The least count of the main scale of a screw gauge is $1mm$. The minimum number of division on its circular scale required to measure $5\mu m$ diameter of wire is:
A) 50
B) 100
C) 200
D) 500
Answer
266.1k+ views
Hint: The smallest value which can be measured by any of the measuring instruments is called the least count. Recall the relation between the least count of screw gauge and the minimum number of the division on the circular scale of the screw gauge.
Formula Used:
$L.C = \dfrac{P}{N}$
Where, $L.C$ is the least count of screw gauge, $P$ is the pitch of screw gauge, $N$ is the minimum number of the division on the circular scale of the screw gauge
Complete step by step answer:
It is given in the question that,
The least count of the main scale of a screw gauge is $1mm$
We know that the pitch of the screw gauge is equal to the least count of the main scale of a screw gauge, that is
$P = 1mm$
On converting it into SI unit that is, in meters, we get
$ \Rightarrow P = {10^{ - 3}}m$
It is also given that the diameter of wire measured with that screw gauge is $5\mu m$.
We have clearly seen that the given diameter of the wire is much smaller than the least count of the main scale of screw gauge. So, for the minimum number of division of circular scale,
Least count of screw gauge is equal to the diameter of wire, that is
$L.C = 5\mu m$
On converting it into SI unit that is, in meters, we get
$ \Rightarrow L.C = 5 \times {10^{ - 6}}m$
Now by applying the formula, we get
$L.C = \dfrac{P}{N}$
Where, $L.C$ is the least count of screw gauge
$P$ is the pitch of screw gauge
$N$ is the minimum number of the division on the circular scale of the screw gauge
On putting the all the values in that formula, we get
$ \Rightarrow 5 \times {10^{ - 6}} = \dfrac{{{{10}^{ - 3}}}}{N}$
On further solving that, we get
$ \Rightarrow N = \dfrac{{1000}}{5}$
So finally we get
$ \Rightarrow N = 200$
Thus, the minimum number of the division on the circular scale of the screw gauge is $200$.
Therefore, the correct answer is C
Note: The screw gauge is an instrument used for measuring exactly the diameter of a thin wire or the width of a sheet of metal. It comprises a U-shaped mount which is fixed with a screwed pin which is fixed to a thimble. A screw gauge of $100$ divisions will move the cap scale along the main scale by $0.01mm$. This is the minimum value up to which a screw gauge can measure and is known as its least count.
Formula Used:
$L.C = \dfrac{P}{N}$
Where, $L.C$ is the least count of screw gauge, $P$ is the pitch of screw gauge, $N$ is the minimum number of the division on the circular scale of the screw gauge
Complete step by step answer:
It is given in the question that,
The least count of the main scale of a screw gauge is $1mm$
We know that the pitch of the screw gauge is equal to the least count of the main scale of a screw gauge, that is
$P = 1mm$
On converting it into SI unit that is, in meters, we get
$ \Rightarrow P = {10^{ - 3}}m$
It is also given that the diameter of wire measured with that screw gauge is $5\mu m$.
We have clearly seen that the given diameter of the wire is much smaller than the least count of the main scale of screw gauge. So, for the minimum number of division of circular scale,
Least count of screw gauge is equal to the diameter of wire, that is
$L.C = 5\mu m$
On converting it into SI unit that is, in meters, we get
$ \Rightarrow L.C = 5 \times {10^{ - 6}}m$
Now by applying the formula, we get
$L.C = \dfrac{P}{N}$
Where, $L.C$ is the least count of screw gauge
$P$ is the pitch of screw gauge
$N$ is the minimum number of the division on the circular scale of the screw gauge
On putting the all the values in that formula, we get
$ \Rightarrow 5 \times {10^{ - 6}} = \dfrac{{{{10}^{ - 3}}}}{N}$
On further solving that, we get
$ \Rightarrow N = \dfrac{{1000}}{5}$
So finally we get
$ \Rightarrow N = 200$
Thus, the minimum number of the division on the circular scale of the screw gauge is $200$.
Therefore, the correct answer is C
Note: The screw gauge is an instrument used for measuring exactly the diameter of a thin wire or the width of a sheet of metal. It comprises a U-shaped mount which is fixed with a screwed pin which is fixed to a thimble. A screw gauge of $100$ divisions will move the cap scale along the main scale by $0.01mm$. This is the minimum value up to which a screw gauge can measure and is known as its least count.
Recently Updated Pages
States of Matter Chapter For JEE Main Chemistry

Circuit Switching vs Packet Switching: Key Differences Explained

Mass vs Weight: Key Differences Explained for Students

JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

[Awaiting the three content sources: Ask AI Response, Competitor 1 Content, and Competitor 2 Content. Please provide those to continue with the analysis and optimization.]

Sign up for JEE Main 2026 Live Classes - Vedantu

Trending doubts
Hybridisation in Chemistry – Concept, Types & Applications

Understanding Atomic Structure for Beginners

Understanding Uniform Acceleration in Physics

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Degree of Dissociation: Meaning, Formula, Calculation & Uses

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Other Pages
JEE Advanced Weightage Chapter Wise 2026 for Physics, Chemistry, and Mathematics

Understanding Displacement and Velocity Time Graphs

Understanding the Angle of Deviation in a Prism

JEE Advanced Chemistry Notes 2026

What Are Alpha, Beta, and Gamma Decay in Nuclear Physics?

Water waves are A Longitudinal B Transverse C Both class 11 physics JEE_Main

