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:

  • [JEE MAIN 2021]
  • A

    Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$

  • B

    $A$ is not correct but $R$ is correct.

  • C

    Both $A$ and $R$ are correct and $R$ is NOT the correct explanation of $A$.

  • D

    ${A}$ is correct but ${R}$ is not correct.

Similar Questions

In an experiment to find out the diameter of wire using screw gauge, the following observation were noted.
$(a)$ Screw moves $0.5\,mm$ on main scale in one complete rotation
$(b)$ Total divisions on circular scale $=50$
$(c)$ Main scale reading is $2.5\,mm$
$(d)$ $45^{\text {th }}$ division of circular scale is in the pitch line
$(e)$ Instrument has $0.03 \;mm$ negative error
Then the diameter of wire is $...........\,mm$

  • [JEE MAIN 2022]

A vernier callipers used by student has $20$ divisions in $1\;cm$ on main scale. $10$ vernier divisions coincide with $9$ main scale divisions. When jaws are closed, zero of main scale is on left of zero of vernier scale and $6\ th$ division of vernier scale coincides with any of main scale divisions. He places a wooden cylinder in between the jaws and measures the length. The zero of vernier scale is on right of $3.20\;cm$ and $8\ th$ vernier division coincides with any main scale division. When he measures thickness of cylinder he finds that zero of vernier scale lies on right of $1.50\;cm$ mark of main scale and $6\ th$ division of vernier scale coincides with any main scale division. The correct values of measured length and diameter are respectively

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 $0.005 \ mm$.

  • [IIT 2015]

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?

  • [IIT 2022]

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  .......... $cm$