$A$ 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 ....... $degree$. (in $^{\circ}$)

  • A
    $59$
  • B
    $58.59$
  • C
    $58.77$
  • D
    $58.65$

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Similar Questions

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

What are the diameter and cross-sectional area of the wire measured using the screw gauge?

$A$ screw gauge gives the following reading when used to measure the diameter of a wire.
Main scale reading : $0 \ mm$
Circular scale reading : $52 \ divisions$
Given that $1 \ mm$ on the main scale corresponds to $100$ divisions of the circular scale. The diameter of the wire from the above data is: (in $cm$)

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

There are $100$ divisions on the circular scale of a screw gauge of pitch $1 \,mm$. With no measuring quantity in between the jaws,the zero of the circular scale lies $5$ divisions below the reference line. The diameter of a wire is then measured using this screw gauge. It is found that $4$ linear scale divisions are clearly visible while $60$ divisions on the circular scale coincide with the reference line. The diameter of the wire is: (in $\,mm$)

In a screw gauge,$5$ complete rotations of the screw cause it to move a linear distance of $0.25\, cm$. There are $100$ circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of $4$ main scale divisions and $30$ circular scale divisions. Assuming negligible zero error,the thickness of the wire is (in $, cm$)

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