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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:
  • 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.
  • 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.
  • 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.
Correct answer is option 'B,C'. Can you explain this answer?
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Consider a Vernier callipers in which each 1 cm on the main scale is d...
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Consider a Vernier callipers in which each 1 cm on the main scale is d...
Vernier calliper:
M.S.D =1/8 cm = 0.125 cm = 1.25 mm
5 V.S.D. = 4 M.S.D
⇒ V.S.D. = 4/5 M.S.D. = 4/5 × 1.25mm = 1 mm
Least count = M.S.D. – V.S.D = 1.25 – 1 = 0.25 mm
Screw gauge:
No. of circular scale divisions = 100
1 rotation (pitch) = 2 linear scale divisions
» If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is –
Pitch = 2 × Least Count of vernier callipers = 2 × 0.25mm = 0.5 mm
Least count of the screw gauge
= Pitch/no. of circular scale divisions = 0.5/100 =0.005mm
Option (B) is correct.
» 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 –
Linear scale division of screw gauge
= 2 × Least count of vernier callipers
= 2 × 0.25 mm
= 0.5 mm
Now,
Pitch = 2 linear scale divisions = 2 × 0.5 = 1mm
Least count of the screw gauge
= Pitch/ No. of circular divisions
= 1/100 mm
0.01 mm
Option (C) is correct.
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Consider a Vernier callipers in which each 1 cm on the main scale is divided into 8 equal divisionsanda 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 scaleand 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.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.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.Correct answer is option 'B,C'. Can you explain this answer?
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Consider a Vernier callipers in which each 1 cm on the main scale is divided into 8 equal divisionsanda 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 scaleand 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.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.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.Correct answer is option 'B,C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Consider a Vernier callipers in which each 1 cm on the main scale is divided into 8 equal divisionsanda 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 scaleand 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.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.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.Correct answer is option 'B,C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider a Vernier callipers in which each 1 cm on the main scale is divided into 8 equal divisionsanda 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 scaleand 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.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.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.Correct answer is option 'B,C'. Can you explain this answer?.
Solutions for Consider a Vernier callipers in which each 1 cm on the main scale is divided into 8 equal divisionsanda 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 scaleand 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.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.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.Correct answer is option 'B,C'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of Consider a Vernier callipers in which each 1 cm on the main scale is divided into 8 equal divisionsanda 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 scaleand 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.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.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.Correct answer is option 'B,C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of Consider a Vernier callipers in which each 1 cm on the main scale is divided into 8 equal divisionsanda 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 scaleand 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.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.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.Correct answer is option 'B,C'. Can you explain this answer?, a detailed solution for Consider a Vernier callipers in which each 1 cm on the main scale is divided into 8 equal divisionsanda 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 scaleand 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.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.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.Correct answer is option 'B,C'. Can you explain this answer? has been provided alongside types of Consider a Vernier callipers in which each 1 cm on the main scale is divided into 8 equal divisionsanda 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 scaleand 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.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.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.Correct answer is option 'B,C'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice Consider a Vernier callipers in which each 1 cm on the main scale is divided into 8 equal divisionsanda 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 scaleand 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.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.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.Correct answer is option 'B,C'. Can you explain this answer? tests, examples and also practice JEE tests.
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