Sigma Percentile
JEE Main 2021
LEVELJEE Main

Animated Solution for Physics - Physics and Measurement: 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.

Select Answer:

Visualized Solution

The Sigma Insight: Vernier Calipers and Screw Gauge

Solution Diagram
The screw gauge is a beautiful piece of engineering that allows us to measure dimensions down to a fraction of a millimeter. But with great precision comes the need for careful calibration. In this problem, we are going to walk through the complete anatomy of a screw gauge measurement, from finding its least count to correcting its inherent zero error.

Understanding the Instrument's Precision

Before we can measure anything, we need to understand the limits of our instrument. The problem states that the main scale moves by for every complete rotation of the circular scale. This distance is known as the pitch of the screw gauge.
The circular scale itself is divided into equal divisions. The Least Count (LC)—the smallest value the instrument can measure—is found by dividing the pitch by the total number of circular divisions.
This tells us that every single step on the circular scale corresponds to a forward or backward movement of exactly .

The Trap of the Zero Error

In an ideal world, when the jaws of the screw gauge are fully closed (using the ratchet to ensure uniform pressure), the zero mark of the circular scale should align perfectly with the reference line of the main scale. However, our instrument has a slight imperfection.
When closed, the division of the circular scale coincides with the reference line. Because the zero mark has already crossed the reference line, the instrument is reading a value even when nothing is between the jaws! This is a positive zero error.
To calculate the exact magnitude of this error, we multiply the coinciding division by the least count:
Keep this value safe. Since the instrument is over-reading by , we will need to subtract this from our final observation to get the true measurement.

Making the Final Measurement

Now, we place our object between the jaws and take the reading. The main scale clearly shows . This is our Main Scale Reading (MSR).
Next, we look at the circular scale to get the fractional part of the measurement. The division perfectly coincides with the reference line. This gives us our Circular Scale Reading (CSR).
The formula for the true reading combines all these elements:
Let's carefully substitute our values into the master equation. Watch out for the minus sign!
And there we have it! By systematically breaking down the instrument's parameters, accounting for its initial calibration error, and carefully combining our readings, we arrive at the precise true reading of .

Similar Questions

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

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)
LEVELJEE Main

The circular scale of a screw gauge has 50 divisions and pitch of 0.5 mm. Find the diameter of sphere. Main scale reading is 2.

(A)
1.2 mm
(B)
1.25 mm
(C)
2.20 mm
(D)
2.25 mm
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

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 Main 2019
LEVELJEE Main

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

(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 Advanced 2015
LEVELJEE Advanced

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

* Multiple Correct Options
(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
(C)
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
JEE Main 2021
LEVELJEE Advanced

Student A and student B used two screw gauges of equal pitch and equal circular divisions to measure the radius of a given wire. The actual value of the radius of the wire is . The absolute value of the difference between the final circular scale readings observed by the students A and B is ......... . [Figure shows position of reference O when jaws of screw gauge are closed] Given, pitch .

JEE Main 2020
LEVELJEE Main

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

(A)
negative,
(B)
positive,
(C)
positive,
(D)
positive,