Given below are two statements:
Statement $I$: In a vernier callipers,one vernier scale division is always smaller than one main scale division.
Statement $II$: The vernier constant is given by one main scale division multiplied by the number of vernier scale division.
In the light of the above statements,choose the correct answer from the options given below.
Statement $I$: In a vernier callipers,one vernier scale division is always smaller than one main scale division.
Statement $II$: The vernier constant is given by one main scale division multiplied by the number of vernier scale division.
In the light of the above statements,choose the correct answer from the options given below.
- ABoth Statement $I$ and Statement $II$ are false.
- BStatement $I$ is true but Statement $II$ is false.
- CBoth Statement $I$ and Statement $II$ are true.
- DStatement $I$ is false but Statement $II$ is true.
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Similar Questions
One main scale division of a Vernier calliper is equal to $1 \text{ mm}$ and the number of divisions on the Vernier scale is $10$. When both the jaws touch each other, the Vernier scale shifts to the left of zero of the main scale in such a way that $4^{th}$ Vernier division coincides with a division of the main scale. If the Vernier calliper measures the length of a wire to be $1 \text{ cm}$, the actual length of the wire is : (in $\text{ cm}$)
DifficultNEET 2026
View SolutionIn an experiment to find out the diameter of a wire using a screw gauge,the following observations were noted:
$(a)$ The 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 $2.5\,mm$.
$(d)$ The $45^{\text{th}}$ division of the circular scale is on the pitch line.
$(e)$ The instrument has a $0.03\,mm$ negative zero error.
Then the diameter of the wire is $...........\,mm$.
$(a)$ The 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 $2.5\,mm$.
$(d)$ The $45^{\text{th}}$ division of the circular scale is on the pitch line.
$(e)$ The instrument has a $0.03\,mm$ negative zero error.
Then the diameter of the wire is $...........\,mm$.
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 SolutionThere are $100$ divisions on the circular scale of a screw gauge of pitch $1 \,mm$. With no measuring quantity in between the jaws,the zero of the circular scale lies $5$ divisions below the reference line. The diameter of a wire is then measured using this screw gauge. It is found that $4$ linear scale divisions are clearly visible while $60$ divisions on the circular scale coincide with the reference line. The diameter of the wire is: (in $\,mm$)
DifficultJEE MAIN 2024
View SolutionThe vernier calliper used for measurement has a positive zero error of $0.3 \ mm$. While taking measurement for the internal diameter of a vessel,it was observed that the zero of the vernier scale lies between $9.5 \ cm$ and $9.6 \ cm$ of the main scale and the $6^{th}$ division of the vernier scale coincides with a main scale division. If the least count of the vernier calliper is $0.01 \ cm$,the correct value of the diameter will be: (in $cm$)
Medium
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