In a screw gauge,$5$ complete rotations of the circular scale give a $1.5 \, mm$ reading on the linear scale. The circular scale has $50$ divisions. The least count of the screw gauge is: (in $, mm$)

  • A
    $0.006$
  • B
    $0.003$
  • C
    $0.015$
  • D
    $0.03$

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Similar Questions

One main scale division of a vernier caliper is equal to $m$ units. If $n^{\text{th}}$ division of the main scale coincides with $(n+1)^{\text{th}}$ division of the vernier scale,the least count of the vernier caliper is:

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In a vernier callipers,$10$ divisions of vernier scale coincide with $9$ divisions of main scale. One division of main scale is of $0.1 \ cm$. If in the measurement of inner diameter of a cylinder,the zero of the vernier scale lies between $1.3 \ cm$ and $1.4 \ cm$ of the main scale and the $2^{\text{nd}}$ division of the vernier scale coincides with a main scale division,then the diameter will be: (in $cm$)

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:

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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