$A$ 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]

  • A
    $6.8$
  • B
    $3.4$
  • C
    $3.9$
  • D
    $2.4$

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Two full turns of the circular scale of a screw gauge cover a distance of $1 \ mm$ on its main scale. The total number of divisions on the circular scale is $50$. The screw gauge has a zero error of $-0.03 \ mm$. While measuring the diameter of a thin wire,a student notes the main scale reading of $3 \ mm$ and the $35^{\text{th}}$ division of the circular scale coincides with the reference line of the main scale. The diameter of the wire is: (in $mm$)

$A$ vernier callipers used by a student has $20$ divisions in $1\;cm$ on the main scale. $10$ vernier divisions coincide with $9$ main scale divisions. When the jaws are closed,the zero of the main scale is to the left of the zero of the vernier scale,and the $6^{th}$ division of the vernier scale coincides with a main scale division. The student places a wooden cylinder between the jaws to measure its length. The zero of the vernier scale is to the right of the $3.20\;cm$ mark,and the $8^{th}$ vernier division coincides with a main scale division. When measuring the thickness (diameter) of the cylinder,the zero of the vernier scale lies to the right of the $1.50\;cm$ mark,and the $6^{th}$ vernier division coincides with a main scale division. The correct values of the measured length and diameter are respectively:

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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 $\text{ cm}$)

Consider the diameter of a spherical object being measured with the help of a Vernier callipers. Suppose its $10$ Vernier Scale Divisions $(V.S.D.)$ are equal to its $9$ Main Scale Divisions $(M.S.D.)$. The least division on the $M.S.$ is $0.1 \ cm$ and the zero of $V.S.$ is at $x=0.1 \ cm$ when the jaws of the Vernier callipers are closed. If the main scale reading for the diameter is $M=5 \ cm$ and the number of the coinciding vernier division is $8$,the measured diameter after zero error correction is: (in $cm$)

The diameter of a wire is measured with a screw gauge having a least count of $0.01\;mm$. Which of the following correctly expresses the diameter?

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