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:

  • A
    $142$
  • B
    $71$
  • C
    $42$
  • D
    $21$

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Similar Questions

The figure below shows a particular position of the Vernier calipers on a centimetre scale. In this position,the value of $x$ shown in the figure is .......... $cm$ (figure is not to scale).

If in a Vernier callipers $10 \,VSD$ coincides with $8 \,MSD$,then the least count of the Vernier calliper is ............ $m$ [given $1 \,MSD = 1 \,mm$].

$A$ specially designed Vernier calliper has the main scale least count of $1 \,mm$. On the Vernier scale,there are $10$ equal divisions and they match with $11$ main scale divisions. Then,the least count of the Vernier calliper is ........... $mm$.

In a screw gauge,there are $100$ divisions on the circular scale and the main scale moves by $0.5\,mm$ on a complete rotation of the circular scale. The zero of the circular scale lies $6$ divisions below the line of graduation when two studs are brought in contact with each other. When a wire is placed between the studs,$4$ linear scale divisions are clearly visible while the $46^{\text{th}}$ division of the circular scale coincides with the reference line. The diameter of the wire is $...........\times 10^{-2}\,mm$.

Which of the following is the most precise device for measuring length:
$A)$ a vernier callipers with $20$ divisions on the sliding scale
$B)$ a screw gauge of pitch $1 \; mm$ and $100$ divisions on the circular scale
$C)$ an optical instrument that can measure length to within a wavelength of light?

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