In a vernier callipers, 10 divisions of vernier scale coincides with 9 divisions of main scale, leas

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1.Units, Dimensions and Measurement

medium

In a vernier callipers, $10$ divisions of vernier scale coincides with $9$ divisions of main scale, the least count of which is $0.1\,cm$. If in the measurement of inner diameter of cylinder zero of vernier scale lies between $1.3\,cm$ and $1.4\, cm$ of main scale and $2^{nd}$ division of vernier scale coincides with main scale division then diameter will be .......... $cm$

$1.30$

$1.34$

$1.32$

$1.36$

Solution

$\mathrm{L} \mathrm{C} =1 \mathrm{MSD}-1 \mathrm{VSD}$

$=1 \mathrm{MSD}-\frac{9}{10} \mathrm{MSD}$

$=\left(1-\frac{9}{10}\right) \times 1 \mathrm{MSD}$ $\left[\begin{array}{l}{\text { Also }} \\ {10 \mathrm{VSD}=9 \mathrm{MSD}} \\ {1 \mathrm{VSD}=\frac{9}{10} \mathrm{MSD}}\end{array}\right]$

$=0.1 \times 0.1 \mathrm{cm}$

$=0.01\,cm$

$Diameter=MSR+LC \times VSR$

$=1.3+0.01 \times 2$

$=1.3+0.02=1.32 \mathrm{cm}$

Standard 11

Physics

$10$ 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 :

medium

(JEE MAIN-2024)

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:

medium

(JEE MAIN-2021)

Least count of a vernier caliper is $\frac{1}{20 \mathrm{~N}} \mathrm{~cm}$. The value of one division on the main scale is $1 \mathrm{~mm}$. Then the number of divisions of main scale that coincide with $\mathrm{N}$ divisions of vernier scale is :

hard

(JEE MAIN-2024)

The least count of the main scale of a vernier callipers is $1\, mm$. Its vernier scale is divided into $10$ divisions and coincide with $9$ divisions of the main scale. When jaws are touching each other, the $7$ th division of vernier scale coincides with a division of main scale and the zero of vernier scale is lying right side of the zero of main scale. When this vernier is used to measure length of a cylinder the zero of the vernier scale between $3.1\, cm$ and $3.2\, cm$ and $4^{th}$ $VSD$ coincides with a main scale division.The length of the cylinder is $…..cm$
($VSD$ is vernier scale division)

hard

(JEE MAIN-2020)

Using screw gauge of pitch $0.1$ $cm$ and $50$ divisions on its circular scale, the thickness of an object is measured. It should correctly be recorded as

medium

(JEE MAIN-2020)

Solution

Similar Questions

$10$ 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 :

Least count of a vernier caliper is $\frac{1}{20 \mathrm{~N}} \mathrm{~cm}$. The value of one division on the main scale is $1 \mathrm{~mm}$. Then the number of divisions of main scale that coincide with $\mathrm{N}$ divisions of vernier scale is :

Using screw gauge of pitch $0.1$ $cm$ and $50$ divisions on its circular scale, the thickness of an object is measured. It should correctly be recorded as

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