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 $cm$)

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)

The diameter of a wire is measured with a screw gauge having a least count of $0.01\;mm$. Which of the following correctly expresses the diameter?

While measuring the diameter of a wire using a screw gauge, the following readings were noted. The main scale reading is $1 \,mm$ and the circular scale reading is equal to $42$ divisions. The pitch of the screw gauge is $1 \,mm$ and it has $100$ divisions on the circular scale. The diameter of the wire is $\frac{x}{50} \,mm$. The value of $x$ is:

In a screw gauge,the fifth division of the circular scale coincides with the reference line when the ratchet is closed. There are $50$ divisions on the circular scale,and the main scale moves by $0.5 \, mm$ on a complete rotation. For a particular observation,the reading on the main scale is $5 \, mm$ and the $20^{th}$ division of the circular scale coincides with the reference line. Calculate the true reading in $mm$.

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 $0.005 \ mm$.