In a screw gauge, there are 100 divisions on the circular scale and the main scale moves by 0.5 mm on a complete rotation of the circular scale. The zero of circular scale lies 6 divisions below the line of graduation when two studs are brought in contact with each other. When a wire is placed between the studs, 4 linear scale divisions are clearly visible while 46 text th division the circular scale coincide with the reference line. The diameter of the wire is × 10-2 mm.

Question

In a screw gauge, there are 100 divisions on the circular scale and the main scale moves by 0.5 mm on a complete rotation of the circular scale. The zero of circular scale lies 6 divisions below the line of graduation when two studs are brought in contact with each other. When a wire is placed between the studs, 4 linear scale divisions are clearly visible while 46 text th division the circular scale coincide with the reference line. The diameter of the wire is × 10-2 mm. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Least count =( text Pitch / text No. of circular divisions ) =(0.5 mm /100) Least count =5 × 10-3 mm Positive Error = MSR + CSR ( LC ) =0 mm +6(5 × 10-3 mm ) Reading of Diameter = MSR + operatornameCSR( LC )- Positive zero error =4 × 0.5 mm +(46(5 × 10-3))-6(5 × 10-3) mm =2 mm +40 × 5 × 10-3 mm =2.2 mm

Least count = No. of circular divisions Pitch ​ =1000.5mm​ Least count =5×10−3mm Positive Error =MSR+CSR(LC) =0mm+6(5×10−3mm) Reading of Diameter = MSR +CSR(LC)− Positive zero error =4×0.5mm+(46(5×10−3))−6(5×10−3)mm =2mm+40×5×10−3mm=2.2mm