The smallest division on the main scale of a Vernier calipers is $0.1 \text{ cm}$. Ten divisions of the Vernier scale correspond to nine divisions of the main scale. The figure below on the left shows the reading of this calipers with no gap between its two jaws. The figure on the right shows the reading with a solid sphere held between the jaws. The correct diameter of the sphere is (in $\text{ cm}$)

  • A
    $3.07$
  • B
    $3.11$
  • C
    $3.15$
  • D
    $3.17$

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$A$ screw gauge has a pitch of $1.5\; mm$ and there is no zero error. The linear scale has markings at $MSD = 1\; mm$ and there are $100$ equal divisions on the circular scale. When the diameter of a sphere is measured with this instrument,the $2\; mm$ mark is visible on the linear scale,but the $3\; mm$ mark is not visible. The $76^{th}$ division of the circular scale is in line with the linear scale. What is the diameter of the sphere in $mm$?

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Which devices are used for the measurement of lengths of different orders?

$A$ screw gauge of pitch $0.5\,mm$ is used to measure the diameter of a uniform wire of length $6.8\,cm$. The main scale reading is $1.5\,mm$ and the circular scale reading is $7$. The calculated curved surface area of the wire to appropriate significant figures is $......\,cm^2$. [Screw gauge has $50$ divisions on the circular scale]

In a screw gauge, the zero of the main scale reference line coincides with the fifth division of the circular scale when two studs are in contact. There are $100$ divisions on the circular scale and the pitch of the screw gauge is $0.1 \text{ mm}$. When the diameter of a sphere is measured, the reading of the main scale is $5 \text{ mm}$ and the $50^{th}$ division of the circular scale coincides with the reference line of the main scale. The diameter of the sphere is . . . . . . $\text{mm}$.

In a vernier calliper,when both jaws touch each other,the zero of the vernier scale shifts towards the left and its $4^{\text{th}}$ division coincides exactly with a certain division on the main scale. If $50$ vernier scale divisions $(VSD)$ are equal to $49$ main scale divisions $(MSD)$ and the zero error in the instrument is $0.04 \text{ mm}$,then how many main scale divisions are there in $1 \text{ cm}$?

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