The pitch and the number of divisions on the circular scale for a given screw gauge are $0.5\,mm$ and $100$ respectively. When the screw gauge is fully tightened without any object,the zero of its circular scale lies $3$ divisions below the mean line. The readings of the main scale and the circular scale for a thin sheet are $5.5\,mm$ and $48$ respectively. The thickness of this sheet is: (in $,mm$)

  • A
    $5.755$
  • B
    $5.950$
  • C
    $5.725$
  • D
    $5.740$

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Similar Questions

In an experiment with Vernier callipers of least count $0.1\,mm$,when two jaws are joined together,the zero of the Vernier scale lies to the right of the zero of the main scale and the $6^{th}$ division of the Vernier scale coincides with a main scale division. While measuring the diameter of a spherical bob,the zero of the Vernier scale lies between $3.2\,cm$ and $3.3\,cm$ marks,and the $4^{th}$ division of the Vernier scale coincides with a main scale division. The diameter of the bob is measured as $.......\,cm$.

In a Vernier Calipers,$10$ divisions of the Vernier scale are equal to $9$ divisions of the main scale. When both jaws of the Vernier calipers touch each other,the zero of the Vernier scale is shifted to the left of the zero of the main scale,and the $4^{\text{th}}$ Vernier scale division exactly coincides with a main scale division. One main scale division is equal to $1\,mm$. While measuring the diameter of a spherical body,the body is held between the two jaws. It is observed that the zero of the Vernier scale lies between $30$ and $31$ divisions of the main scale,and the $6^{\text{th}}$ Vernier scale division exactly coincides with a main scale division. The diameter of the spherical body is $.......\,cm$.

In a vernier callipers, $50$ vernier scale divisions are equal to $48$ main scale divisions. If one main scale division $= 0.05 \ mm$, then the least count of the vernier callipers is . . . . . . $mm$.

The least count of a vernier caliper is $\frac{1}{20N} \text{ cm}$. The value of one division on the main scale is $1 \text{ mm}$. Then the number of divisions of the main scale that coincide with $N$ divisions of the vernier scale is:

When the gap is closed without placing any object in a screw gauge whose least count is $0.005 \ mm$,the $5^{th}$ division on its circular scale coincides with the reference line on the main scale. When a small sphere is placed,the reading on the main scale advances by $4$ divisions,whereas the circular scale reading advances by five times the corresponding reading when no object was placed. There are $200$ divisions on the circular scale. The radius of the sphere is .......... $mm$.

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