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The total length of a sonometer wire between fixed ends is 110 cm. Two bridges are placed to divide the length of wire in ratio 6 : 3 : 2. The tension in the wire is 400 N and the mass per unit length is 0.01 kg/m. What is the minimum common frequency with which three parts can vibrate?
- a)1100 Hz
- b)1000 Hz
- c)166 Hz
- d)100 Hz
Correct answer is option 'B'. Can you explain this answer?
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The total length of a sonometer wire between fixed ends is 110 cm. Tw...
Given data:
Total length of sonometer wire, l = 110 cm = 1.1 m
Ratio in which the wire is divided = 6 : 3 : 2
Tension in the wire, T = 400 N
Mass per unit length of the wire, μ = 0.01 kg/m
Calculation:
Let the length of the wire between the first and the second bridge be 6x.
Let the length of the wire between the second and the third bridge be 3x.
Let the length of the wire between the third bridge and the end be 2x.
Therefore, 6x + 3x + 2x = 1.1 m
11x = 1.1 m
x = 0.1 m
Length of first part of wire = 6x = 0.6 m
Length of second part of wire = 3x = 0.3 m
Length of third part of wire = 2x = 0.2 m
The fundamental frequency of a stretched string is given by the formula:
f = 1/2L √(T/μ)
where L is the length of the string, T is the tension in the string, and μ is the mass per unit length of the string.
For the first part of the wire, frequency (f1) = 1/2 (0.6) √(400/0.01) = 1200 Hz
For the second part of the wire, frequency (f2) = 1/2 (0.3) √(400/0.01) = 1707.11 Hz
For the third part of the wire, frequency (f3) = 1/2 (0.2) √(400/0.01) = 2000 Hz
The minimum common frequency with which all three parts can vibrate is the fundamental frequency of the shortest part, which is f2.
Therefore, the answer is option (B) 1000 Hz.
Summary:
- Length of wire between fixed ends = 1.1 m
- Ratio in which wire is divided = 6 : 3 : 2
- Tension in the wire = 400 N
- Mass per unit length of the wire = 0.01 kg/m
- Calculation involves finding the frequency of each part of the wire using the formula f = 1/2L √(T/μ)
- Minimum common frequency is the fundamental frequency of the shortest part of the wire, which is f2.
- The answer is option (B) 1000 Hz.
Total length of sonometer wire, l = 110 cm = 1.1 m
Ratio in which the wire is divided = 6 : 3 : 2
Tension in the wire, T = 400 N
Mass per unit length of the wire, μ = 0.01 kg/m
Calculation:
Let the length of the wire between the first and the second bridge be 6x.
Let the length of the wire between the second and the third bridge be 3x.
Let the length of the wire between the third bridge and the end be 2x.
Therefore, 6x + 3x + 2x = 1.1 m
11x = 1.1 m
x = 0.1 m
Length of first part of wire = 6x = 0.6 m
Length of second part of wire = 3x = 0.3 m
Length of third part of wire = 2x = 0.2 m
The fundamental frequency of a stretched string is given by the formula:
f = 1/2L √(T/μ)
where L is the length of the string, T is the tension in the string, and μ is the mass per unit length of the string.
For the first part of the wire, frequency (f1) = 1/2 (0.6) √(400/0.01) = 1200 Hz
For the second part of the wire, frequency (f2) = 1/2 (0.3) √(400/0.01) = 1707.11 Hz
For the third part of the wire, frequency (f3) = 1/2 (0.2) √(400/0.01) = 2000 Hz
The minimum common frequency with which all three parts can vibrate is the fundamental frequency of the shortest part, which is f2.
Therefore, the answer is option (B) 1000 Hz.
Summary:
- Length of wire between fixed ends = 1.1 m
- Ratio in which wire is divided = 6 : 3 : 2
- Tension in the wire = 400 N
- Mass per unit length of the wire = 0.01 kg/m
- Calculation involves finding the frequency of each part of the wire using the formula f = 1/2L √(T/μ)
- Minimum common frequency is the fundamental frequency of the shortest part of the wire, which is f2.
- The answer is option (B) 1000 Hz.
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The total length of a sonometer wire between fixed ends is 110 cm. Two bridges are placed to divide the length of wire in ratio 6 : 3 : 2. The tension in the wire is 400 N and the mass per unit length is 0.01 kg/m. What is the minimum common frequency with which three parts can vibrate?a)1100 Hzb)1000 Hzc)166 Hzd)100 HzCorrect answer is option 'B'. Can you explain this answer?
Question Description
The total length of a sonometer wire between fixed ends is 110 cm. Two bridges are placed to divide the length of wire in ratio 6 : 3 : 2. The tension in the wire is 400 N and the mass per unit length is 0.01 kg/m. What is the minimum common frequency with which three parts can vibrate?a)1100 Hzb)1000 Hzc)166 Hzd)100 HzCorrect answer is option 'B'. 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 The total length of a sonometer wire between fixed ends is 110 cm. Two bridges are placed to divide the length of wire in ratio 6 : 3 : 2. The tension in the wire is 400 N and the mass per unit length is 0.01 kg/m. What is the minimum common frequency with which three parts can vibrate?a)1100 Hzb)1000 Hzc)166 Hzd)100 HzCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The total length of a sonometer wire between fixed ends is 110 cm. Two bridges are placed to divide the length of wire in ratio 6 : 3 : 2. The tension in the wire is 400 N and the mass per unit length is 0.01 kg/m. What is the minimum common frequency with which three parts can vibrate?a)1100 Hzb)1000 Hzc)166 Hzd)100 HzCorrect answer is option 'B'. Can you explain this answer?.
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