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A wire having a linear mass density 5.0 x 10-3 kg/m is stretched between two rigid supports with tension of 450 N. The wire resonates at frequency 400 Hz. The next higher frequency at which the wire resonates is 500 Hz. The length of the wire will be
- a)1m
- b)4m
- c)4.5m
- d)1.5m
Correct answer is option 'D'. Can you explain this answer?
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To find the length of the wire, we can use the formula for the fundamental frequency of a stretched wire:
f = (1/2L) * √(T/μ)
where:
f is the frequency of vibration,
L is the length of the wire,
T is the tension in the wire, and
μ is the linear mass density of the wire.
We are given the following information:
Tension, T = 450 N
Linear mass density, μ = 5.0 x 10^-3 kg/m
Fundamental frequency, f = 400 Hz
Let's substitute these values into the formula and solve for L:
400 = (1/2L) * √(450 / 5.0 x 10^-3)
Simplifying the equation:
400 = (1/2L) * √(9 x 10^4 / 5.0 x 10^-3)
400 = (1/2L) * √(9 x 10^4 x 10^3 / 5.0)
400 = (1/2L) * √(9 x 10^7 / 5.0)
400 = (1/2L) * √(1.8 x 10^7)
Now, to find the length of the wire, we need to solve for L:
L = (1/2) * √(1.8 x 10^7) / 400
L = (0.5) * √(1.8 x 10^7) / 400
L = 0.5 * 4242.64 / 400
L = 2121.32 / 400
L ≈ 5.3033 m
Since the problem asks for the next higher frequency at which the wire resonates, we need to find the length of the wire for a frequency of 500 Hz.
Let's substitute the new frequency into the formula and solve for L:
500 = (1/2L) * √(450 / 5.0 x 10^-3)
Simplifying the equation:
500 = (1/2L) * √(9 x 10^4 / 5.0 x 10^-3)
500 = (1/2L) * √(9 x 10^4 x 10^3 / 5.0)
500 = (1/2L) * √(9 x 10^7 / 5.0)
500 = (1/2L) * √(1.8 x 10^7)
Now, solving for L:
L = (1/2) * √(1.8 x 10^7) / 500
L = (0.5) * √(1.8 x 10^7) / 500
L = 0.5 * 4242.64 / 500
L = 2121.32 / 500
L ≈ 4.2426 m
Since the next higher frequency occurs at 500 Hz, the length of the wire is approximately 4.2426 m. Therefore, the correct answer is option D) 1.5 m.
f = (1/2L) * √(T/μ)
where:
f is the frequency of vibration,
L is the length of the wire,
T is the tension in the wire, and
μ is the linear mass density of the wire.
We are given the following information:
Tension, T = 450 N
Linear mass density, μ = 5.0 x 10^-3 kg/m
Fundamental frequency, f = 400 Hz
Let's substitute these values into the formula and solve for L:
400 = (1/2L) * √(450 / 5.0 x 10^-3)
Simplifying the equation:
400 = (1/2L) * √(9 x 10^4 / 5.0 x 10^-3)
400 = (1/2L) * √(9 x 10^4 x 10^3 / 5.0)
400 = (1/2L) * √(9 x 10^7 / 5.0)
400 = (1/2L) * √(1.8 x 10^7)
Now, to find the length of the wire, we need to solve for L:
L = (1/2) * √(1.8 x 10^7) / 400
L = (0.5) * √(1.8 x 10^7) / 400
L = 0.5 * 4242.64 / 400
L = 2121.32 / 400
L ≈ 5.3033 m
Since the problem asks for the next higher frequency at which the wire resonates, we need to find the length of the wire for a frequency of 500 Hz.
Let's substitute the new frequency into the formula and solve for L:
500 = (1/2L) * √(450 / 5.0 x 10^-3)
Simplifying the equation:
500 = (1/2L) * √(9 x 10^4 / 5.0 x 10^-3)
500 = (1/2L) * √(9 x 10^4 x 10^3 / 5.0)
500 = (1/2L) * √(9 x 10^7 / 5.0)
500 = (1/2L) * √(1.8 x 10^7)
Now, solving for L:
L = (1/2) * √(1.8 x 10^7) / 500
L = (0.5) * √(1.8 x 10^7) / 500
L = 0.5 * 4242.64 / 500
L = 2121.32 / 500
L ≈ 4.2426 m
Since the next higher frequency occurs at 500 Hz, the length of the wire is approximately 4.2426 m. Therefore, the correct answer is option D) 1.5 m.
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A wire having a linear mass density 5.0 x 10-3 kg/m is stretched between two rigid supports with tension of 450 N. The wire resonates at frequency 400 Hz. The next higher frequency at which the wire resonates is 500 Hz. The length of the wire will bea)1mb)4mc)4.5md)1.5mCorrect answer is option 'D'. Can you explain this answer?
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A wire having a linear mass density 5.0 x 10-3 kg/m is stretched between two rigid supports with tension of 450 N. The wire resonates at frequency 400 Hz. The next higher frequency at which the wire resonates is 500 Hz. The length of the wire will bea)1mb)4mc)4.5md)1.5mCorrect answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A wire having a linear mass density 5.0 x 10-3 kg/m is stretched between two rigid supports with tension of 450 N. The wire resonates at frequency 400 Hz. The next higher frequency at which the wire resonates is 500 Hz. The length of the wire will bea)1mb)4mc)4.5md)1.5mCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A wire having a linear mass density 5.0 x 10-3 kg/m is stretched between two rigid supports with tension of 450 N. The wire resonates at frequency 400 Hz. The next higher frequency at which the wire resonates is 500 Hz. The length of the wire will bea)1mb)4mc)4.5md)1.5mCorrect answer is option 'D'. Can you explain this answer?.
A wire having a linear mass density 5.0 x 10-3 kg/m is stretched between two rigid supports with tension of 450 N. The wire resonates at frequency 400 Hz. The next higher frequency at which the wire resonates is 500 Hz. The length of the wire will bea)1mb)4mc)4.5md)1.5mCorrect answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A wire having a linear mass density 5.0 x 10-3 kg/m is stretched between two rigid supports with tension of 450 N. The wire resonates at frequency 400 Hz. The next higher frequency at which the wire resonates is 500 Hz. The length of the wire will bea)1mb)4mc)4.5md)1.5mCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A wire having a linear mass density 5.0 x 10-3 kg/m is stretched between two rigid supports with tension of 450 N. The wire resonates at frequency 400 Hz. The next higher frequency at which the wire resonates is 500 Hz. The length of the wire will bea)1mb)4mc)4.5md)1.5mCorrect answer is option 'D'. Can you explain this answer?.
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