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A steel wire of diameter 0.5 mm and Young’s modulus 2 ×1011 N m−2 carries a load of mass M. The length of the wire with the load is 1.0 m. A vernier scale with 10 divisions is attached to the end of this wire. Next to the steel wire is a reference wire to which a main scale, of least count 1.0 mm, is attached. The 10 divisions of the vernier scale correspond to 9 divisions of the main scale. Initially, the zero of vernier scale coincides with the zero of main scale. If the load on the steel wire is increased by 1.2 kg, the vernier scale division which coincides with a main scale division is __________. Take g = 10 ms−2 and π = 3.2 .
Correct answer is '3'. Can you explain this answer?
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Verified Answer
A steel wire of diameter 0.5 mm and Young’s modulus 2 ×101...
According to Hooke’s law,
The least count of the vernier caliper is
Therefore, the third division of vernier scale coincides with the main scale.
The least count of the vernier caliper is
Therefore, the third division of vernier scale coincides with the main scale.
Most Upvoted Answer
A steel wire of diameter 0.5 mm and Young’s modulus 2 ×101...
's modulus 2.1 x 10^11 Pa is stretched by a force of 500 N. Calculate the elongation of the wire.
Solution:
The formula for calculating the elongation (ΔL) of a wire under tension is:
ΔL = FL / AY
Where F is the force applied to the wire, L is the original length of the wire, A is the cross-sectional area of the wire, and Y is the Young's modulus of the wire.
To apply this formula, we need to find the original length and cross-sectional area of the wire.
The cross-sectional area of the wire can be calculated using the formula:
A = πr^2
Where r is the radius of the wire. Since the diameter of the wire is given as 0.5 mm, the radius is 0.25 mm or 0.00025 m.
Therefore, A = π(0.00025)^2 = 1.96 x 10^-7 m^2.
The original length of the wire is not given, so we can assume it to be 1 meter for simplicity.
Now, we can substitute the values in the formula for ΔL:
ΔL = FL / AY = (500 N)(1 m) / (1.96 x 10^-7 m^2)(2.1 x 10^11 Pa) = 1.14 x 10^-3 m
Therefore, the elongation of the wire is 1.14 mm.
Solution:
The formula for calculating the elongation (ΔL) of a wire under tension is:
ΔL = FL / AY
Where F is the force applied to the wire, L is the original length of the wire, A is the cross-sectional area of the wire, and Y is the Young's modulus of the wire.
To apply this formula, we need to find the original length and cross-sectional area of the wire.
The cross-sectional area of the wire can be calculated using the formula:
A = πr^2
Where r is the radius of the wire. Since the diameter of the wire is given as 0.5 mm, the radius is 0.25 mm or 0.00025 m.
Therefore, A = π(0.00025)^2 = 1.96 x 10^-7 m^2.
The original length of the wire is not given, so we can assume it to be 1 meter for simplicity.
Now, we can substitute the values in the formula for ΔL:
ΔL = FL / AY = (500 N)(1 m) / (1.96 x 10^-7 m^2)(2.1 x 10^11 Pa) = 1.14 x 10^-3 m
Therefore, the elongation of the wire is 1.14 mm.
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A steel wire of diameter 0.5 mm and Young’s modulus 2 ×1011 N m−2 carries a load of mass M. The length of the wire with the load is 1.0 m. A vernier scale with 10 divisions is attached to the end of this wire. Next to the steel wire is a reference wire to which a main scale, of least count 1.0 mm, is attached. The 10 divisions of the vernier scale correspond to 9 divisions of the main scale. Initially, the zero of vernier scale coincides with the zero of main scale. If the load on the steel wire is increased by 1.2 kg, the vernier scale division which coincides with a main scale division is __________. Take g = 10 ms−2 and π = 3.2 .Correct answer is '3'. Can you explain this answer?
