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

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Similar Questions

$A$ tiny metallic rectangular sheet has a length and breadth of $5 \ mm$ and $2.5 \ mm$,respectively. Using a specially designed screw gauge which has a pitch of $0.75 \ mm$ and $15$ divisions on the circular scale,you are asked to find the area of the sheet. In this measurement,the maximum fractional error will be $\frac{x}{100}$ where $x$ is . . . . . .

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 callipers,$10$ divisions of vernier scale coincide with $9$ divisions of main scale. One division of main scale is of $0.1 \ cm$. If in the measurement of inner diameter of a cylinder,the zero of the vernier scale lies between $1.3 \ cm$ and $1.4 \ cm$ of the main scale and the $2^{\text{nd}}$ division of the vernier scale coincides with a main scale division,then the diameter will be: (in $cm$)

Assertion $A$: If in five complete rotations of the circular scale,the distance travelled on the main scale of the screw gauge is $5 \, mm$ and there are $50$ total divisions on the circular scale,then the least count is $0.001 \, cm$.
Reason $R$: $\text{Least Count} = \frac{\text{Pitch}}{\text{Total divisions on circular scale}}$
In the light of the above statements,choose the most appropriate answer from the options given below:

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

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