The internal and external diameters of a hollow cylinder measured with vernier calipers are $(5.73 \pm 0.01) \text{ cm}$ and $(6.01 \pm 0.01) \text{ cm}$ respectively. Then the thickness of the cylinder wall is

  • A
    $(0.28 \pm 0.01) \text{ cm}$
  • B
    $(0.28 \pm 0.02) \text{ cm}$
  • C
    $(0.14 \pm 0.02) \text{ cm}$
  • D
    $(0.14 \pm 0.01) \text{ cm}$

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In an experiment to find out the diameter of a wire using a screw gauge,the following observations were noted:
$(A)$ Screw moves $0.5 \ mm$ on the main scale in one complete rotation.
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$(C)$ Main scale reading is $2.5 \ mm$.
$(D)$ $45^{\text{th}}$ division of the circular scale is in the reference line.
$(E)$ Instrument has $0.03 \ mm$ negative zero error.
Then the diameter of the wire is: (in $mm$)

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The main scale division of a vernier caliper is $1 \ mm$. The vernier scale divisions are in an arithmetic progression $(A.P.)$; the $1^{st}$ division is $0.95 \ mm$, the $2^{nd}$ division is $0.90 \ mm$, and so on. When an object is placed between the jaws of the vernier caliper, the zero of the vernier scale lies between $3.1 \ cm$ and $3.2 \ cm$, and the $4^{th}$ division of the vernier scale coincides with a main scale division. The reading of the vernier caliper is .......... $cm$.

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Given below are two statements: one is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$.
Assertion $(A)$: In a Vernier calliper,if a positive zero error exists,then while taking measurements,the reading taken will be more than the actual reading.
Reason $(R)$: The zero error in a Vernier calliper might have happened due to a manufacturing defect or due to rough handling.
In the light of the above statements,choose the correct answer from the options given below:

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