Two full turns of the circular scale of screw gauge cover a distance of $1\,mm$ on scale. The total number of divisions on circular scale is $50$. Further, it is found that screw gauge has a zero error of $+0.03\,mm$. While measuring the diameter of a thin wire a student notes the main scale reading of $3\,mm$ and the number of circular scale division in line, with the main scale is $35$. The diameter of the wire is .......... $mm$

- A
$3.32$

- B
$3.73$

- C
$3.67$

- D
$3.38$

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]

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

- [JEE MAIN 2024]

An experiment is performed to obtain the value of acceleration due to gravity $g$ by using a simple pendulum of length $L$. In this experiment time for $100\, oscillations$ is measured by using a watch of $1\, second$ least count and the value is $90.0\, seconds$. The length $L$ is measured by using a meter scale of least count $1\, mm$ and the value is $20.0\, cm$. The error in the determination of $g$ would be ........... $\%$

- [JEE MAIN 2014]

A travelling microscope is used to determine the refractive index of a glass slab. If $40$ divisions are there in $1 \; cm$ on main scale and $50$ Vernier scale divisions are equal to $49$ main scale divisions, then least count of the travelling microscope is $\dots \; \times 10^{-6} \; m$

- [JEE MAIN 2022]

In a vernier callipers, each $cm$ on the main scale is divided into $20$ equal parts. If tenth vernier scale division coincides with nineth main scale division. Then the value of vernier constant will be $\dots \; \times 10^{-2} \;mm$

- [JEE MAIN 2022]