
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
170.1k+ 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
Molarity vs Molality: Definitions, Formulas & Key Differences

Preparation of Hydrogen Gas: Methods & Uses Explained

Polymers in Chemistry: Definition, Types, Examples & Uses

P Block Elements: Definition, Groups, Trends & Properties for JEE/NEET

Order of Reaction in Chemistry: Definition, Formula & Examples

Hydrocarbons: Types, Formula, Structure & Examples Explained

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Displacement-Time Graph and Velocity-Time Graph for JEE

Uniform Acceleration

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Instantaneous Velocity - Formula based Examples for JEE

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Other Pages
NCERT Solution for Class 11 Physics Chapter 1 Units and Measurements - 2025-26

NCERT Solution for Class 11 Physics Chapter 2 Motion In A Straight Line - 2025-26

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement
