In a screw gauge, there are $100$ divisions on the circular scale and the main scale moves by $0.5\,mm$ on a complete rotation of the circular scale. The zero of circular scale lies $6$ divisions below the line of graduation when two studs are brought in contact with each other. When a wire is placed between the studs, $4$ linear scale divisions are clearly visible while $46^{\text {th }}$ division the circular scale coincide with the reference line. The diameter of the wire is $...........\times 10^{-2}\,mm$
In a screw gauge, there are $100$ divisions on the circular scale and the main scale moves by $0.5\,mm$ on a complete rotation of the circular scale. The zero of circular scale lies $6$ divisions below the line of graduation when two studs are brought in contact with each other. When a wire is placed between the studs, $4$ linear scale divisions are clearly visible while $46^{\text {th }}$ division the circular scale coincide with the reference line. The diameter of the wire is $...........\times 10^{-2}\,mm$
- [JEE MAIN 2023]
- A
$23$
- B
$20$
- C
$21$
- D
$22$
Similar Questions
The main scale of vernier callipers reads in millimetre and its vernier is divided into $8$ divisions, which coincides with $5$ divisions of main scale. When two jaws of instrument touch each other, the zero of the vernier coincides with the zero of main scale. A rod is tightly placed along its length between both jaws. It is observed that the zero of vernier scale lies just left to $36^{th}$ division of main scale and fourth division of vernier scale coincides with the main scale Then the measured value is .......... $cm$
In finding out refractive index of glass slab the following observations were made through travelling microscope $50$ vernier scale division $=$ $49 \mathrm{MSD} ; 20$ divisions on main scale in each $\mathrm{cm}$ For mark on paper
$\mathrm{MSR}=8.45 \mathrm{~cm}, \mathrm{VC}=26$
For mark on paper seen through slab
$\mathrm{MSR}=7.12 \mathrm{~cm}, \mathrm{VC}=41$
For powder particle on the top surface of the glass slab
$\mathrm{MSR}=4.05 \mathrm{~cm}, \mathrm{VC}=1$
$(\mathrm{MSR}=$ Main Scale Reading, $\mathrm{VC}=$ Vernier Coincidence)
Refractive index of the glass slab is:
In finding out refractive index of glass slab the following observations were made through travelling microscope $50$ vernier scale division $=$ $49 \mathrm{MSD} ; 20$ divisions on main scale in each $\mathrm{cm}$ For mark on paper
$\mathrm{MSR}=8.45 \mathrm{~cm}, \mathrm{VC}=26$
For mark on paper seen through slab
$\mathrm{MSR}=7.12 \mathrm{~cm}, \mathrm{VC}=41$
For powder particle on the top surface of the glass slab
$\mathrm{MSR}=4.05 \mathrm{~cm}, \mathrm{VC}=1$
$(\mathrm{MSR}=$ Main Scale Reading, $\mathrm{VC}=$ Vernier Coincidence)
Refractive index of the glass slab is:
- [JEE MAIN 2024]
In a Vernier Calipers. $10$ divisions of Vernier scale is equal to the $9$ divisions of main scale. When both jaws of Vernier calipers touch each other, the zero of the Vernier scale is shifted to the left of $zero$ of the main scale and $4^{\text {th }}$ Vernier scale division exactly coincides with the main scale reading. One main scale division is equal to $1\,mm$. While measuring diameter of a spherical body, the body is held between two jaws. It is now observed that zero of the Vernier scale lies between $30$ and $31$ divisions of main scale reading and $6^{\text {th }}$ Vernier scale division exactly. coincides with the main scale reading. The diameter of the spherical body will be $.......cm$
- [JEE MAIN 2022]
A screw gauge gives the following reading when used to measure the diameter of a wire.
Main scale reading : $0\ mm$
Circular scale reading : $52\ divisions$
Given that $1\ mm$ on main scale corresponds to $100$ divisions of the circular scale. The diameter of wire from the above data is:
A screw gauge gives the following reading when used to measure the diameter of a wire.
Main scale reading : $0\ mm$
Circular scale reading : $52\ divisions$
Given that $1\ mm$ on main scale corresponds to $100$ divisions of the circular scale. The diameter of wire from the above data is:
- [AIEEE 2011]
Asseretion $A:$ If in five complete rotations of the circular scale, the distance travelled on main scale of the screw gauge is $5\, {mm}$ and there are $50$ total divisions on circular scale, then least count is $0.001\, {cm}$.
Reason $R:$ Least Count $=\frac{\text { Pitch }}{\text { Total divisions on circular scale }}$
In the light of the above statements, choose the most appropriate answer from the options given below:
Asseretion $A:$ If in five complete rotations of the circular scale, the distance travelled on main scale of the screw gauge is $5\, {mm}$ and there are $50$ total divisions on circular scale, then least count is $0.001\, {cm}$.
Reason $R:$ Least Count $=\frac{\text { Pitch }}{\text { Total divisions on circular scale }}$
In the light of the above statements, choose the most appropriate answer from the options given below:
- [JEE MAIN 2021]