(C) $1$. Vernier Callipers: $1$ main scale division $(MSD)$ $= 1/8 \ cm = 0.125 \ cm$. Given $5$ $VSD$ $= 4$ $MSD$,so $1$ $VSD$ $= 4/5$ $MSD$ $= 0.8 \

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(C) $1$. Vernier Callipers: $1$ main scale division $(MSD)$ $= 1/8 \ cm = 0.125 \ cm$. Given $5$ $VSD$ $= 4$ $MSD$,so $1$ $VSD$ $= 4/5$ $MSD$ $= 0.8 \times 0.125 \ cm = 0.1 \ cm$. Least Count $(LC)$ $= 1$ $MSD$ $- 1$ $VSD$ $= 0.125 - 0.1 = 0.025 \ cm = 0.25 \ mm$.$2$. Screw Gauge: One rotation moves it by $2$ divisions on the linear scale. Let $1$ linear scale division $= x$. Pitch $P = 2x$. $LC$ $= P / 100 = 2x / 100 = x / 50$.$3$. Case $1$: Pitch $P = 2 \times$ $LC$ of Vernier $= 2 \times 0.25 \ mm = 0.5 \ mm$. Then $LC$ of screw gauge $= 0.5 \ mm / 100 = 0.005 \ mm$. (Option $B$ is correct).$4$. Case $2$: Linear scale division $x = 2 \times$ $LC$ of Vernier $= 2 \times 0.25 \ mm = 0.5 \ mm$. Then $LC$ of screw gauge $= x / 50 = 0.5 \ mm / 50 = 0.01 \ mm$. (Option $C$ is correct).

Class 11 Physics Units, Dimensions and Measurement Vernier Calipers, Micrometer screw gauge

Consider a Vernier callipers in which each $1 \ cm$ on the main scale is divided into $8$ equal divisions and a screw gauge with $100$ divisions on its circular scale. In the Vernier callipers,$5$ divisions of the Vernier scale coincide with $4$ divisions on the main scale and in the screw gauge,one complete rotation of the circular scale moves it by two divisions on the linear scale. Then:
$(A)$ If the pitch of the screw gauge is twice the least count of the Vernier callipers,the least count of the screw gauge is $0.01 \ mm$.
$(B)$ If the pitch of the screw gauge is twice the least count of the Vernier callipers,the least count of the screw gauge is $0.005 \ mm$.
$(C)$ If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers,the least count of the screw gauge is $0.01 \ mm$.
$(D)$ If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers,the least count of the screw gauge is $0.005 \ mm$.

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$(A, D)$
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$(B, D)$
C
$(B, C)$
D
$(C, D)$

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Units, Dimensions and Measurement

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Vernier Calipers, Micrometer screw gauge

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When both jaws of vernier callipers touch each other, the zero mark of the vernier scale is to the right of the zero mark of the main scale, and the $4^{\text{th}}$ mark on the vernier scale coincides with a certain mark on the main scale. While measuring the length of a cylinder, the observer observes $15$ divisions on the main scale and the $5^{\text{th}}$ division of the vernier scale coincides with a main scale division. The measured length of the cylinder is . . . . . . $mm$. (Least count of Vernier calliper $= 0.1 \ mm$)

DifficultJEE MAIN 2026

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In a Vernier calipers, when both jaws touch each other, the zero of the Vernier scale is shifted to the right of the zero of the main scale and the $7^{th}$ Vernier division coincides with a main scale division. If the value of $1$ main scale division is $1 \text{ mm}$ and there are $10$ Vernier scale divisions, then the Vernier caliper has

MediumJEE MAIN 2026

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The 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

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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 $mm$)

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The least count of a screw gauge is $0.01 \ mm$. If the pitch is increased by $75\%$ and the number of divisions on the circular scale is reduced by $50\%$,the new least count will be . . . . . . $\times 10^{-3} \ mm$.

MediumJEE MAIN 2025

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The least count of a screw gauge is $0.01 \ mm$. If the pitch is increased by $75\%$ and the number of divisions on the circular scale is reduced by $50\%$,the new least count will be . . . . . . $\times 10^{-3} \ mm$.

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