$A$ 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 $45^{th}$ division coincides with the main scale line and the zero of the main scale is barely visible. What is the thickness of the sheet (in $mm$) if the main scale reading is $0.5 \ mm$ and the $25^{th}$ division coincides with the main scale line?
- A$0.70$
- B$0.50$
- C$0.75$
- D$0.80$
Explore More
Similar Questions
In an experiment to find out the diameter of a wire using a screw gauge,the following observations were noted:
$(A)$ Screw moves $0.5 \ mm$ on the main scale in one complete rotation.
$(B)$ Total divisions on the circular scale $= 50$.
$(C)$ Main scale reading is $2.5 \ mm$.
$(D)$ $45^{\text{th}}$ division of the circular scale is in the reference line.
$(E)$ Instrument has $0.03 \ mm$ negative zero error.
Then the diameter of the wire is: (in $mm$)
$(A)$ Screw moves $0.5 \ mm$ on the main scale in one complete rotation.
$(B)$ Total divisions on the circular scale $= 50$.
$(C)$ Main scale reading is $2.5 \ mm$.
$(D)$ $45^{\text{th}}$ division of the circular scale is in the reference line.
$(E)$ Instrument has $0.03 \ mm$ negative zero error.
Then the diameter of the wire is: (in $mm$)
Difficult
View Solution$10$ divisions on the main scale of a Vernier calliper coincide with $11$ divisions on the Vernier scale. If each division on the main scale is of $5$ units,the least count of the instrument is :
The pitch of a screw gauge is $0.5 \ mm$ and there are $100$ divisions on its circular scale. The instrument reads $+2$ divisions on the circular scale when nothing is placed between its jaws. In measuring the diameter of a wire,there are $8$ divisions on the main scale and the $83^{rd}$ division of the circular scale coincides with the reference line. Then the diameter of the wire is (in $mm$)
Medium
View SolutionThe diameter of a cylinder is measured using a vernier callipers with no zero error. It is found that the zero of the vernier scale lies between $5.10 \ cm$ and $5.15 \ cm$ of the main scale. The vernier scale has $50$ divisions equivalent to $2.45 \ cm$. The $24^{\text{th}}$ division of the vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is: (in $cm$)
DifficultIIT 2013
View Solution$A$ travelling microscope is used to determine the refractive index of a glass slab. If $40$ divisions are there in $1 \; cm$ on the main scale and $50$ Vernier scale divisions are equal to $49$ main scale divisions,then the least count of the travelling microscope is $\dots \times 10^{-6} \; m$.
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo