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:
Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$
$A$ is not correct but $R$ is correct.
Both $A$ and $R$ are correct and $R$ is NOT the correct explanation of $A$.
${A}$ is correct but ${R}$ is not correct.
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 $0.005 \ mm$.
Area of the cross-section of a wire is measured using a screw gauge. The pitch of the main scale is $0.5 mm$. The circular scale has 100 divisions and for one full rotation of the circular scale, the main scale shifts by two divisions. The measured readings are listed below.
Measurement condition | Main scale reading | Circular scale reading |
Two arms of gauge touching each other without wire | $0$ division | $4$ division |
Attempt-$1$: With wire | $4$ division | $20$ division |
Attempt-$2$: With wire | $4$ division | $16$ division |
What are the diameter and cross-sectional area of the wire measured using the screw gauge?
Main scale division of vernier calliper is $1\ mm$ and vernier scale division are in $A.P. ; 1^{st}$ division is $0.95\ mm$ ; $2^{nd}$ division is $0.9\ mm$ and so on. When an object is placed between jaws of vernier calliper, zero of vernier lies between $3.1\ cm$ and $3.2\ cm$ and $4^{th}$ division of vernier coincide with main scale division. Reading of vernier is .......... $cm$