JEE Main 2021
LEVELJEE Advanced
Visualized Solution
The Sigma Insight: Vernier Calipers and Screw Gauge
The Tale of Two Screw Gauges
Unraveling the Zero Error Trap
Imagine you are holding a precision instrument like a screw gauge. You close the jaws completely, expecting the circular scale to read a perfect zero. But alas, the real world is messy! The zero mark is slightly off. This is what we call a Zero Error, and mastering its sign convention is the key to unlocking this beautiful problem.
In this scenario, we have two students, A and B, using two different screw gauges to measure the exact same wire. The true radius of the wire is given as . Both gauges have a pitch of and circular divisions.
Before we dive into the readings, let's establish our fundamental tool—the Least Count (LC).
Analyzing Screw Gauge A
Let's look closely at the first screw gauge. When the jaws are fully closed, the reference line aligns perfectly with the division on the circular scale. Because the numbers on the circular scale increase downwards, the zero mark has already crossed the reference line.
This means the screw has advanced past the true zero. It is reading a positive value even when there is nothing between the jaws! This is a positive zero error.
The golden rule of measurements states that the true value is always the measured value minus the zero error.
Let's substitute our known values for student A:
We know that the measured value is the sum of the Main Scale Reading (MSR) and the Circular Scale Reading (CSR). Since the pitch is , the MSR must be a multiple of . The largest multiple of that fits into is .
This leaves for the circular scale. Dividing this by the least count gives us the exact number of divisions.
Analyzing Screw Gauge B
Now, let's turn our attention to gauge B. When closed, the reference line points to the division. The zero mark hasn't even reached the reference line yet! It is short by exactly divisions ().
Because it hasn't reached zero, it is reading a negative value. This is a negative zero error.
Let's apply our golden rule once again for student B:
Subtracting a negative is the same as adding, so we get:
Just like before, the MSR must be , leaving for the circular scale.
The Final Calculation
We now have the circular scale readings for both students. Student A reads divisions, and student B reads divisions. The question asks for the absolute difference between these two readings.
The Sign Convention Trap
There is a massive catch here that traps many students (and even some textbooks!). Some resources use a flawed formula where they define "Error" as the "Zero Correction" and add it to the measured value instead of subtracting it.
If you were to make that mistake, you would calculate and . Miraculously, the absolute difference is still exactly ! The math is forgiving in this specific case because the sign error is applied consistently to both gauges. However, as a true physicist, you must always stick to the correct physical reality: True Value = Measured Value - Zero Error.
Similar Questions
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Two full turns of the circular scale of a screw gauge cover a distance of on its main scale. The total number of divisions on the circular scale is . Further, it is found that the screw gauge has a zero error of . While measuring the diameter of a thin wire, a student notes the main scale reading of and the number of circular scale divisions in line with the main scale as . The diameter of the wire is
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Area of the cross-section of a wire is measured using a screw gauge. The pitch of the main scale is . The circular scale has divisions and for one full rotation of the circular scale, the main scale shifts by two divisions. The measured readings are listed below. \begin{array}{|l|c|c|} \hline \textbf{Measurement condition} & \textbf{Main scale reading} & \textbf{Circular scale reading} \\ \hline \text{Two arms of gauge touching} & & \\ \text{each other without wire} & 0\text{ division} & 4\text{ division} \\ \hline \text{Attempt-1: With wire} & 4\text{ divisions} & 20\text{ divisions} \\ \hline \text{Attempt-2: With wire} & 4\text{ divisions} & 16\text{ divisions} \\ \hline \end{array} What are the diameter and cross-sectional area of the wire measured using the screw gauge?
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Two thin wires, Wire-1 of diameter and Wire-2 of unknown diameter are given. To obtain the value of , the diameters of the two wires are measured with a screw gauge. The screw gauge has a pitch of and there are divisions on the circular scale (CS). The smallest division on the linear scale (LS) is . The table shows the readings of LS and CS for the measurements. The value of (in ) is:
JEE Main 2021
LEVELJEE Main
The pitch of the screw gauge is and there are divisions on the circular scale. When nothing is put in between the jaws, the zero of the circular scale lies divisions below the reference line. When a wire is placed between the jaws, the first linear scale division is clearly visible while division on circular scale coincides with the reference line. The radius of the wire is
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In a screw gauge, 5th division of the circular scale coincides with the reference line when the ratchet is closed. There are 50 divisions on the circular scale, and the main scale moves by 0.5 mm on a complete rotation. For a particular observation the reading on the main scale is 5 mm and the 20th division of the circular scale coincides with reference line. Calculate the true reading.
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A screw gauge gives the following reading when used to measure the diameter of a wire. Main scale reading Circular scale reading divisions Given that on main scale corresponds to divisions of the circular scale. The diameter of wire from the above data is
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The pitch and the number of divisions, on the circular scale for a given screw gauge are and , respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies divisions below the mean line. The readings of the main scale and the circular scale for a thin sheet are and respectively, the thickness of this sheet is
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Using screw gauge of pitch and divisions on its circular scale, the thickness of an object is measured. It should correctly be recorded as
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Consider a vernier caliper in which each on the main scale is divided into equal divisions and a screw gauge with divisions on its circular scale. In the vernier callipers, divisions of the vernier scale coincide with divisions on the main scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then
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(A)
if the pitch of the screw gauge is twice the least count of the vernier caliper, the least count of the screw gauge is
(B)
if the pitch of the screw gauge is twice the least count of the Vernier caliper, the least count of the screw gauge is
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if the least count of the linear scale of the screw gauge is twice the least count of the Vernier calipers, the least count of the screw gauge is
(D)
if the least count of the linear scale of the screw gauge is twice the least count of the vernier caliper, the least count of the screw gauge is
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A screw gauge has divisions on its circular scale. The circular scale is units ahead of the pitch scale marking, prior to use. Upon one complete rotation of the circular scale, a displacement of is noticed on the pitch scale. The nature of zero error involved and the least count of the screw gauge, are respectively
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negative,
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positive,
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positive,
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positive,