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Standard XII

Physics

Deviation in Prism

A spectromete...

Question

A spectrometer gives the following reading when used to measure the angle of a prism.
Main scale reading : 58.5 degree
Vernier scale reading : 09 divisions
Given that 1 division on main scale corresponds to 0.5 degree. Total divisions on the vernier scale is 30 and match with 29 divisions of the main scale. The angle of prism from the above data is

58.59 degree

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58.77 degree

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58.65 degree

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59 degree

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Solution

The correct option is C 58.65 degree
Reading = M.S.R. + V.S.R. $\times$ L.C.
$= 58.5 + 9 \times \frac{0.5}{30}$
= 58.65 degree

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Similar questions

Q. A spectrometer gives the following reading when used to measure the angle of a prism. Main scale reading: 58.5 degree, Vernier scale reading : 9 divisions. Given that 1 division on main scale corresponds to 0.5 degree. Total divisions on the vernier scale is 30 and match with 29 divisions of the main scale. The angle of the prism from the above data:

Q. A vernier caliper has no zero error, one main scale division =

1 mm

and

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main scale division =

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vernier scale divisions. A spring is held between jaws without exerting force on jaws. Main scale reading is

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, and 5th vernier scale division coincides with a main scale division. When jaws are pressed by force of

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on both sides, main scale reading remains

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, but 1st division of vernier scale coincides with a main scale division. Now spring is removed, cut into two pieces and when inserted between jaws without any force, main scale reading =

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and 5th vernier scale division coincides with a main scale. If we apply force of

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on both the jaws, main scale reading gets

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and

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vernier scale division coincides with a main scale division. Find

^{'} x^{'}

Q. 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 main scale corresponds to 100 divisions of the circular scale. The diameter of wire from the above data is :

Q. Consider the following data for a vernier calliper:
10 main scale division = 1 cm.
10 vernier division matches with 9 Main scale division.

While measuring the zero error of the vernier we see that zero of vernier scale is to the right of the zero marking of the main scale with

6^{t h}

vernier division coinciding with main scale divisions.

The actual reading for length measurement is 4.3 cm with 2nd vernier division coinciding with main scale graduation.

Estimate the length.

Consider the following data: 10 main scale division = 1cm, $10^{t h}$ vernier division match with $9^{t h}$ Main scale division. While measuring the zero error of the vernier we see that zero of vernier scale is to the right of the zero marking of the main scale with $6^{t h}$ vernier division coinciding with main scale divisions. The actual reading for length measurement is 4.3 cm with 2nd vernier division coinciding with main scale graduation. Estimate the length.

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The correct option is C 58.65 degree Reading = M.S.R. + V.S.R. × L.C. =58.5+9×0.530 = 58.65 degree

The correct option is C 58.65 degree
Reading = M.S.R. + V.S.R. $\times$ L.C.
$= 58.5 + 9 \times \frac{0.5}{30}$
= 58.65 degree