JEE Main 2019
LEVELJEE Main
Visualized Solution
The Sigma Insight: Vernier Calipers and Screw Gauge
Decoding the Screw Gauge
Measuring instruments like the screw gauge are absolute favorites in JEE because they test your ability to visualize physical rotations and handle tiny decimal calculations without making silly mistakes. In this problem, we are tasked with finding the exact thickness of a thin sheet. But before we jump into the final reading, we need to establish the foundational parameters of our instrument.
The Master Key
Least Count
The Least Count (LC) is the absolute smallest value our screw gauge can measure. It is the physical distance the spindle moves when the circular scale is rotated by exactly one division.
We calculate it using the formula:
The problem gives us a pitch of and divisions on the circular scale. Substituting these values:
This is our multiplier for any reading on the circular scale.
The Trap
Analyzing the Zero Error
Here is where most students make a fatal error. The problem states: "When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line."
Imagine you are closing the jaws of the screw gauge. As you turn the thimble, the zero mark approaches the central reference line (the mean line). If the jaws completely close and the zero mark is still below the mean line, it means the thimble hasn't rotated enough to reach the true zero. The instrument is lagging behind.
Because it is lagging, any reading you take will be artificially low by exactly divisions. This is known as a negative zero error. To correct an artificially low reading, we must add the missing divisions back to our observed reading.
The Final Calculation
Now, let's bring it all together. The total thickness is the sum of the Main Scale Reading (MSR) and the corrected Circular Scale Reading (CSR), multiplied by the Least Count.
We are given an MSR of and an observed CSR of . Let's substitute our values:
First, we apply our zero error correction to the circular scale:
Next, we convert these divisions into a physical length by multiplying with the Least Count:
Finally, we add this to our main scale reading:
And there we have it! The exact thickness of the sheet is , which corresponds perfectly to option (c). Always remember to visualize the physical state of the instrument before blindly applying formulas!
Similar Questions
JEE Main 2016
LEVELJEE Advanced
A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of aluminium. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet, if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line?
(A)
0.75 mm
(B)
0.80 mm
(C)
0.70 mm
(D)
0.50 mm
LEVELJEE Main
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
(A)
(B)
(C)
(D)
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
(A)
(B)
(C)
(D)
JEE Main 2020
LEVELJEE Main
Using screw gauge of pitch and divisions on its circular scale, the thickness of an object is measured. It should correctly be recorded as
(A)
(B)
(C)
(D)
JEE Advanced 2022
LEVELJEE Advanced
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?
(A)
(B)
(C)
(D)
JEE Main 2019
LEVELJEE Main
The least count of the main scale of a screw gauge is . The minimum number of divisions on its circular scale required to measure diameter of a wire is
(A)
(B)
(C)
(D)
JEE Main 2020
LEVELJEE Main
If the screw on a screw gauge is given six rotations, it moves by 3 mm on the main scale. If there are 50 divisions on the circular scale, the least count of the screw gauge is
(A)
0.001 cm
(B)
0.01 cm
(C)
0.02 cm
(D)
0.001 cm
LEVELJEE Main
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
(A)
(B)
(C)
(D)
JEE Advanced 2026
LEVELJEE Main
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
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.
(A)
5.00 mm
(B)
5.25 mm
(C)
5.15 mm
(D)
5.20 mm