The pitch of the screw gauge is $1\, mm$ and there are $100$ divisions on the circular scale. When nothing is put in between the jaws,the zero of the circular scale lies $8$ divisions below the reference line. When a wire is placed between the jaws,the first linear scale division is clearly visible while $72^{nd}$ division on the circular scale coincides with the reference line. The radius of the wire is.........$mm$
- A$1.64$
- B$0.82$
- C$1.80$
- D$0.90$
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Similar Questions
$A$ spectrometer gives the following reading when used to measure the angle of a prism.
Main scale reading : $58.5^{\circ}$
Vernier scale reading : $09$ divisions
Given that $1$ division on the main scale corresponds to $0.5^{\circ}$. The total number of divisions on the Vernier scale is $30$,which matches $29$ divisions of the main scale. The angle of the prism from the above data is ....... . (in $^{\circ}$)
Main scale reading : $58.5^{\circ}$
Vernier scale reading : $09$ divisions
Given that $1$ division on the main scale corresponds to $0.5^{\circ}$. The total number of divisions on the Vernier scale is $30$,which matches $29$ divisions of the main scale. The angle of the prism from the above data is ....... . (in $^{\circ}$)
MediumAIEEE 2012
View Solution$A$ screw gauge has a least count of $0.01\, mm$ and there are $50$ divisions on its circular scale. The pitch of the screw gauge is $........\, mm$.
$N$ divisions on the main scale of a vernier calliper coincide with $(N + 1)$ divisions of the vernier scale. If each division of the main scale is $a$ units,then the least count of the instrument is:
DifficultAIEEE 2012
View SolutionIn an experiment,the angles are required to be measured using an instrument in which $29$ divisions of the main scale exactly coincide with the $30$ divisions of the vernier scale. If the smallest division of the main scale is half a degree $(=0.5^{\circ})$,then the least count of the instrument is
DifficultTS EAMCET 2020
View SolutionThe vernier constant of Vernier callipers is $0.1 \,mm$ and it has zero error of $(-0.05) \,cm$. While measuring the diameter of a sphere,the main scale reading is $1.7 \,cm$ and the coinciding vernier division is $5$. The corrected diameter will be ........... $\times 10^{-2} \,cm$.
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