Problem 5 - Master Resource Book in JEE Main Physics

Question

On measuring diameter of a wire with help of screw gauge, main scale reading is \(1 \mathrm{~mm}\) and 6 th division of circular scale lying over reference line. On measuring zero error, it is found that zero of circular scale has advanced from reference line by 3 divisions on circular scale, then corrected diameter is (a) \(1.09 \mathrm{~mm}\) (b) \(1.06 \mathrm{~mm}\) (c) \(1.03 \mathrm{~mm}\) (d) \(1.60 \mathrm{~mm}\)

Step-by-Step Solution

Answer

The corrected diameter of the wire is (c) 1.03 mm.

Step 1: Understand the Problem

We have a measured main scale reading of the wire's diameter as \(1 \text{ mm}\), and the circular scale shows the 6th division over the reference line. There's a zero error of 3 divisions, where the zero of the circular scale has advanced. We need to determine the corrected diameter of the wire.

Step 2: Determine the Circular Scale Reading

Each division on the circular scale corresponds to a certain value. This value is typically related to how precise the screw gauge is. In this problem, let's assume the least count (or the value of each division on the circular scale) is 0.01 mm (a common value), thus 6 divisions on that scale means \(6 \times 0.01 \text{ mm} = 0.06 \text{ mm}\).

Step 3: Calculate the Raw Reading

Add the main scale reading to the circular scale reading to get the total: \[ \text{Reading} = 1 \text{ mm} + 0.06 \text{ mm} = 1.06 \text{ mm}. \] This is the uncorrected diameter of the wire.

Step 4: Account for Zero Error

The zero error indicates the scale has advanced by 3 divisions. Accounting for zero error requires adjusting the raw reading. The zero error adjustment is \(3 \times 0.01 \text{ mm} = 0.03 \text{ mm}\). Since it is an advanced zero error, you subtract it: \(1.06 \text{ mm} - 0.03 \text{ mm} = 1.03 \text{ mm}\).

Step 5: Select the Correct Answer

According to the corrected reading of \(1.03 \text{ mm}\), we find that the correct option is (c) \(1.03 \text{ mm}\).

Key Concepts

Zero Error Correction

When using a screw gauge for precise measurements, one common error to be aware of is zero error. This occurs when the zero mark on the circular scale does not align perfectly with that of the main scale, even when the measurement device is fully closed. In this exercise, the circular scale's zero has shifted by 3 divisions. Zero error can either be positive or negative, depending on whether the scale advances or lags behind. Correcting this error is vital to ensure accurate readings.

• First, determine the value of zero error by noting how many divisions the circular scale's zero is offset.
• Here, the zero error is 3 divisions, with the least count being 0.01 mm per division, resulting in a total zero error of 0.03 mm.
• Subtract this error from the raw measurement when the error is positive (advanced zero), as with this wire's diameter measurement.

Adjusting for zero error refines our measurement from 1.06 mm to the corrected value of 1.03 mm.

Diameter Measurement

The primary function of a screw gauge is to measure the diameter of small objects with precision. This is achieved through both a main scale and a circular scale. Initially, the main scale reading provides the larger, more apparent dimension, while the circular scale offers the more minute adjustments required for precision.
In this scenario, the initial main scale reading was 1 mm. The circular scale, which improves precision, indicated the 6th division was in line with the reference line. Accounting for this, the contribution to the measurement from the circular scale amounted to 0.06 mm because the least count is 0.01 mm. Thus, you add these two values: the main scale (1 mm) and the circular scale (0.06 mm), resulting in a preliminary raw diameter measurement of 1.06 mm.
This precise approach allows us to combine coarse and fine measurements efficiently to determine an object's diameter accurately.

Least Count Calculation

Least count is the smallest possible measurement a device can reliably record and is crucial in understanding how accurate your measurements are. For a screw gauge, this is determined by the relationship between the pitch and the number of divisions on the circular scale.

• Pitch is the length the spindle moves per full rotation of the circular scale, typically provided as 0.5 mm for many screw gauges.
• With 50 divisions on the circular scale, the least count is derived by dividing pitch by the number of divisions: \( \text{Least count} = \frac{\text{Pitch}}{\text{Number of divisions}} = \frac{0.5 \text{ mm}}{50} = 0.01 \text{ mm} \).

The least count of 0.01 mm thus informs us how each division on the circular scale equates to an additional 0.01 mm in measurement. This knowledge enabled us to calculate that six divisions corresponded to 0.06 mm. Finally, understanding and applying the least count correctly ensures precise and accurate measurements in experiments.