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A sonometer wire of length 1.5 m is made of steel. The tension in it produces an elastic strain of 1 %.
What is the fundamental frequency of steel if density and elasticity of steel are 7.7 × 103 kg/m3 and
2.2 × 1011N/m2 respectively?
What is the fundamental frequency of steel if density and elasticity of steel are 7.7 × 103 kg/m3 and
2.2 × 1011N/m2 respectively?
- a)178.2 Hz
- b)200.5 Hz
- c)770 Hz
- d)188.5 Hz
Correct answer is option 'A'. Can you explain this answer?
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A sonometer wire of length 1.5 m is made of steel. The tension in it p...
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A sonometer wire of length 1.5 m is made of steel. The tension in it p...
To find the fundamental frequency of the steel wire, we need to use the formula for the fundamental frequency of a string:
f = (1/2L) * sqrt(T/μ)
Where:
f = fundamental frequency
L = length of the wire
T = tension in the wire
μ = linear mass density of the wire
Given:
L = 1.5 m
Tension produces an elastic strain of 1%
Density (μ) = 7.7 x 10^3 kg/m^3 (taking it as 7.7 g/cm^3 and converting to kg/m^3)
First, we need to find the tension in the wire. The elastic strain can be defined as the change in length divided by the original length:
Elastic strain = (change in length) / (original length)
1% = (change in length) / (1.5 m)
(change in length) = (1%) * (1.5 m)
(change in length) = 0.01 * 1.5 m
(change in length) = 0.015 m
Now, we can find the tension using Hooke's Law:
Tension = (Young's Modulus) * (Elastic strain) * (Original length)
Let's assume the Young's Modulus for steel is 200 GPa (2 x 10^11 N/m^2):
Tension = (2 x 10^11 N/m^2) * (0.015 m) * (1.5 m)
Tension = 4.5 x 10^9 N
Now, we can substitute the values into the formula for the fundamental frequency:
f = (1/2L) * sqrt(T/μ)
f = (1/2 * 1.5 m) * sqrt((4.5 x 10^9 N) / (7.7 x 10^3 kg/m^3))
f = (1/3) * sqrt(5.84 x 10^5 Hz^2)
f = (1/3) * 764.26 Hz
f ≈ 254.75 Hz
Therefore, the fundamental frequency of the steel wire is approximately 254.75 Hz.
f = (1/2L) * sqrt(T/μ)
Where:
f = fundamental frequency
L = length of the wire
T = tension in the wire
μ = linear mass density of the wire
Given:
L = 1.5 m
Tension produces an elastic strain of 1%
Density (μ) = 7.7 x 10^3 kg/m^3 (taking it as 7.7 g/cm^3 and converting to kg/m^3)
First, we need to find the tension in the wire. The elastic strain can be defined as the change in length divided by the original length:
Elastic strain = (change in length) / (original length)
1% = (change in length) / (1.5 m)
(change in length) = (1%) * (1.5 m)
(change in length) = 0.01 * 1.5 m
(change in length) = 0.015 m
Now, we can find the tension using Hooke's Law:
Tension = (Young's Modulus) * (Elastic strain) * (Original length)
Let's assume the Young's Modulus for steel is 200 GPa (2 x 10^11 N/m^2):
Tension = (2 x 10^11 N/m^2) * (0.015 m) * (1.5 m)
Tension = 4.5 x 10^9 N
Now, we can substitute the values into the formula for the fundamental frequency:
f = (1/2L) * sqrt(T/μ)
f = (1/2 * 1.5 m) * sqrt((4.5 x 10^9 N) / (7.7 x 10^3 kg/m^3))
f = (1/3) * sqrt(5.84 x 10^5 Hz^2)
f = (1/3) * 764.26 Hz
f ≈ 254.75 Hz
Therefore, the fundamental frequency of the steel wire is approximately 254.75 Hz.
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A sonometer wire of length 1.5 m is made of steel. The tension in it produces an elastic strain of 1 %.What is the fundamental frequency of steel if density and elasticity of steel are 7.7 × 103 kg/m3 and2.2 × 1011N/m2 respectively?a)178.2 Hzb)200.5 Hzc)770 Hzd)188.5 HzCorrect answer is option 'A'. Can you explain this answer?
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A sonometer wire of length 1.5 m is made of steel. The tension in it produces an elastic strain of 1 %.What is the fundamental frequency of steel if density and elasticity of steel are 7.7 × 103 kg/m3 and2.2 × 1011N/m2 respectively?a)178.2 Hzb)200.5 Hzc)770 Hzd)188.5 HzCorrect answer is option 'A'. 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 A sonometer wire of length 1.5 m is made of steel. The tension in it produces an elastic strain of 1 %.What is the fundamental frequency of steel if density and elasticity of steel are 7.7 × 103 kg/m3 and2.2 × 1011N/m2 respectively?a)178.2 Hzb)200.5 Hzc)770 Hzd)188.5 HzCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A sonometer wire of length 1.5 m is made of steel. The tension in it produces an elastic strain of 1 %.What is the fundamental frequency of steel if density and elasticity of steel are 7.7 × 103 kg/m3 and2.2 × 1011N/m2 respectively?a)178.2 Hzb)200.5 Hzc)770 Hzd)188.5 HzCorrect answer is option 'A'. Can you explain this answer?.
A sonometer wire of length 1.5 m is made of steel. The tension in it produces an elastic strain of 1 %.What is the fundamental frequency of steel if density and elasticity of steel are 7.7 × 103 kg/m3 and2.2 × 1011N/m2 respectively?a)178.2 Hzb)200.5 Hzc)770 Hzd)188.5 HzCorrect answer is option 'A'. 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 A sonometer wire of length 1.5 m is made of steel. The tension in it produces an elastic strain of 1 %.What is the fundamental frequency of steel if density and elasticity of steel are 7.7 × 103 kg/m3 and2.2 × 1011N/m2 respectively?a)178.2 Hzb)200.5 Hzc)770 Hzd)188.5 HzCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A sonometer wire of length 1.5 m is made of steel. The tension in it produces an elastic strain of 1 %.What is the fundamental frequency of steel if density and elasticity of steel are 7.7 × 103 kg/m3 and2.2 × 1011N/m2 respectively?a)178.2 Hzb)200.5 Hzc)770 Hzd)188.5 HzCorrect answer is option 'A'. Can you explain this answer?.
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