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# Consider a Vernier calipers 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

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 0.01 mm.
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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 0.005 mm.
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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.
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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.
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Solution

## The correct options are 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 0.005 mm. 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.Vernier Calipers: 1 cm is divided into 8 divisions, 1 main scale division is given as 18cm Given 5 Vernier scale coincides with 4 main scale divisions 5V.S.D=4M.S.D L.C= 1M.S.D−45M.S.D L.C= 1M.S.D−45M.S.D=15M.S.D=140cm Given pitch of Screw gauge is 2 times L.C. of V.S p=2×140cm=120cm No. of divisions on circular scale =100 The least count of screw gauge is =pitchno.ofdivisionsoncircularscale =120100=0.0005 cm=0.005 mm Hence, option B is correct. Given least count of the linear scale of the screw gauge is twice the least count of the Vernier calipers. One complete rotation of the circular scale moves it by two divisions on the linear scale, so pitch is 2 times linear scale division of screw gauge, Pitch =2 × linear scale division of screw gauge =4× L.C of V.S =4×140=110 The least count of screw gauge is pitchno.ofdivisionsoncircularscale No. of divisions on circular scale =100 L.C=110100=0.001 cm=0.01 mm Option C is correct . Hence, option B and C is correct.  Suggest Corrections  4      Similar questions  Related Videos   Significant Figures
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