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Question: In a screw gauge, there are 100 divisions in circular scale and each main scale division is of 1 mm....

In a screw gauge, there are 100 divisions in circular scale and each main scale division is of 1 mm. When there is no gap between the jaws, 97th97^{th} divisions coincides with the main scale zero and zero of main scale is not visible. While measuring the diameter of a ball, the circular scale is between 3 mm mark and 4 mm mark such that the 76th76^{th} division of circular scale coincides with the reference line. Select the correct alternative

A

the least count of the micrometer is 0.01 cm

B

the zero error is -0.04 mm

C

the diameter of the ball is 3.79 cm

D

the main scale reading is 4 mm

Answer

the diameter of the ball is 3.79 cm

Explanation

Solution

Solution:

  1. Least count (LC):
    Since one main scale division = 1 mm and there are 100 divisions on the circular scale,
    LC=1mm100=0.01mm=0.001cmLC = \frac{1 mm}{100} = 0.01 mm = 0.001 cm.

  2. Zero error determination:
    When the jaws touch there should be no gap, so the correct reading ought to be 0. However, the 97th division of the circular scale is seen at the reference line.
    Using the standard rule for a “non‐visible zero” (case II), the zero error is given by
    Zero error=[(Total divisionsobserved division)×LC]Zero\ error = –[(Total\ divisions – observed\ division) × LC]
    =[(10097)×0.01]=0.03 mm= –[(100 – 97) × 0.01] = –0.03\ mm.

  3. Measuring the ball’s diameter:
    When measuring the ball, the reading is taken as:
    Measured Reading=Main Scale Reading+(Circular Scale Reading×LC)Measured\ Reading = Main\ Scale\ Reading + (Circular\ Scale\ Reading × LC).
    Since the reading “lies between the 3 mm mark and 4 mm mark,” the main scale reading is taken as 3 mm.
    Circular reading = 76 divisions, so
    Contribution=76×0.01 mm=0.76 mmContribution = 76 × 0.01\ mm = 0.76\ mm.
    Thus, observed reading = 3 + 0.76 = 3.76 mm.
    Now, correcting for zero error:
    True reading=3.76(0.03)=3.76+0.03=3.79 mm=0.379cmTrue\ reading = 3.76 – (–0.03) = 3.76 + 0.03 = 3.79\ mm = 0.379 cm.

  4. Checking Options:

    • Option 1 states “least count is 0.01 cm” – Incorrect.
    • Option 2 gives “zero error is –0.04 mm” which does not agree with the computed –0.03 mm.
    • Option 3 gives “the diameter of the ball is 3.79 cm” which exactly matches our corrected reading.
    • Option 4 “the main scale reading is 4 mm” is not consistent with the reading between 3 mm and 4 mm.

Thus, the only correct alternative is option 3.