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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 :

  • Option 1)

    0.52 cm

  • Option 2)

    0.052 cm

  • Option 3)

    0.026 cm

  • Option 4)

    0.005 cm

 

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

    

 

 

 

 

 

   Least count of screw gauge=\frac{1}{100}mm=0.01 mm

Diameter= \: Division \: \: on \: \: circular \: scale \times least \: \: count+main \: \: scale \: \: reading

                  =52\times \frac{1}{100}+100= 0.52mm

                 diameter=0.052 cm


Option 1)

0.52 cm

This option is incorrect

Option 2)

0.052 cm

This option is correct

Option 3)

0.026 cm

This option is incorrect

Option 4)

0.005 cm

This option is incorrect

Posted by

Sabhrant Ambastha

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