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A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1 m and its cross-sectional area is 4.9i * 0 ^ - 7 * m ^ 2 If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency 140 rad s ^ - 1 If the Young's modulus of the material of the wire is n * 10 ^ 9 * N * m ^ - 2 the value of nis a) 4 2 (ל c) 4.5?
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A 0.1 kg mass is suspended from a wire of negligible mass. The length ...
Introduction
To find the Young's modulus (E) of the wire, we start with the information provided. The mass (m) is 0.1 kg, the length of the wire (L) is 1 m, the cross-sectional area (A) is 4.9 × 10^-7 m^2, and the angular frequency (ω) is 140 rad/s.
Simple Harmonic Motion and Angular Frequency
- The angular frequency of a mass-spring system can be expressed as:
ω = √(k/m)
- Here, k is the spring constant, which can be related to Young's modulus (E) by the formula:
k = (E * A) / L
Combining the Formulas
- By substituting the expression for k into the angular frequency formula, we get:
ω = √((E * A) / (m * L))
- Rearranging yields:
E = (ω^2 * m * L) / A
Substituting Values
- Now substituting the known values:
- m = 0.1 kg
- L = 1 m
- A = 4.9 × 10^-7 m^2
- ω = 140 rad/s
- Calculation gives:
E = (140^2 * 0.1 * 1) / (4.9 × 10^-7)
Final Calculation
- Evaluating the above expression leads to:
E = (19600 * 0.1) / (4.9 × 10^-7)
E = 1960 / (4.9 × 10^-7) ≈ 4.0 × 10^9 N/m²
- Thus, n = 4.
Conclusion
The value of n is 4.
To find the Young's modulus (E) of the wire, we start with the information provided. The mass (m) is 0.1 kg, the length of the wire (L) is 1 m, the cross-sectional area (A) is 4.9 × 10^-7 m^2, and the angular frequency (ω) is 140 rad/s.
Simple Harmonic Motion and Angular Frequency
- The angular frequency of a mass-spring system can be expressed as:
ω = √(k/m)
- Here, k is the spring constant, which can be related to Young's modulus (E) by the formula:
k = (E * A) / L
Combining the Formulas
- By substituting the expression for k into the angular frequency formula, we get:
ω = √((E * A) / (m * L))
- Rearranging yields:
E = (ω^2 * m * L) / A
Substituting Values
- Now substituting the known values:
- m = 0.1 kg
- L = 1 m
- A = 4.9 × 10^-7 m^2
- ω = 140 rad/s
- Calculation gives:
E = (140^2 * 0.1 * 1) / (4.9 × 10^-7)
Final Calculation
- Evaluating the above expression leads to:
E = (19600 * 0.1) / (4.9 × 10^-7)
E = 1960 / (4.9 × 10^-7) ≈ 4.0 × 10^9 N/m²
- Thus, n = 4.
Conclusion
The value of n is 4.
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A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1 m and its cross-sectional area is 4.9i * 0 ^ - 7 * m ^ 2 If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency 140 rad s ^ - 1 If the Young's modulus of the material of the wire is n * 10 ^ 9 * N * m ^ - 2 the value of nis a) 4 2 (ל c) 4.5?
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A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1 m and its cross-sectional area is 4.9i * 0 ^ - 7 * m ^ 2 If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency 140 rad s ^ - 1 If the Young's modulus of the material of the wire is n * 10 ^ 9 * N * m ^ - 2 the value of nis a) 4 2 (ל c) 4.5? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1 m and its cross-sectional area is 4.9i * 0 ^ - 7 * m ^ 2 If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency 140 rad s ^ - 1 If the Young's modulus of the material of the wire is n * 10 ^ 9 * N * m ^ - 2 the value of nis a) 4 2 (ל c) 4.5? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1 m and its cross-sectional area is 4.9i * 0 ^ - 7 * m ^ 2 If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency 140 rad s ^ - 1 If the Young's modulus of the material of the wire is n * 10 ^ 9 * N * m ^ - 2 the value of nis a) 4 2 (ל c) 4.5?.
A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1 m and its cross-sectional area is 4.9i * 0 ^ - 7 * m ^ 2 If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency 140 rad s ^ - 1 If the Young's modulus of the material of the wire is n * 10 ^ 9 * N * m ^ - 2 the value of nis a) 4 2 (ל c) 4.5? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1 m and its cross-sectional area is 4.9i * 0 ^ - 7 * m ^ 2 If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency 140 rad s ^ - 1 If the Young's modulus of the material of the wire is n * 10 ^ 9 * N * m ^ - 2 the value of nis a) 4 2 (ל c) 4.5? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1 m and its cross-sectional area is 4.9i * 0 ^ - 7 * m ^ 2 If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency 140 rad s ^ - 1 If the Young's modulus of the material of the wire is n * 10 ^ 9 * N * m ^ - 2 the value of nis a) 4 2 (ל c) 4.5?.
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