Consider a Vernier callipers in which each 1 cm on the main scale is divided into 8 equal divisions and a screw gauge with 100 divisions on its circular scale. In the Vernier callipers, 5 divisions of the Vernier scale coincide with 4 divisions on the main scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then:

Engineering

Physics

Vernier Calipers and Travelling Microscope

Question

If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01 mm.

If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01 mm.

If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm.

If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm.

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$1 MSD = (\frac{1 cm}{8})$

$5VSD = 4 MSD = \frac{4}{8} cm = \frac{1}{2} cm$

1 VSD = $\frac{1}{10}$ cm = 1 mm

So, LC = 1 MSD – 1 VSD = $(\frac{10}{8} - 1) mm = (\frac{5}{4} - 1) mm = \frac{1}{4}$ = 0.25 mm

For Screw Gauge

(B) Pitch = 2 × 0.25 mm = 0.5 mm

100 circular parts = pitch = 0.5 mm

So, 1 circular part = $\frac{0.5}{100}$ mm = 0.005 mm

100 circular parts = 2 × 0.5 mm = 1 mm

So 1 circular division = 0.01 mm