Sigma Percentile
JEE Main 2020
LEVELJEE Main

Animated Solution for Physics - Physics and Measurement: Using screw gauge of pitch and divisions on its circular scale, the thickness of an object is measured. It should correctly be recorded as

Select Answer:

Visualized Solution

The Sigma Insight: Vernier Calipers and Screw Gauge

Solution Diagram

The Anatomy of a Screw Gauge

Imagine you are holding a precision instrument like a screw gauge. It consists of two primary scales: a stationary main scale and a rotating circular scale. When you rotate the circular scale by exactly one full turn, it advances linearly along the main scale. This linear distance covered in one complete rotation is known as the pitch of the screw gauge.
In our problem, the pitch is given as . This means every time you turn the circular scale , the jaws of the screw gauge open or close by exactly .

Finding the Least Count

The true power of a screw gauge lies in its ability to measure tiny fractions of the pitch. This is achieved by dividing the circular scale into multiple equal parts. The smallest measurement the instrument can accurately record is called its Least Count (LC).
The formula for the least count is beautifully simple:
Let's substitute the values provided in the question. We have a pitch of and divisions on the circular scale.
This result, , is the absolute minimum distance the screw gauge can measure.

The Integer Multiple Rule

Here is where many students make a silly mistake. A measuring instrument is like a staircase; you can only stand on a specific step, never floating between two steps. Therefore, any valid reading taken by this screw gauge must be an exact integer multiple of its least count.
Mathematically, a valid reading can be expressed as:
where is a whole number (integer).

Evaluating the Options

To find the correct measurement among the given options, we simply divide each option by our least count (). The correct option will yield a perfect integer.
Let's test them: - Option (a): (Not an integer) - Option (b): (Perfect Integer!) - Option (c): (Not an integer) - Option (d): (Not an integer)
Since is the only value that perfectly divides by the least count without leaving a remainder, it is the only physically possible reading for this specific screw gauge.
The correct option is (b).

Similar Questions

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 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 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 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 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
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 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:

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