The least count of the main scale of a vernier callipers is $1\, mm$. Its vernier scale is divided into $10$ divisions and coincides with $9$ divisions of the main scale. When jaws are touching each other,the $7^{th}$ division of the vernier scale coincides with a division of the main scale and the zero of the vernier scale lies to the right side of the zero of the main scale. When this vernier is used to measure the length of a cylinder,the zero of the vernier scale is between $3.1\, cm$ and $3.2\, cm$ and the $4^{th}$ $VSD$ coincides with a main scale division. The length of the cylinder is $.....\, cm$. ($VSD$ is vernier scale division)
- A$3.21$
- B$2.99$
- C$3.2$
- D$3.07$
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Similar Questions
In a vernier callipers,$10$ divisions of vernier scale coincide with $9$ divisions of main scale,the least count of which is $0.1\,cm$. If in the measurement of inner diameter of a cylinder,the zero of the vernier scale lies between $1.3\,cm$ and $1.4\,cm$ of the main scale and the $2^{nd}$ division of the vernier scale coincides with a main scale division,then the diameter will be .......... $cm$.
Medium
View SolutionIf $50$ Vernier divisions are equal to $49$ main scale divisions of a travelling microscope and one smallest reading of main scale is $0.5 \,mm$, the Vernier constant of travelling microscope is:
The diameter of a spherical bob is measured using a vernier callipers. $9$ divisions of the main scale,in the vernier callipers,are equal to $10$ divisions of vernier scale. One main scale division is $1\, {mm}$. The main scale reading is $10\, {mm}$ and $8^{\text{th}}$ division of vernier scale was found to coincide exactly with one of the main scale division. If the given vernier callipers has positive zero error of $0.04\, {cm}$,then the radius of the bob is $...... \,\times 10^{-2} \,{cm}$.
DifficultJEE MAIN 2021
View SolutionThe circular divisions of a screw gauge are $50$. It moves $0.5 \ mm$ on the main scale in one rotation. When the diameter of a wire is measured,the main scale reading is $3.5 \ mm$ and the circular scale reading is $32$. If the zero error (positive) in the screw gauge is $0.06 \ mm$,then the diameter of the wire is:
Medium
View SolutionIn a screw gauge, the zero of the circular scale lies $3$ divisions above the horizontal pitch line when their metallic studs are brought in contact. Using this instrument, the thickness of a sheet is measured. If the pitch scale reading is $1 \ mm$ and the circular scale reading is $51$, then the correct thickness of the sheet is . . . . . . $mm$. [Assume least count is $0.01 \ mm$]
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