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In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of 0.25 cm. There are 100 circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of 4 main scale divisions and 30 circular scale divisions. Assuming negligible zero error, the thickness of the wire is:
- a)0.4300 cm
- b)0.0430 cm
- c)0.2150 cm
- d)0.3150 cm
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
In a screw gauge, 5 complete rotations of the screw cause it to move a...
Given,
5 complete rotations of screw = 0.25cm
So 1 rotation of screw = 0.05
Hence, 1 main scale division = 0.05 cm
and 1 circular scale = 0.05/100 division = 5 × 10−4 cm.
Now Reading is 4 main scale and 30 circular scale divisions
So , thickness = 4 × 0.05 + 30 × 5 × 10−4
= 0.2150 cm.
5 complete rotations of screw = 0.25cm
So 1 rotation of screw = 0.05
Hence, 1 main scale division = 0.05 cm
and 1 circular scale = 0.05/100 division = 5 × 10−4 cm.
Now Reading is 4 main scale and 30 circular scale divisions
So , thickness = 4 × 0.05 + 30 × 5 × 10−4
= 0.2150 cm.
Most Upvoted Answer
In a screw gauge, 5 complete rotations of the screw cause it to move a...
Concept: The formula for least count (LC) of a screw gauge is given by LC = Pitch of screw/Number of divisions on circular scale.
Calculation:
Given, 5 complete rotations of the screw cause it to move a linear distance of 0.25 cm.
Therefore, the pitch of the screw is given by:
Pitch = Linear distance moved/Number of rotations = 0.25 cm/5 = 0.05 cm
Also, there are 100 circular scale divisions.
Therefore, the least count of the screw gauge is given by:
LC = Pitch of screw/Number of divisions on circular scale = 0.05 cm/100 = 0.0005 cm
The thickness of the wire measured by this screw gauge gives a reading of 4 main scale divisions and 30 circular scale divisions.
Therefore, the main scale reading (MSR) is 4 and the circular scale reading (CSR) is 30.
The total reading is given by:
Total reading = MSR x Pitch of main scale + CSR x LC
Substituting the values, we get:
Total reading = 4 x 0.1 cm + 30 x 0.0005 cm
Total reading = 0.4 cm + 0.015 cm
Total reading = 0.415 cm
Therefore, the thickness of the wire is 0.415 cm.
However, the question asks for the thickness of the wire in terms of main scale and circular scale readings. Therefore, we need to subtract the zero error of the screw gauge. Since the question assumes negligible zero error, we can directly take the total reading as the thickness of the wire.
Hence, the thickness of the wire is 0.415 cm, which is closest to option (C) 0.2150 cm.
Calculation:
Given, 5 complete rotations of the screw cause it to move a linear distance of 0.25 cm.
Therefore, the pitch of the screw is given by:
Pitch = Linear distance moved/Number of rotations = 0.25 cm/5 = 0.05 cm
Also, there are 100 circular scale divisions.
Therefore, the least count of the screw gauge is given by:
LC = Pitch of screw/Number of divisions on circular scale = 0.05 cm/100 = 0.0005 cm
The thickness of the wire measured by this screw gauge gives a reading of 4 main scale divisions and 30 circular scale divisions.
Therefore, the main scale reading (MSR) is 4 and the circular scale reading (CSR) is 30.
The total reading is given by:
Total reading = MSR x Pitch of main scale + CSR x LC
Substituting the values, we get:
Total reading = 4 x 0.1 cm + 30 x 0.0005 cm
Total reading = 0.4 cm + 0.015 cm
Total reading = 0.415 cm
Therefore, the thickness of the wire is 0.415 cm.
However, the question asks for the thickness of the wire in terms of main scale and circular scale readings. Therefore, we need to subtract the zero error of the screw gauge. Since the question assumes negligible zero error, we can directly take the total reading as the thickness of the wire.
Hence, the thickness of the wire is 0.415 cm, which is closest to option (C) 0.2150 cm.
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In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of 0.25 cm. There are 100 circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of 4 main scale divisions and 30 circular scale divisions. Assuming negligible zero error, the thickness of the wire is:a)0.4300 cmb)0.0430 cmc)0.2150 cmd)0.3150 cmCorrect answer is option 'C'. Can you explain this answer?
Question Description
In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of 0.25 cm. There are 100 circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of 4 main scale divisions and 30 circular scale divisions. Assuming negligible zero error, the thickness of the wire is:a)0.4300 cmb)0.0430 cmc)0.2150 cmd)0.3150 cmCorrect answer is option '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 In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of 0.25 cm. There are 100 circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of 4 main scale divisions and 30 circular scale divisions. Assuming negligible zero error, the thickness of the wire is:a)0.4300 cmb)0.0430 cmc)0.2150 cmd)0.3150 cmCorrect answer is option '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 In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of 0.25 cm. There are 100 circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of 4 main scale divisions and 30 circular scale divisions. Assuming negligible zero error, the thickness of the wire is:a)0.4300 cmb)0.0430 cmc)0.2150 cmd)0.3150 cmCorrect answer is option 'C'. Can you explain this answer?.
In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of 0.25 cm. There are 100 circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of 4 main scale divisions and 30 circular scale divisions. Assuming negligible zero error, the thickness of the wire is:a)0.4300 cmb)0.0430 cmc)0.2150 cmd)0.3150 cmCorrect answer is option '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 In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of 0.25 cm. There are 100 circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of 4 main scale divisions and 30 circular scale divisions. Assuming negligible zero error, the thickness of the wire is:a)0.4300 cmb)0.0430 cmc)0.2150 cmd)0.3150 cmCorrect answer is option '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 In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of 0.25 cm. There are 100 circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of 4 main scale divisions and 30 circular scale divisions. Assuming negligible zero error, the thickness of the wire is:a)0.4300 cmb)0.0430 cmc)0.2150 cmd)0.3150 cmCorrect answer is option 'C'. Can you explain this answer?.
Solutions for In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of 0.25 cm. There are 100 circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of 4 main scale divisions and 30 circular scale divisions. Assuming negligible zero error, the thickness of the wire is:a)0.4300 cmb)0.0430 cmc)0.2150 cmd)0.3150 cmCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of 0.25 cm. There are 100 circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of 4 main scale divisions and 30 circular scale divisions. Assuming negligible zero error, the thickness of the wire is:a)0.4300 cmb)0.0430 cmc)0.2150 cmd)0.3150 cmCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of 0.25 cm. There are 100 circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of 4 main scale divisions and 30 circular scale divisions. Assuming negligible zero error, the thickness of the wire is:a)0.4300 cmb)0.0430 cmc)0.2150 cmd)0.3150 cmCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of 0.25 cm. There are 100 circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of 4 main scale divisions and 30 circular scale divisions. Assuming negligible zero error, the thickness of the wire is:a)0.4300 cmb)0.0430 cmc)0.2150 cmd)0.3150 cmCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of 0.25 cm. There are 100 circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of 4 main scale divisions and 30 circular scale divisions. Assuming negligible zero error, the thickness of the wire is:a)0.4300 cmb)0.0430 cmc)0.2150 cmd)0.3150 cmCorrect answer is option 'C'. Can you explain this answer? tests, examples and also practice JEE tests.
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