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Units and Measurements question

2020 · 3 Sep · Shift 1 · Q65
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Units and Measurements question

2020 · 3 Sep · Shift 1 · Q65

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
Using screw gauge of pitch 0.1 cm and 50 divisions on its circular scale, the thickness of an object is measured. It should correctly be recorded as
  1. A
    2.123 cm
  2. B
    2.124 cm
  3. C
    2.125 cm
  4. D
    2.121 cm
View written solutionFree

Correct answer: B

  1. Find the least count of the screw gauge

Given:

  • Pitch =0.1 cm= 0.1\ \text{cm}=0.1 cm
  • Number of circular scale divisions =50= 50=50

Least count is

L.C.=PitchNumber of divisions=0.150=0.002 cm\text{L.C.} = \frac{\text{Pitch}}{\text{Number of divisions}} = \frac{0.1}{50} = 0.002\ \text{cm}L.C.=Number of divisionsPitch​=500.1​=0.002 cm
  1. Rule for recording a measurement

A screw gauge measurement should be recorded up to its least count. Since the least count is 0.002 cm0.002\ \text{cm}0.002 cm, the measured value must differ in steps of 0.002 cm0.002\ \text{cm}0.002 cm.

So valid readings can be like:

…, 2.120, 2.122, 2.124, 2.126, …\ldots,\ 2.120,\ 2.122,\ 2.124,\ 2.126,\ \ldots…, 2.120, 2.122, 2.124, 2.126, …
  1. Check each option

  • A: 2.123 cm2.123\ \text{cm}2.123 cm

    Not a multiple of 0.002 cm0.002\ \text{cm}0.002 cm from a proper base reading, so not correctly recorded.

  • B: 2.124 cm2.124\ \text{cm}2.124 cm

    This matches the least count step of 0.002 cm0.002\ \text{cm}0.002 cm. Correct.

  • C: 2.125 cm2.125\ \text{cm}2.125 cm

    Not consistent with least count 0.002 cm0.002\ \text{cm}0.002 cm. Incorrect.

  • D: 2.121 cm2.121\ \text{cm}2.121 cm

    Also not consistent with least count 0.002 cm0.002\ \text{cm}0.002 cm. Incorrect.

  1. Final answer

Therefore, the thickness should be recorded as

2.124 cm\boxed{2.124\ \text{cm}}2.124 cm​

So the correct option is B.

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