
The smallest division on the main scale of a vernier calipers is $1mm$, and $10$ vernier division coincide with $9$ main scale division. while measuring the diameter of a sphere, the zero mark of the vernier. scale lies between $2.0$ and $2.1\,cm$ the fifth vernier scale coincides with a scale division. Then diameter of the sphere is:
A) $2.05\,cm$
B) $3.05\,cm$
C) $2.50\,cm$
D) $\text{None of these}$
Answer
216k+ views
Hint: Here we find the reading of the scale, the above statement considers that we know the magnitude of the scale, and then we begin to measure the diameter of the sphere, we proceed with the formula for measuring the diameter. By using the method of the vernier.
Formula used:
$LC = \dfrac{{Smaller\,reading\,on\,main\,scale}}{{No.of\,division\,on\,vernier\,scale}}$
$\Rightarrow$ $MSD$ is the main scale division
$\Rightarrow$ $VSD$ is the vernier scale division
$\Rightarrow$ $LC$ is the least count
Complete step by step solution:
Given by,
Main scale division,
$1MSD = 1mm$
Vernier scale division,
$1VSD = 9MSD$
We find the diameter of the sphere,
Therefore,
$1VSD = \dfrac{9}{{10}}MSD$
On simplifying,
$\Rightarrow$ $1VSD = .9MSD$
Now,
We find the least count,
We know that,
$Least\,Count\,of\,the\,vernier\,calipers = \dfrac{{Smaller\,reading\,on\,main\,scale}}{{No.of\,division\,on\,vernier\,scale}}$
Rearranging the above formula,
We get,
$Least\,Count = 1Main\,Scale\,Division - 1Vernier\,Scale\,Division$
$\Rightarrow$ $LC = 1 - .9 = .1mm$
Here,
$\Rightarrow$ $LC = .01\,cm$
Then we find the diameter of the sphere,
$\Rightarrow$ $Diameter = 2.0 + VSR$
$VSR$ vernier scale reading is $5$ and multiplying with least count of vernier calipers
We get,
$\Rightarrow$ $Diameter = 2.0 + 5 \times .01$
On simplifying,
$\Rightarrow$ $Diameter = 2.05\,cm$
Hence, the option A is the correct answer, the diameter of the sphere is $2.05\,cm$.
Note: Whenever this kind of problem It slides parallel to the main scale and allows for readings on the main scale to be rendered to a fraction of a division. With the assistance of a meter scale, the duration is measured. It is graduated in cm and mm such that one small division's value is one millimeter, so a meter scale can be used to accurately measure a length of up to one millimeter.
Formula used:
$LC = \dfrac{{Smaller\,reading\,on\,main\,scale}}{{No.of\,division\,on\,vernier\,scale}}$
$\Rightarrow$ $MSD$ is the main scale division
$\Rightarrow$ $VSD$ is the vernier scale division
$\Rightarrow$ $LC$ is the least count
Complete step by step solution:
Given by,
Main scale division,
$1MSD = 1mm$
Vernier scale division,
$1VSD = 9MSD$
We find the diameter of the sphere,
Therefore,
$1VSD = \dfrac{9}{{10}}MSD$
On simplifying,
$\Rightarrow$ $1VSD = .9MSD$
Now,
We find the least count,
We know that,
$Least\,Count\,of\,the\,vernier\,calipers = \dfrac{{Smaller\,reading\,on\,main\,scale}}{{No.of\,division\,on\,vernier\,scale}}$
Rearranging the above formula,
We get,
$Least\,Count = 1Main\,Scale\,Division - 1Vernier\,Scale\,Division$
$\Rightarrow$ $LC = 1 - .9 = .1mm$
Here,
$\Rightarrow$ $LC = .01\,cm$
Then we find the diameter of the sphere,
$\Rightarrow$ $Diameter = 2.0 + VSR$
$VSR$ vernier scale reading is $5$ and multiplying with least count of vernier calipers
We get,
$\Rightarrow$ $Diameter = 2.0 + 5 \times .01$
On simplifying,
$\Rightarrow$ $Diameter = 2.05\,cm$
Hence, the option A is the correct answer, the diameter of the sphere is $2.05\,cm$.
Note: Whenever this kind of problem It slides parallel to the main scale and allows for readings on the main scale to be rendered to a fraction of a division. With the assistance of a meter scale, the duration is measured. It is graduated in cm and mm such that one small division's value is one millimeter, so a meter scale can be used to accurately measure a length of up to one millimeter.
Recently Updated Pages
Alpha, Beta, and Gamma Decay Explained

Alpha Particle Scattering and Rutherford Model Explained

Angular Momentum of a Rotating Body: Definition & Formula

Apparent Frequency Explained: Formula, Uses & Examples

Applications of Echo in Daily Life and Science

Atomic Structure Explained: Key Concepts for Students

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

JEE Main Correction Window 2026 Session 1 Dates Announced - Edit Form Details, Dates and Link

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

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

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

NCERT Solutions For Class 11 Physics Chapter 8 Mechanical Properties Of Solids

Motion in a Straight Line Class 11 Physics Chapter 2 CBSE Notes - 2025-26

NCERT Solutions for Class 11 Physics Chapter 7 Gravitation 2025-26

How to Convert a Galvanometer into an Ammeter or Voltmeter

