Student A and student B used two screw gauges of equal pitch and 100 equal circular divisions to measure the radius of a given wire. The actual value of the radius of the wire is 0.322 cm. The absolute value of the difference between the final circular scale readings observed by the students A and B is ___.

Given pitch = 0.1 cm.

We need to determine the absolute value of the difference between the final circular scale readings observed by student A and student B when measuring the radius of a wire.

1. Calculate the Least Count (LC) of the Screw Gauges

From the problem page, both screw gauges have the same pitch and total number of circular divisions:

• Pitch = $$0.1\text{ cm}$$
• Number of circular divisions = $$100$$

The least count ($$\text{LC}$$) is calculated as:

$$\text{LC} = \frac{\text{Pitch}}{\text{Number of circular divisions}} = \frac{0.1\text{ cm}}{100} = 0.001\text{ cm}$$

2. Determine the Initial Zero Errors

Looking at the provided reference diagrams, we can determine the initial zero error for each student's instrument based on where the zero mark lies relative to the reference line:

• Screw Gauge A:
  The $$5^{\text{th}}$$ division coincides with the reference line when the studs are closed, meaning it has a positive zero error:

  $$\text{Zero Error}_A = +5 \times \text{LC} = +5 \times 0.001\text{ cm} = +0.005\text{ cm}$$

• Screw Gauge B:
  The $$92^{\text{nd}}$$ division coincides with the reference line. Since it is below the zero mark ($$100 - 92 = 8$$ divisions behind), it has a negative zero error:

  $$\text{Zero Error}_B = -(100 - 92) \times \text{LC} = -8 \times 0.001\text{ cm} = -0.008\text{ cm}$$

3. Relate Observed Readings to Actual Value

The true radius value is given as $$\text{True Reading} = 0.322\text{ cm}$$. The general formula for a measurement is:

$$\text{True Reading} = \text{Observed Reading} - \text{Zero Error}$$

$$\text{True Reading} = (\text{Main Scale Reading} + n \times \text{LC}) - \text{Zero Error}$$

Where $$n$$ represents the final circular scale reading. Let's express this for both students:

• For Student A:
  

  $$0.322 = \text{MSR}_A + (n_A \times 0.001) - 0.005$$

  $$n_A \times 0.001 = 0.327 - \text{MSR}_A$$

• For Student B:
  

  $$0.322 = \text{MSR}_B + (n_B \times 0.001) - (-0.008)$$

  $$n_B \times 0.001 = 0.314 - \text{MSR}_B$$

Since both gauges have a pitch of $$0.1\text{ cm}$$, the pitch/main scale reading ($$\text{MSR}$$) must be identical for measuring the same object ($$\text{MSR}_A = \text{MSR}_B = 0.3\text{ cm}$$). Let's solve for the exact circular scale divisions ($n_A$$ and $$n_B$$):

$$n_A \times 0.001 = 0.327 - 0.3 = 0.027 \implies n_A = 27$$

$$n_B \times 0.001 = 0.314 - 0.3 = 0.014 \implies n_B = 14$$

4. Calculate the Absolute Difference

Finally, find the absolute difference between the two circular scale readings ($$n_A$$ and $$n_B$$):

$$\text{Difference} = |n_A - n_B| = |27 - 14| = 13$$

Conclusion

The absolute value of the difference between the final circular scale readings observed by the students is 13.