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A steel wire of diameter 0.5 mm and Young’s modulus 2 x 1011 Nm-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 is π = 3.2.
    Correct answer is '3.00'. Can you explain this answer?
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    A steel wire of diameter 0.5 mm and Young’s modulus 2 x1011Nm-2c...
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    A steel wire of diameter 0.5 mm and Young’s modulus 2 x1011Nm-2c...
    's modulus 200 GPa is stretched by a force of 1000 N. Calculate the elongation of the wire.

    To calculate the elongation of the wire, we can use Hooke's Law, which states that the elongation of a material is directly proportional to the force applied and inversely proportional to its cross-sectional area.

    First, we need to calculate the cross-sectional area of the wire. The formula for the cross-sectional area of a wire is:

    Area = π * (diameter/2)^2

    Plugging in the values, we have:

    Area = π * (0.5 mm/2)^2
    = π * (0.25 mm)^2
    = 0.19635 mm^2

    Now, we can use Hooke's Law to calculate the elongation. The formula for elongation is:

    Elongation = (Force * Length) / (Area * Young's Modulus)

    Plugging in the values, we have:

    Elongation = (1000 N * Length) / (0.19635 mm^2 * 200 GPa)

    Since the units are not consistent, we need to convert them.

    1 mm^2 = 1 x 10^-6 m^2
    1 GPa = 1 x 10^9 Pa

    Elongation = (1000 N * Length) / (0.19635 mm^2 * 200 x 10^9 Pa)
    = (1000 N * Length) / (0.19635 x 10^-6 m^2 * 200 x 10^9 Pa)
    = (1000 N * Length) / (0.03927 m^2 * 200 x 10^9 Pa)
    = (1000 N * Length) / (7.854 x 10^9 N/m^2)
    = (1000 N * Length) / (7.854 GPa)

    To find the elongation, we need to know the length of the wire. Without that information, we cannot calculate the elongation.
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    A steel wire of diameter 0.5 mm and Young’s modulus 2 x1011Nm-2carries 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-2and is π= 3.2.Correct answer is '3.00'. Can you explain this answer?
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    A steel wire of diameter 0.5 mm and Young’s modulus 2 x1011Nm-2carries 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-2and is π= 3.2.Correct answer is '3.00'. 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 x1011Nm-2carries 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-2and is π= 3.2.Correct answer is '3.00'. 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 x1011Nm-2carries 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-2and is π= 3.2.Correct answer is '3.00'. Can you explain this answer?.
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