Q

# Solve this problem 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 scaledivisions. The thickness of a wire measured by this screw gauge gives a reading of 4 main scale division

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

776 Views

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

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

$least count =\frac{0.25}{5\times 100}cm = 5\times 10^{-4}cm$

$Reading = 4\times 0.05 cm +30\times 5\times 10^{-4}cm$

$=(0.2+0.0150)cm$

= 0.2150cm

Option 1)

0.4300 cm

Option 2)

0.2150 cm

Option 3)

0.3150 cm

Option 4)

0.0430 cm

Exams
Articles
Questions