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NVAMediumJEE 2023Vernier Calipers & Screw Gauge

JEE Physics 2023 Question with Solution

In a screw gauge, there are 100100 divisions on the circular scale and the main scale moves by 0.5 mm0.5 \, \text{mm} on a complete rotation of the circular scale. The zero of circular scale lies 66 divisions below the line of graduation when two studs are brought in contact with each other. When a wire is placed between the studs, 44 linear scale divisions are clearly visible while the **4646**th division the circular scale coincide with the reference line. The diameter of the wire is _____ ×10−2\times 10^{-2} mm.

Answer

Correct answer:220

Step-by-step solution

Standard Method

Given: The screw gauge has 100100 circular scale divisions and pitch 0.5 mm0.5 \, \text{mm}. Zero error is positive because the zero of circular scale lies 66 divisions below the reference line when the studs touch. With the wire inserted, main scale reading is 44 linear divisions and circular scale reading is 4646 divisions.

Find: The diameter of the wire.

First find the least count:

Least count=PitchNumber of circular divisions=0.5100 mm=5×10−3 mm\text{Least count} = \frac{\text{Pitch}}{\text{Number of circular divisions}} = \frac{0.5}{100} \, \text{mm} = 5 \times 10^{-3} \, \text{mm}

The positive zero error is:

Zero error=6×5×10−3 mm\text{Zero error} = 6 \times 5 \times 10^{-3} \, \text{mm}

Observed reading of the wire is:

Observed reading=MSR+CSR×LC\text{Observed reading} = \text{MSR} + \text{CSR} \times \text{LC} =4×0.5 mm+46×5×10−3 mm= 4 \times 0.5 \, \text{mm} + 46 \times 5 \times 10^{-3} \, \text{mm} =2 mm+0.23 mm= 2 \, \text{mm} + 0.23 \, \text{mm}

Since the zero error is positive, subtract it:

Diameter=2 mm+(46−6)×5×10−3 mm\text{Diameter} = 2 \, \text{mm} + (46-6) \times 5 \times 10^{-3} \, \text{mm} =2 mm+40×5×10−3 mm= 2 \, \text{mm} + 40 \times 5 \times 10^{-3} \, \text{mm} =2.2 mm= 2.2 \, \text{mm}

The solution states: The correct answer is 2.22.2. Since the question asks for the value in the form _____ ×10−2\times 10^{-2} mm,

2.2 mm=220×10−2 mm2.2 \, \text{mm} = 220 \times 10^{-2} \, \text{mm}

Therefore, the required numerical answer is 220220.

Reading and zero-correction interpretation

Given: The main scale moves 0.5 mm0.5 \, \text{mm} in one complete rotation, so that is the pitch. There are 100100 circular divisions.

Find: How the final answer is obtained in the asked format.

A complete rotation corresponds to 0.5 mm0.5 \, \text{mm}, so one circular division corresponds to:

0.5100 mm=0.005 mm\frac{0.5}{100} \, \text{mm} = 0.005 \, \text{mm}

When the screw is closed, the zero lies below the graduation line by 66 divisions, so the instrument shows an extra reading. This is a positive zero error of:

6×0.005=0.03 mm6 \times 0.005 = 0.03 \, \text{mm}

With the wire in place, 44 linear scale divisions are visible. Using the working shown in the source solution, this gives main scale reading:

4×0.5=2.0 mm4 \times 0.5 = 2.0 \, \text{mm}

The circular scale contribution is:

46×0.005=0.23 mm46 \times 0.005 = 0.23 \, \text{mm}

Hence observed reading is:

2.0+0.23=2.23 mm2.0 + 0.23 = 2.23 \, \text{mm}

Corrected diameter is:

2.23−0.03=2.20 mm2.23 - 0.03 = 2.20 \, \text{mm}

Now convert to the requested form:

2.20 mm=220×10−2 mm2.20 \, \text{mm} = 220 \times 10^{-2} \, \text{mm}

So the final answer is 220220.

Common mistakes

  • Using the zero error with the wrong sign. Here the zero lies below the reference line when the studs touch, so the zero error is positive and must be subtracted from the observed reading.

  • Taking least count as 1000.5\frac{100}{0.5} instead of 0.5100\frac{0.5}{100}. Least count is pitch divided by the number of circular divisions, not the reverse.

  • Reporting 2.22.2 directly as the answer and ignoring the requested format. The question asks for the number multiplying 10−2 mm10^{-2} \, \text{mm}, so 2.2 mm=220×10−2 mm2.2 \, \text{mm} = 220 \times 10^{-2} \, \text{mm}.

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