level, numerical answer, medium

24 of 70

Question 24 of 70

In a screw gauge, there are 100100 divisions on the circular scale and the main scale moves by 0.5mm0.5mm on a complete rotation of the circular scale. The zero of circular scale lies 66 divisions below the line of graduation when two studs are brought in contact with each other. When a wire is placed between the studs, 44 linear scale divisions are clearly visible while 46th division the circular scale coincide with the reference line. The diameter of the wire is ______ ×10−2mm.\times {10}^{-2}mm.

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Answer: 220

Least count of screw gauge is=PitchNo.ofcirculardivisions=\dfrac{Pitch}{No.ofcirculardivisions}
=0.5mm100=\dfrac{0.5mm}{100}
Least count, LC=5×10−3mmLC=5\times {10}^{-3}mm
The zero of circular scale lies 66 divisions below the line of graduation. So, zero error is positive.
Positive zero error =MSR+CSR(LC)=MSR+CSR(LC)
=0mm+6(5×10−3mm)=0mm+6\left(5\times {10}^{-3}mm\right)
Diameter of wire, d=MSR+CSR(LC)–Positivezeroerrord=MSR+CSR(LC)–Positivezeroerror
=4×0.5mm+46(5×10−3)−6(5×10−3)mm=4\times 0.5mm+46\left(5\times {10}^{-3}\right)-6\left(5\times {10}^{-3}\right)mm
=2mm+40×5×10−3mm=2.2mm=2mm+40\times 5\times {10}^{-3}mm=2.2mm
Hence, diameter of wire is 220×10−2mm220\times {10}^{-2}mm.

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