If the screw on a screw gauge is given six rotations,it moves by $3\; mm$ on the main scale. If there are $50$ divisions on the circular scale,the least count of the screw gauge is:

  • A
    $0.001\; mm$
  • B
    $0.001\; cm$
  • C
    $0.02\; mm$
  • D
    $0.01\; cm$

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Similar Questions

Answer the following:
$(a)$ You are given a thread and a metre scale. How will you estimate the diameter of the thread?
$(b)$ $A$ screw gauge has a pitch of $1.0\; mm$ and $200$ divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale?
$(c)$ The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of $100$ measurements of the diameter expected to yield a more reliable estimate than a set of $5$ measurements only?

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)

$A$ screw gauge has some zero error but its value is unknown. We have two identical rods. When the first rod is inserted in the screw gauge,the state of the instrument is shown by diagram $(I)$. When both the rods are inserted together in series,the state is shown by diagram $(II)$. What is the zero error of the instrument in $mm$? Given: $1 \, M.S.D. = 100 \, C.S.D. = 1 \, mm$.

Identify the physical quantity that cannot be measured using a spherometer:

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

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