Question 293: Units , Dimensions and Error Analysis

Question

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:

Accepted Answer

In given Vernier callipers, each 1 cm is equally divided into 8 main scale divisions (MSD). Thus, 1 MSD = $${1 \over 8}$$ = 0.125 cm. Further, 4 main scale divisions coincide with 5 Vernier scale divisions (VSD) i.e., 4 MSD = 5 VSD. Thus, 1 VSD = 4/5 MSD = 0.8 $$	imes$$ 0.125 = 0.1 cm. The least count of the Vernier callipers is given by
LC = 1 MSD $$-$$ 1 VSD = 0.125 $$-$$ 0.1 = 0.025 cm.
In screw gauge, let l be the distance between two adjacent divisions on the linear scale. The pitch p of the screw gauge is the distance travelled on the linear scale when it makes one complete rotation. Since circular scale moves by two divisions on the linear scale when it makes one complete rotation, we get p = 2l. The least count of the screw gauge is defined as ratio of the pitch to the number of divisions on the circular scale (n) i.e.,
$$LC' = {p \over n} = {{2l} \over {100}} = {l \over {50}}$$ ....... (1)
If p = 2 LC = 2(0.025) = 0.05 cm, then $$l = {p \over 2} = 0.025$$ cm. Substitute l in equation (1) to get the least count of the screw gauge
$$LC' = {{0.025} \over {50}} = 5 	imes {10^{ - 4}}$$ cm = 0.005 mm.
If l = 2 LC = 2(0.025) = 0.05 cm then  equation (1) gives
$$LC' = {{0.05} \over {50}} = 1 	imes {10^{ - 3}}$$ cm = 0.01 mm.

Answer

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.

Answer

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.

Answer

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.

Answer

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.

Units , Dimensions and Error Analysis

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