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  •  A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it ?Option: 1  A meter scale.

 A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it ?

Option: 1

 A meter scale.
 


Option: 2

A vernier calliper where the 10 divisions in vernier scale matches with 9 division in main scale and main scale has 10 divisions in 1 cm.


Option: 3

 A screw gauge having 100 divisions in the circular scale and pitch as 1 mm.

 


Option: 4

 A screw gauge having 50 divisions in the circular scale and pitch as 1 mm.


Answers (1)

As we learnt in concept

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

 

     He used to measure it by a vernier callipier where 10 divisions in the vernier scale and main scale has 10 divisions in 1cm

 

 

 

 

Posted by

Ramraj Saini

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