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In an experiment, sonometer wire vibrates in 4 loops when a 50 g weight is placed in the pan of weight 15 g. To make the wire vibrates in 6 loops the mass that has to be removed from the pan is approximately [frequency of vibration remains constant]
- a)12 g
- b)24 g
- c)36 g
- d)42 g
Correct answer is option 'C'. Can you explain this answer?
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In an experiment, sonometer wire vibrates in 4 loops when a 50 g weig...
To understand why the mass that has to be removed from the pan is approximately 36 g, let's break down the problem into smaller steps.
Given information:
- The sonometer wire vibrates in 4 loops when a 50 g weight is placed in the pan.
- The weight of the pan is 15 g.
- The frequency of vibration remains constant.
Step 1: Determine the effective mass added to the wire.
The effective mass added to the wire is the sum of the weight added to the pan and the weight of the pan itself.
Effective mass = weight added + weight of the pan
Effective mass = 50 g + 15 g
Effective mass = 65 g
Step 2: Determine the effective length of the wire for 4 loops.
The effective length of the wire for 4 loops can be calculated using the formula:
Length = (n * λ) / 2
where n is the number of loops and λ is the wavelength.
For 4 loops, we have:
Length = (4 * λ) / 2
Length = 2λ
Step 3: Determine the effective length of the wire for 6 loops.
Since the frequency of vibration remains constant, the wavelength will change when the number of loops changes. Let's assume the effective length of the wire for 6 loops is L.
For 6 loops, we have:
L = (6 * λ) / 2
L = 3λ
Step 4: Calculate the mass that needs to be removed.
The ratio of the effective lengths for 4 loops and 6 loops is given by:
L_4 / L_6 = 2λ / 3λ
L_4 / L_6 = 2/3
Since the frequency of vibration remains constant, the ratio of the effective masses is equal to the inverse of the ratio of the effective lengths:
m_4 / m_6 = L_6 / L_4
m_4 / m_6 = 3/2
We know that the effective mass for 4 loops is 65 g. Let's assume the mass that needs to be removed for 6 loops is m.
m_4 / m_6 = 65 g / m
3/2 = 65 g / m
Solving for m, we get:
m = (2 * 65 g) / 3
m ≈ 43.33 g
Therefore, the mass that needs to be removed from the pan is approximately 43.33 g.
Given information:
- The sonometer wire vibrates in 4 loops when a 50 g weight is placed in the pan.
- The weight of the pan is 15 g.
- The frequency of vibration remains constant.
Step 1: Determine the effective mass added to the wire.
The effective mass added to the wire is the sum of the weight added to the pan and the weight of the pan itself.
Effective mass = weight added + weight of the pan
Effective mass = 50 g + 15 g
Effective mass = 65 g
Step 2: Determine the effective length of the wire for 4 loops.
The effective length of the wire for 4 loops can be calculated using the formula:
Length = (n * λ) / 2
where n is the number of loops and λ is the wavelength.
For 4 loops, we have:
Length = (4 * λ) / 2
Length = 2λ
Step 3: Determine the effective length of the wire for 6 loops.
Since the frequency of vibration remains constant, the wavelength will change when the number of loops changes. Let's assume the effective length of the wire for 6 loops is L.
For 6 loops, we have:
L = (6 * λ) / 2
L = 3λ
Step 4: Calculate the mass that needs to be removed.
The ratio of the effective lengths for 4 loops and 6 loops is given by:
L_4 / L_6 = 2λ / 3λ
L_4 / L_6 = 2/3
Since the frequency of vibration remains constant, the ratio of the effective masses is equal to the inverse of the ratio of the effective lengths:
m_4 / m_6 = L_6 / L_4
m_4 / m_6 = 3/2
We know that the effective mass for 4 loops is 65 g. Let's assume the mass that needs to be removed for 6 loops is m.
m_4 / m_6 = 65 g / m
3/2 = 65 g / m
Solving for m, we get:
m = (2 * 65 g) / 3
m ≈ 43.33 g
Therefore, the mass that needs to be removed from the pan is approximately 43.33 g.
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Community Answer
In an experiment, sonometer wire vibrates in 4 loops when a 50 g weig...
∴ Mass removed from the pan
M1 - M2 = 65 - 28.8 = 36.2 g
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In an experiment, sonometer wire vibrates in 4 loops when a 50 g weight is placed in the pan of weight 15 g. To make the wire vibrates in 6 loops the mass that has to be removed from the pan is approximately [frequency of vibration remains constant]a)12 gb)24 gc)36 gd)42 gCorrect answer is option 'C'. Can you explain this answer?
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In an experiment, sonometer wire vibrates in 4 loops when a 50 g weight is placed in the pan of weight 15 g. To make the wire vibrates in 6 loops the mass that has to be removed from the pan is approximately [frequency of vibration remains constant]a)12 gb)24 gc)36 gd)42 gCorrect answer is option 'C'. 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 In an experiment, sonometer wire vibrates in 4 loops when a 50 g weight is placed in the pan of weight 15 g. To make the wire vibrates in 6 loops the mass that has to be removed from the pan is approximately [frequency of vibration remains constant]a)12 gb)24 gc)36 gd)42 gCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In an experiment, sonometer wire vibrates in 4 loops when a 50 g weight is placed in the pan of weight 15 g. To make the wire vibrates in 6 loops the mass that has to be removed from the pan is approximately [frequency of vibration remains constant]a)12 gb)24 gc)36 gd)42 gCorrect answer is option 'C'. Can you explain this answer?.
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