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Units and Measurements question

2021 · 25 Feb · Shift 1 · Q56
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Units and Measurements question

2021 · 25 Feb · Shift 1 · Q56

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
The pitch of the screw gauge is 1 mm and there are 100 divisions on the circular scale. When nothing is put in between the jaws, the zero of the circular scale lies 8 divisions below the reference line. When a wire is placed between the jaws, the first linear scale division is clearly visible while 72nd division on circular scale coincides with the reference line. The radius of the wire is :
  1. A
    1.80 mm
  2. B
    0.90 mm
  3. C
    0.82 mm
  4. D
    1.64 mm
View written solutionFree

Correct answer: C

  1. Find the least count of the screw gauge

Given:

  • Pitch =1 mm= 1\text{ mm}=1 mm
  • Number of circular scale divisions =100= 100=100

So, least count is

L.C.=PitchNo. of divisions=1100 mm=0.01 mm\text{L.C.} = \frac{\text{Pitch}}{\text{No. of divisions}} = \frac{1}{100}\text{ mm} = 0.01\text{ mm}L.C.=No. of divisionsPitch​=1001​ mm=0.01 mm
  1. Determine the zero error

When nothing is between the jaws, the zero of the circular scale lies 8 divisions below the reference line.

This means the instrument shows a positive reading even when the true reading is zero, so the zero error is positive.

Magnitude of zero error:

8×0.01=0.08 mm8 \times 0.01 = 0.08\text{ mm}8×0.01=0.08 mm

Hence, zero correction is

−0.08 mm-0.08\text{ mm}−0.08 mm
  1. Find the observed reading for the wire
  • The first linear scale division is clearly visible ⇒\Rightarrow⇒ main scale reading =1.00 mm= 1.00\text{ mm}=1.00 mm
  • Circular scale reading =72= 72=72 divisions

So circular scale contribution is

72×0.01=0.72 mm72 \times 0.01 = 0.72\text{ mm}72×0.01=0.72 mm

Observed diameter of wire:

1.00+0.72=1.72 mm1.00 + 0.72 = 1.72\text{ mm}1.00+0.72=1.72 mm
  1. Apply zero correction

True diameter:

1.72−0.08=1.64 mm1.72 - 0.08 = 1.64\text{ mm}1.72−0.08=1.64 mm
  1. Find the radius

Radius is half of diameter:

r=1.642=0.82 mmr = \frac{1.64}{2} = 0.82\text{ mm}r=21.64​=0.82 mm
  1. Check options
  • A: 1.80 mm1.80\text{ mm}1.80 mm ✗
  • B: 0.90 mm0.90\text{ mm}0.90 mm ✗
  • C: 0.82 mm0.82\text{ mm}0.82 mm ✓
  • D: 1.64 mm1.64\text{ mm}1.64 mm ✗

Therefore, the correct option is C.

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