(only for genius,.) Two full turns of a circular scale of a screw Guage covers a distance 1mm on its main scale.the total number on divisions on the circular scale is 50 . further it is found that the screw gauge has a zero error of - 0.03 while measuring the diameter of a thin wire a student observes the main scale reading 3mm and the number of a circular scale division in the line with the main scale divisions as 35,,,then find diameter of wire...
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Two full turns of the circular scale of a screw gauge cover a distance of 1 mm on its main scale. The total number of divisions on the circular scale is 50. Further, it is found that the screw gauge has a zero error of – 0.03 mm. While measuring the diameter of a thin wire, a student notes the main scale reading of 3 mm and the number of circular scale divisions in line with the main scale as 35. The diameter of the wire is
Option 1) 3.38 mm Option 2) 3.32 mm Option 3) 3.73 mm Option 4) 3.67 mm.
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P prateek
Answered 1 year, 4 months ago
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
Diameter = M.S.R + C.S.R x LC + Z.E
3+35\times ( \frac{0.5}{50})+0.03 =3.38mm
thnx bahia for giving my answers
and you draw very beautiful
good artist like me
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