$A$ screw gauge has some zero error but its value is unknown. We have two identical rods. When the first rod is inserted in the screw gauge,the state of the instrument is shown by diagram $(I)$. When both the rods are inserted together in series,the state is shown by diagram $(II)$. What is the zero error of the instrument in $mm$? Given: $1 \, M.S.D. = 100 \, C.S.D. = 1 \, mm$.

  • A
    $-0.16$
  • B
    $+0.16$
  • C
    $+0.14$
  • D
    $-0.14$

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Similar Questions

$A$ Vernier calipers has $1 \,mm$ marks on the main scale. It has $20$ equal divisions on the Vernier scale which match with $16$ main scale divisions. For this Vernier calipers, the least count is (in $\,mm$)

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Area of the cross-section of a wire is measured using a screw gauge. The pitch of the main scale is $0.5 \text{ mm}$. The circular scale has $100$ divisions and for one full rotation of the circular scale, the main scale shifts by two divisions. The measured readings are listed below.
Measurement conditionMain scale readingCircular scale reading
Two arms of gauge touching each other without wire$0$ division$4$ division
Attempt-$1$: With wire$4$ division$20$ division
Attempt-$2$: With wire$4$ division$16$ division

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In a vernier callipers,$(N+1)$ divisions of vernier scale coincide with $N$ divisions of main scale. If $1 \text{ MSD}$ represents $0.1 \text{ mm}$,the vernier constant (in $\text{cm}$) is:

Answer the following:
$(a)$ You are given a thread and a metre scale. How will you estimate the diameter of the thread?
$(b)$ $A$ screw gauge has a pitch of $1.0\; mm$ and $200$ divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale?
$(c)$ The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of $100$ measurements of the diameter expected to yield a more reliable estimate than a set of $5$ measurements only?

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