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 $50$. Further,it is found that 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 number of circular scale divisions in line with the main scale as $35$. The diameter of the wire is ....... $mm$

  • A
    $3.38$
  • B
    $3.32$
  • C
    $3.73$
  • D
    $3.67$

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Similar Questions

In a vernier callipers,$50$ vernier scale divisions are equal to $48$ main scale divisions. If one main scale division $= 0.05 \ mm$,then the least count of the vernier callipers is . . . . . . $mm$.

In a screw gauge,there are $100$ divisions on the circular scale and the main scale moves by $0.5\,mm$ on a complete rotation of the circular scale. The zero of the circular scale lies $6$ divisions below the line of graduation when two studs are brought in contact with each other. When a wire is placed between the studs,$4$ linear scale divisions are clearly visible while the $46^{\text{th}}$ division of the circular scale coincides with the reference line. The diameter of the wire is $...........\times 10^{-2}\,mm$.

In a screw gauge,the fifth division of the circular scale coincides with the reference line when the ratchet is closed. There are $50$ divisions on the circular scale,and the main scale moves by $0.5 \, mm$ on a complete rotation. For a particular observation,the reading on the main scale is $5 \, mm$ and the $20^{th}$ division of the circular scale coincides with the reference line. Calculate the true reading in $mm$.

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 diameter of a sphere is measured using a vernier caliper whose $9$ divisions of the main scale are equal to $10$ divisions of the vernier scale. The shortest division on the main scale is equal to $1 \,mm$. The main scale reading is $2 \,cm$ and the second division of the vernier scale coincides with a division on the main scale. If the mass of the sphere is $8.635 \,g$, the density of the sphere is:

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