JEE Advanced2022PhysicsExperimental PhysicsActual
Area of the cross-section of a wire is measured using a screw gauge. The pitch of the main scale is 0 . 5 mm . The circular scale has 100 divisions and for one full rotation of the circular scale, the main scale shifts by two divisions. The measured readings are listed below. Measurement condition Main scale reading Circular scale reading Two arms of gauge touching each other without wire 0 division 4 divisions Attem
Options
- A2 . 22 ± 0 . 02 mm ,   π 1 . 23   ±   0 . 02 mm 2
- B2 . 22 ± 0 . 01 mm , π 1 . 23 ± 0 . 01 mm 2
- C2 . 14 ± 0 . 02 mm , π 1 . 14 ± 0 . 02 mm 2
- D2 . 14 ± 0 . 01 mm , π 1 . 14 ± 0 . 01 mm 2
Correct answer
C. 2 . 14 ± 0 . 02 mm , π 1 . 14 ± 0 . 02 mm 2
Step-by-step solution
Note: This question is given bonus by JEE council. In one rotation 2 divisions of the main scale are crossed. Therefore, the least count of the screw gauge is L C = 2 × 0 . 5 100 = 0 . 01   mm And the zero error is 4 × L C = 0 . 04   mm Reading-1 R 1 = MSR + LC × CSR - Zero   error = 2 + 0 . 20 - 0 . 04 mm = 2 . 16 mm Reading-2 R 2 = 2 + 0 . 16 - 0 . 04 mm = 2 . 12 mm Therefore, average reading R m = R 1 + R 2 2 = 2 . 14   mm Average mean error = R m - R 1 + R m - R 2 2 = 0 . 02 &