2020 A/L Physics Past Paper · Question 02 (English Medium)

Micrometer screw gauge: the diameter of a sphere with a zero error

The scale of a micrometer screw gauge when its spindle and anvil just touch is shown in figure (a). The scale when a metal sphere is placed exactly between the spindle and the anvil is shown in figure (b). The pitch of the screw is 0.50.5 mm, and the circular scale is divided into 5050 equal divisions.

What is the correct diameter of the metal sphere?

  1. (1) 3.283.28 mm
  2. (2) 3.313.31 mm
  3. (3) 3.783.78 mm
  4. (4) 3.813.81 mm
  5. (5) 3.843.84 mm
Answer and explanation

Answer: (3) 3.783.78 mm

Given information

We know:

The pitch is p=0.5p = 0.5 mm.

The circular scale has N=50N = 50 divisions.

Figure (a): at the touch, the circular scale stands at its 33rd division on the main line.

Figure (b): the main scale shows 3.53.5 mm (the 33 mm mark is visible above, the next half-millimetre mark below).

The circular scale stands at its 3131st division.

What the question asks

We want: the correct diameter of the metal sphere, once the zero error is corrected.

Relevant theory

By rotating one division of the circular scale the distance moved by the screw is the least count.

This error is called the zero error of the screw gauge.

for the correction this value should be deducted from the relevant reading

Required equations

LC=pNLC = \frac{p}{N}

R=m+c×LCR = m + c \times LC

e=z×LCe = z \times LC

D=R−eD = R - e

Applying it to this question

Plan:

First find the least count.

Then find the zero error from figure (a).

Then find the reading on the sphere from figure (b).

Finally subtract the zero error from the reading.

Reasoning and calculation

0.010.01

LC=0.550LC = \frac{0.5}{50}

+0.03+0.03

e=3×0.01e = 3 \times 0.01

3.53.5

m=3.5 mmm = 3.5\ \text{mm}

0.310.31

c×LC=31×0.01c \times LC = 31 \times 0.01

3.813.81

R=3.5+0.31R = 3.5 + 0.31

3.783.78

D=3.81−0.03D = 3.81 - 0.03

Checks

A quick check. Every in-between value is in mm.

LC=0.01LC = 0.01 mm, e=0.03e = 0.03 mm, R=3.81R = 3.81 mm, D=3.78D = 3.78 mm.

The units match, so we can subtract directly.

Does it make sense?

The zero error is positive, so the gauge over-reads every measurement by 0.030.03 mm.

So the correct diameter should be smaller than the reading.

3.78<3.813.78 < 3.81 ✓.

m+c∗p/N−z∗p/Nm + c * p / N - z * p / N

Conclusion

D=3.81−0.03=3.78 mmD = 3.81 - 0.03 = 3.78\ \text{mm}

We subtract the zero error of +0.03+0.03 mm from the reading of 3.813.81 mm.

The sphere's correct diameter is 3.783.78 mm.

The answer is (3).

Reading the main scale as 33 mm without seeing the half-millimetre mark on the lower row of figure (b).

The upper marks are full millimetres, the lower ones are half-millimetres. The last mark showing before the edge of the thimble is 3.53.5 mm.

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