#### In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of 0.25 cm. There are 100 circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of 4 main scale divisions and 30 circular scale divisions. Assuming negligible zero error, the thickness of the wire is : Option 1) 0.4300 cm Option 2) 0.2150 cm Option 3) 0.3150 cm Option 4) 0.0430 cm

As we learned

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

- wherein

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.

1 V.S.D. =  M.S.D.

Vernier Constant (L.C.) = 1 M.S.D. - 1 V.S.D. = 1 M.S.D.  M.S.D.

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

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

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

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

= 0.2150cm

Option 1)

0.4300 cm

Option 2)

0.2150 cm

Option 3)

0.3150 cm

Option 4)

0.0430 cm