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A steel wire of diameter 0.5 mm and Young’s modulus 2 ×1011 N m−2 carries a load of mass M. The length of the wire with the load is 1.0 m. A vernier scale with 10 divisions is attached to the end of this wire. Next to the steel wire is a reference wire to which a main scale, of least count 1.0 mm, is attached. The 10 divisions of the vernier scale correspond to 9 divisions of the main scale. Initially, the zero of vernier scale coincides with the zero of main scale. If the load on the steel wire is increased by 1.2 kg, the vernier scale division which coincides with a main scale division is __________. Take g = 10 ms−2 and π = 3.2 .Correct answer is '3'. 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 A steel wire of diameter 0.5 mm and Young’s modulus 2 ×1011 N m−2 carries a load of mass M. The length of the wire with the load is 1.0 m. A vernier scale with 10 divisions is attached to the end of this wire. Next to the steel wire is a reference wire to which a main scale, of least count 1.0 mm, is attached. The 10 divisions of the vernier scale correspond to 9 divisions of the main scale. Initially, the zero of vernier scale coincides with the zero of main scale. If the load on the steel wire is increased by 1.2 kg, the vernier scale division which coincides with a main scale division is __________. Take g = 10 ms−2 and π = 3.2 .Correct answer is '3'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A steel wire of diameter 0.5 mm and Young’s modulus 2 ×1011 N m−2 carries a load of mass M. The length of the wire with the load is 1.0 m. A vernier scale with 10 divisions is attached to the end of this wire. Next to the steel wire is a reference wire to which a main scale, of least count 1.0 mm, is attached. The 10 divisions of the vernier scale correspond to 9 divisions of the main scale. Initially, the zero of vernier scale coincides with the zero of main scale. If the load on the steel wire is increased by 1.2 kg, the vernier scale division which coincides with a main scale division is __________. Take g = 10 ms−2 and π = 3.2 .Correct answer is '3'. Can you explain this answer?.
A steel wire of diameter 0.5 mm and Young’s modulus 2 ×1011 N m−2 carries a load of mass M. The length of the wire with the load is 1.0 m. A vernier scale with 10 divisions is attached to the end of this wire. Next to the steel wire is a reference wire to which a main scale, of least count 1.0 mm, is attached. The 10 divisions of the vernier scale correspond to 9 divisions of the main scale. Initially, the zero of vernier scale coincides with the zero of main scale. If the load on the steel wire is increased by 1.2 kg, the vernier scale division which coincides with a main scale division is __________. Take g = 10 ms−2 and π = 3.2 .Correct answer is '3'. 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 A steel wire of diameter 0.5 mm and Young’s modulus 2 ×1011 N m−2 carries a load of mass M. The length of the wire with the load is 1.0 m. A vernier scale with 10 divisions is attached to the end of this wire. Next to the steel wire is a reference wire to which a main scale, of least count 1.0 mm, is attached. The 10 divisions of the vernier scale correspond to 9 divisions of the main scale. Initially, the zero of vernier scale coincides with the zero of main scale. If the load on the steel wire is increased by 1.2 kg, the vernier scale division which coincides with a main scale division is __________. Take g = 10 ms−2 and π = 3.2 .Correct answer is '3'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A steel wire of diameter 0.5 mm and Young’s modulus 2 ×1011 N m−2 carries a load of mass M. The length of the wire with the load is 1.0 m. A vernier scale with 10 divisions is attached to the end of this wire. Next to the steel wire is a reference wire to which a main scale, of least count 1.0 mm, is attached. The 10 divisions of the vernier scale correspond to 9 divisions of the main scale. Initially, the zero of vernier scale coincides with the zero of main scale. If the load on the steel wire is increased by 1.2 kg, the vernier scale division which coincides with a main scale division is __________. Take g = 10 ms−2 and π = 3.2 .Correct answer is '3'. Can you explain this answer?.
Solutions for A steel wire of diameter 0.5 mm and Young’s modulus 2 ×1011 N m−2 carries a load of mass M. The length of the wire with the load is 1.0 m. A vernier scale with 10 divisions is attached to the end of this wire. Next to the steel wire is a reference wire to which a main scale, of least count 1.0 mm, is attached. The 10 divisions of the vernier scale correspond to 9 divisions of the main scale. Initially, the zero of vernier scale coincides with the zero of main scale. If the load on the steel wire is increased by 1.2 kg, the vernier scale division which coincides with a main scale division is __________. Take g = 10 ms−2 and π = 3.2 .Correct answer is '3'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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ample number of questions to practice A steel wire of diameter 0.5 mm and Young’s modulus 2 ×1011 N m−2 carries a load of mass M. The length of the wire with the load is 1.0 m. A vernier scale with 10 divisions is attached to the end of this wire. Next to the steel wire is a reference wire to which a main scale, of least count 1.0 mm, is attached. The 10 divisions of the vernier scale correspond to 9 divisions of the main scale. Initially, the zero of vernier scale coincides with the zero of main scale. If the load on the steel wire is increased by 1.2 kg, the vernier scale division which coincides with a main scale division is __________. Take g = 10 ms−2 and π = 3.2 .Correct answer is '3'. Can you explain this answer? tests, examples and also practice JEE tests.
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