The length of a cylinder is measured with the help of vernier callipers whose nine divisions of the main scale are equal to ten divisions of the vernier scale. The smallest division on the main scale is $0.5 \ mm$. It is observed that the zero of the vernier scale has just crossed the $78^{th}$ division of the main scale,and the fourth division of the vernier scale coincides with a main scale division. The length of the cylinder is $..... \ mm$.

  • A
    $78.40$
  • B
    $78.20$
  • C
    $39.40$
  • D
    $39.20$

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Similar Questions

The diagrams show readings of a screw gauge. Figure $(i)$ shows the zero error reading when the screw gauge is closed,and figure $(ii)$ shows the reading when the screw gauge is being used to measure the diameter of a ball-bearing. What is the diameter of the ball-bearing in $mm$? There are $50$ divisions on the circular scale.

$A$ 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]

Consider a Vernier callipers in which each $1 \ cm$ on the main scale is divided into $8$ equal divisions and a screw gauge with $100$ divisions on its circular scale. In the Vernier callipers,$5$ divisions of the Vernier scale coincide with $4$ divisions on the main scale and in the screw gauge,one complete rotation of the circular scale moves it by two divisions on the linear scale. Then:
$(A)$ If the pitch of the screw gauge is twice the least count of the Vernier callipers,the least count of the screw gauge is $0.01 \ mm$.
$(B)$ If the pitch of the screw gauge is twice the least count of the Vernier callipers,the least count of the screw gauge is $0.005 \ mm$.
$(C)$ If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers,the least count of the screw gauge is $0.01 \ mm$.
$(D)$ If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers,the least count of the screw gauge is $0.005 \ mm$.

The circular divisions of a screw gauge are $50$. It moves $0.5 \ mm$ on the main scale in one rotation. When the diameter of a wire is measured,the main scale reading is $3.5 \ mm$ and the circular scale reading is $32$. If the zero error (positive) in the screw gauge is $0.06 \ mm$,then the diameter of the wire is:

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 $\text{ cm}$)

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