The pitch of a screw gauge is $0.5mm$ and the head scale is divided in $100$ parts. What is the least count of the screw gauge?
A) $0.5mm$ or $0.05cm$
B) $0.05mm$ or $0.005cm$
C) $0.005mm$ or $0.0005cm$
D) $0.0005mm$ or $0.00005cm$
Answer
300.6k+ views
Hint: To solve this question we should know what the least count of any instrument is. It is the smallest measurement that can be made directly from the instrument. We simply have to find the length for each head scale division where the total length of the head scale is $0.5mm$.
Formulae used:
$L.C = \dfrac{{pitch}}{{divisio{n_{c.s}}}}$
Here $L.C$ is the least count of the screw gauge, $pitch$ is the distance moved by the circular scale in one rotation, and $divisio{n_{c.s}}$ is the number of divisions on the circular head scale.
Complete step by step answer:
To start with, let’s see what the least count of a screw gauge means.
Least count can be defined as the smallest, most accurate, direct measurement we can take from any screw gauge. The less is the least count the more precise is the measurement. The least count can be calculated using the formulae,
$L.C = \dfrac{{pitch}}{{divisio{n_{c.s}}}}$
Here $L.C$ is the least count of the screw gauge, $pitch$ is the distance moved by the circular scale in one rotation, and $divisio{n_{c.s}}$ is the number of divisions on the circular head scale.
In the question, the pitch is given to be $0.5mm$ and the number of divisions on the circular head scale is $100$. So the least count of the given screw gauge is,
$ \Rightarrow L.C = \dfrac{{0.5mm}}{{100}} = 0.005mm = 0.0005cm$
So option (C) is the correct answer.
Additional Information:
A screw gauge is an instrument that is used to measure the diameter of thin sheets, wires, or plates. Its pitch can be defined as the distance moved by the circular scale, in one rotation, on the main scale.
Note: While solving questions related to least count, be cautious of the formulae. Always use the correct formulae. Also, the unit is very important. The use of wrong units gives an incorrect answer. So the units must be used properly.
Formulae used:
$L.C = \dfrac{{pitch}}{{divisio{n_{c.s}}}}$
Here $L.C$ is the least count of the screw gauge, $pitch$ is the distance moved by the circular scale in one rotation, and $divisio{n_{c.s}}$ is the number of divisions on the circular head scale.
Complete step by step answer:
To start with, let’s see what the least count of a screw gauge means.
Least count can be defined as the smallest, most accurate, direct measurement we can take from any screw gauge. The less is the least count the more precise is the measurement. The least count can be calculated using the formulae,
$L.C = \dfrac{{pitch}}{{divisio{n_{c.s}}}}$
Here $L.C$ is the least count of the screw gauge, $pitch$ is the distance moved by the circular scale in one rotation, and $divisio{n_{c.s}}$ is the number of divisions on the circular head scale.
In the question, the pitch is given to be $0.5mm$ and the number of divisions on the circular head scale is $100$. So the least count of the given screw gauge is,
$ \Rightarrow L.C = \dfrac{{0.5mm}}{{100}} = 0.005mm = 0.0005cm$
So option (C) is the correct answer.
Additional Information:
A screw gauge is an instrument that is used to measure the diameter of thin sheets, wires, or plates. Its pitch can be defined as the distance moved by the circular scale, in one rotation, on the main scale.
Note: While solving questions related to least count, be cautious of the formulae. Always use the correct formulae. Also, the unit is very important. The use of wrong units gives an incorrect answer. So the units must be used properly.
Recently Updated Pages
Four persons A B C and D initially at the corners of class 11 physics JEE_Main

What is the difference between Conduction and conv class 11 physics JEE_Main

Moment of inertia of solid sphere about its diameter class 11 physics JEE_Main

If a piece of ice floating on the surface of water class 11 physics JEE_Main

At what temperature speed of sound in air will be doubled class 11 physics JEE_Main

A closed organ pipe and an open organ pipe are tuned class 11 physics JEE_Main

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

Understanding Atomic Structure for Beginners

Understanding Uniform Acceleration in Physics

Other Pages
CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2026-27 PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 1 Units And Measurements - 2026-27 Free PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 2 Motion In A Straight Line - 2026-27 Free PDF Download (Login Required)

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2026-27 Free PDF Download (Sign-in Required)

CBSE Notes Class 11 Physics Chapter 2 - Motion in a Straight Line - 2026-27 PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 3 Motion In A Plane - 2026-27 Free PDF Download (Sign-in Required)

