A student measured diameter of a small steel ball using a screw gauge of least count 0.001, cm

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1.Units, Dimensions and Measurement

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A student measured the diameter of a small steel ball using a screw gauge of least count $0.001\, cm.$ The main scale reading is $5\, mm$ and zero of circular scale division coincides with $25$ divisions above the reference level. If screw gauge has a zero error of $-0.004 \,cm,$ the correct diameter of the ball is

$0.521 \,cm$

$\;$$0.525\, cm$

$0.529\, cm$

$\;$ $0.053\, cm$

(NEET-2018)

Solution

Diameter of the ball
$= MSR \,+ \,CSR \times$ (Least count) $- Zero\, error$
$= 5\ mm + 25 \times 0.001\ cm – (-0.004)\ cm$
$= 0.5\ cm + 25 \times 0.001\ cm – (-0.004)\ cm = 0.529\ cm.$

Standard 11

Physics

In a screw gauge, $5$ complete rotations of circular scale give $1.5\, mm$ reading on linear scale. Circular scale has $50$ divisions. Least count of the screw gauge is

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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, the state of the instrument is shown by diagram $(I).$ When both the rods are inserted together in series then the state is shown by the diagram $(II).$ What is the zero error of the instrument ? ………. $mm$

$1\,M.S.D. = 100\, C.S.D. = 1\, mm $

medium

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

easy

One main scale division of a vernier callipers is $a$ $cm$ and $n ^{\text {th }}$ division of the vernier scale coincide with $( n -1)^{\text {th }}$ division of the main scale. The least count of the callipers in $mm$ is

medium

(JEE MAIN-2021)

If $50$ Vernier divisions are equal to $49$ main scale divisions of a travelling microscope and one smallest reading of main scale is $0.5 \mathrm{~mm}$, the Vernier constant of travelling microscope is:

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(JEE MAIN-2024)

Solution

Similar Questions

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