Consider a vernier callipers in which each 1 cm on the main scale is divided into 8 equal divisions and a screw gauge ,

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asked Mar 30, 2018 in Physics by shabnam praween (19,050 points) 5 6 8

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 division 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,
(a) If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01mm.
(b) If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005mm.
(c) 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.
(d) 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

1 Answer

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answered Mar 30, 2018 by santoshjha (25,550 points) 4 5 6
selected Apr 1, 2018 by Vikash Kumar
 
Best answer

For Vernier calipers

(C) & (D) Least count of linear scale of screw gauge = 0.05 cm
pitch = 0.05 x 2cm = 0.1 cm

Leastcount of screw gauge =0.1/100=0.01mm

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