In an experiment the angles are required to be measured using an instrument. 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°), then the least count of the instrument is

  • Option 1)

    one minute

  • Option 2)

    half minute

  • Option 3)

    one degree

  • Option 4)

    half degree

 

Answers (1)

As we learnt in

To measure the diameter of small spherical cylindrical body using Vernier Callipers -

Vernier Constant

= 1 Main scale division - 1 V.S. Division 

V.C= 1 M.S.D - 1 V.S.D

M.S.D=  Main Scale Reading

V.S.D= Vernier Scale Reading

 

- wherein

Total observed reading = N+n \times V.C

N= Nth division 

Observations: 

1.    Vernier constant (least count) of the Vernier Callipers: 

                    1 M.S.D. = 1 mm

            10 vernier scale divisions = 9 main scale divisions

                i.e.     10 V.S.D. = 9 M.S.D.

    \therefore        1 V.S.D. = =\frac{9}{10} M.S.D.

                Vernier Constant (L.C.) = 1 M.S.D. - 1 V.S.D. = 1 M.S.D. -\frac{9}{10} M.S.D.

                            =\left(1-\frac{9}{10} \right )M.S.D.=\frac{1}{10}\times1M.S.D.

                            =\frac{1}{10}\times1mm=0.1mm=0.01cm

2.    Zero error: (i) ........ cm (ii) ............cm (iii) ..........cm 

        Mean Zero Error (e) = ............ cm 

        Mean Zero Correction (c) = - (Mean Zero Error)

                                                = .......... cm

    

 

 

 

 

 

 V.S.D. = \frac{29}{30} M.S.D.

Least count = 1 M.S.D. - 1 V.S.D.

  = 1M.S.D. - \frac{29}{30} M.S.D. = \frac{1}{30}M.S.D.

    =\frac{1}{30}\times 0.5\ degree = 1\ minute

 


Option 1)

one minute

Correct

Option 2)

half minute

Incorrect

Option 3)

one degree

Incorrect

Option 4)

half degree

Incorrect

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