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When 5V potential difference is applied across a wire length 0.1 m, the drift speed of electron is 2.5 x 10-4 ms-1 . If the electron density in the wire is 8 x 1028 m-3 , the resistivity of the material is close to
- a)1.6 x 10-8 Ω
- b)1.6 x 10-7 Ω
- c)1.6 x 10-6 Ω
- d)1.6 x 10-5 Ω
Correct answer is option 'D'. Can you explain this answer?
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Given data:
- Potential difference (V) = 5V
- Wire length (L) = 0.1m
- Drift speed of electron (v_d) = 2.5 x 10^-4 m/s
- Electron density (n) = 8 x 10^28 m^-3
Calculating current density:
- Current density (J) is given by the formula J = n * e * v_d, where e is the charge of an electron.
- The charge of an electron (e) is approximately 1.6 x 10^-19 C.
- Substituting the given values, we get J = 8 x 10^28 * 1.6 x 10^-19 * 2.5 x 10^-4 = 3.2 A/m^2.
Calculating resistivity:
- Resistivity (ρ) is given by the formula ρ = E / J * A, where E is the potential difference, J is the current density, and A is the cross-sectional area of the wire.
- The cross-sectional area A can be calculated using the length of the wire.
- A = π * r^2, where r is the radius of the wire.
- Assuming the wire is cylindrical, the radius can be calculated as r = √(A / π).
- Substituting the values, we get A = π * (0.1 / (2 * π))^2 ≈ 2.5 x 10^-3 m^2.
- Now, substituting the values of E, J, and A, we get ρ = 5 / 3.2 * 2.5 x 10^-3 ≈ 1.6 x 10^-5 Ω.
Therefore, the resistivity of the material is close to 1.6 x 10^-5 Ω, which corresponds to option 'D'.
- Potential difference (V) = 5V
- Wire length (L) = 0.1m
- Drift speed of electron (v_d) = 2.5 x 10^-4 m/s
- Electron density (n) = 8 x 10^28 m^-3
Calculating current density:
- Current density (J) is given by the formula J = n * e * v_d, where e is the charge of an electron.
- The charge of an electron (e) is approximately 1.6 x 10^-19 C.
- Substituting the given values, we get J = 8 x 10^28 * 1.6 x 10^-19 * 2.5 x 10^-4 = 3.2 A/m^2.
Calculating resistivity:
- Resistivity (ρ) is given by the formula ρ = E / J * A, where E is the potential difference, J is the current density, and A is the cross-sectional area of the wire.
- The cross-sectional area A can be calculated using the length of the wire.
- A = π * r^2, where r is the radius of the wire.
- Assuming the wire is cylindrical, the radius can be calculated as r = √(A / π).
- Substituting the values, we get A = π * (0.1 / (2 * π))^2 ≈ 2.5 x 10^-3 m^2.
- Now, substituting the values of E, J, and A, we get ρ = 5 / 3.2 * 2.5 x 10^-3 ≈ 1.6 x 10^-5 Ω.
Therefore, the resistivity of the material is close to 1.6 x 10^-5 Ω, which corresponds to option 'D'.
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When5Vpotential difference is applied across a wire length 0.1 m,the drift speed of electron is 2.5 x 10-4 ms-1.If the electron density in the wire is 8 x 1028 m-3 ,the resistivity of the material is close toa)1.6 x 10-8Ωb)1.6 x 10-7Ωc)1.6 x 10-6Ωd)1.6 x 10-5ΩCorrect answer is option 'D'. Can you explain this answer?
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When5Vpotential difference is applied across a wire length 0.1 m,the drift speed of electron is 2.5 x 10-4 ms-1.If the electron density in the wire is 8 x 1028 m-3 ,the resistivity of the material is close toa)1.6 x 10-8Ωb)1.6 x 10-7Ωc)1.6 x 10-6Ωd)1.6 x 10-5ΩCorrect answer is option 'D'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about When5Vpotential difference is applied across a wire length 0.1 m,the drift speed of electron is 2.5 x 10-4 ms-1.If the electron density in the wire is 8 x 1028 m-3 ,the resistivity of the material is close toa)1.6 x 10-8Ωb)1.6 x 10-7Ωc)1.6 x 10-6Ωd)1.6 x 10-5ΩCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When5Vpotential difference is applied across a wire length 0.1 m,the drift speed of electron is 2.5 x 10-4 ms-1.If the electron density in the wire is 8 x 1028 m-3 ,the resistivity of the material is close toa)1.6 x 10-8Ωb)1.6 x 10-7Ωc)1.6 x 10-6Ωd)1.6 x 10-5ΩCorrect answer is option 'D'. Can you explain this answer?.
When5Vpotential difference is applied across a wire length 0.1 m,the drift speed of electron is 2.5 x 10-4 ms-1.If the electron density in the wire is 8 x 1028 m-3 ,the resistivity of the material is close toa)1.6 x 10-8Ωb)1.6 x 10-7Ωc)1.6 x 10-6Ωd)1.6 x 10-5ΩCorrect answer is option 'D'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about When5Vpotential difference is applied across a wire length 0.1 m,the drift speed of electron is 2.5 x 10-4 ms-1.If the electron density in the wire is 8 x 1028 m-3 ,the resistivity of the material is close toa)1.6 x 10-8Ωb)1.6 x 10-7Ωc)1.6 x 10-6Ωd)1.6 x 10-5ΩCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When5Vpotential difference is applied across a wire length 0.1 m,the drift speed of electron is 2.5 x 10-4 ms-1.If the electron density in the wire is 8 x 1028 m-3 ,the resistivity of the material is close toa)1.6 x 10-8Ωb)1.6 x 10-7Ωc)1.6 x 10-6Ωd)1.6 x 10-5ΩCorrect answer is option 'D'. Can you explain this answer?.
Solutions for When5Vpotential difference is applied across a wire length 0.1 m,the drift speed of electron is 2.5 x 10-4 ms-1.If the electron density in the wire is 8 x 1028 m-3 ,the resistivity of the material is close toa)1.6 x 10-8Ωb)1.6 x 10-7Ωc)1.6 x 10-6Ωd)1.6 x 10-5ΩCorrect answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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When5Vpotential difference is applied across a wire length 0.1 m,the drift speed of electron is 2.5 x 10-4 ms-1.If the electron density in the wire is 8 x 1028 m-3 ,the resistivity of the material is close toa)1.6 x 10-8Ωb)1.6 x 10-7Ωc)1.6 x 10-6Ωd)1.6 x 10-5ΩCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for When5Vpotential difference is applied across a wire length 0.1 m,the drift speed of electron is 2.5 x 10-4 ms-1.If the electron density in the wire is 8 x 1028 m-3 ,the resistivity of the material is close toa)1.6 x 10-8Ωb)1.6 x 10-7Ωc)1.6 x 10-6Ωd)1.6 x 10-5ΩCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of When5Vpotential difference is applied across a wire length 0.1 m,the drift speed of electron is 2.5 x 10-4 ms-1.If the electron density in the wire is 8 x 1028 m-3 ,the resistivity of the material is close toa)1.6 x 10-8Ωb)1.6 x 10-7Ωc)1.6 x 10-6Ωd)1.6 x 10-5ΩCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
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