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A close-coiled helical spring, made out of a 10 mm diameter steel wire, has 20 complete coils each of mean diameter of 100 mm. A weight of 10 kg is dropped onto the spring, compresses it 60 mm, then height from which the weight is dropped (in mm) will be (take G = 85000 MPa)
- a)37
- b)38
Correct answer is between '37,38'. Can you explain this answer?
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A close-coiled helical spring, made out of a 10 mm diameter steel wir...
Applying the deflection equation of close coiled helical spring
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Now, by energy balance
⇒ h = 37.48mm
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A close-coiled helical spring, made out of a 10 mm diameter steel wir...
The given problem can be solved using the concept of spring stiffness and Hooke's law. Here are the steps to solve the problem:
1. Determine the spring constant:
The spring constant (k) can be calculated using the formula:
k = (Gd^4) / (8D^3n)
where G is the shear modulus of the material (given as 85000 MPa),
d is the diameter of the wire (10 mm),
D is the mean diameter of the spring (100 mm),
and n is the number of coils (20).
Substituting the given values, we get:
k = (85000 * 10^3 * (10^(-3))^4) / (8 * (100 * 10^(-3))^3 * 20)
Simplifying this, we get:
k ≈ 212.5 N/mm
2. Determine the force applied on the spring:
The force applied on the spring can be calculated using the formula:
F = mg
where m is the mass of the weight (10 kg) and g is the acceleration due to gravity (9.81 m/s^2).
Substituting the given values, we get:
F = 10 * 9.81
F ≈ 98.1 N
3. Determine the compression of the spring:
The compression of the spring (x) can be calculated using Hooke's law:
F = kx
where F is the force applied on the spring and k is the spring constant.
Substituting the given values, we get:
98.1 = 212.5 * x
Solving for x, we get:
x ≈ 0.461 mm
4. Determine the height from which the weight is dropped:
The potential energy (PE) of the weight can be calculated using the formula:
PE = mgh
where m is the mass of the weight, g is the acceleration due to gravity, and h is the height from which the weight is dropped.
When the weight is dropped, it compresses the spring by 60 mm, which is equal to x (as calculated in step 3).
Therefore, the potential energy is converted to the elastic potential energy stored in the compressed spring.
Equating the potential energy to the elastic potential energy, we have:
mgh = 0.5kx^2
Substituting the given values, we get:
10 * 9.81 * h = 0.5 * 212.5 * (0.461)^2
Solving for h, we get:
h ≈ 37.56 mm
Therefore, the height from which the weight is dropped is approximately 37.56 mm.
1. Determine the spring constant:
The spring constant (k) can be calculated using the formula:
k = (Gd^4) / (8D^3n)
where G is the shear modulus of the material (given as 85000 MPa),
d is the diameter of the wire (10 mm),
D is the mean diameter of the spring (100 mm),
and n is the number of coils (20).
Substituting the given values, we get:
k = (85000 * 10^3 * (10^(-3))^4) / (8 * (100 * 10^(-3))^3 * 20)
Simplifying this, we get:
k ≈ 212.5 N/mm
2. Determine the force applied on the spring:
The force applied on the spring can be calculated using the formula:
F = mg
where m is the mass of the weight (10 kg) and g is the acceleration due to gravity (9.81 m/s^2).
Substituting the given values, we get:
F = 10 * 9.81
F ≈ 98.1 N
3. Determine the compression of the spring:
The compression of the spring (x) can be calculated using Hooke's law:
F = kx
where F is the force applied on the spring and k is the spring constant.
Substituting the given values, we get:
98.1 = 212.5 * x
Solving for x, we get:
x ≈ 0.461 mm
4. Determine the height from which the weight is dropped:
The potential energy (PE) of the weight can be calculated using the formula:
PE = mgh
where m is the mass of the weight, g is the acceleration due to gravity, and h is the height from which the weight is dropped.
When the weight is dropped, it compresses the spring by 60 mm, which is equal to x (as calculated in step 3).
Therefore, the potential energy is converted to the elastic potential energy stored in the compressed spring.
Equating the potential energy to the elastic potential energy, we have:
mgh = 0.5kx^2
Substituting the given values, we get:
10 * 9.81 * h = 0.5 * 212.5 * (0.461)^2
Solving for h, we get:
h ≈ 37.56 mm
Therefore, the height from which the weight is dropped is approximately 37.56 mm.
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A close-coiled helical spring, made out of a 10 mm diameter steel wire, has 20 complete coils each of mean diameter of 100 mm. A weight of 10 kg is dropped onto the spring, compresses it 60 mm, then height from which the weight is dropped (in mm) will be (take G = 85000 MPa)a)37b)38Correct answer is between '37,38'. Can you explain this answer?
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A close-coiled helical spring, made out of a 10 mm diameter steel wire, has 20 complete coils each of mean diameter of 100 mm. A weight of 10 kg is dropped onto the spring, compresses it 60 mm, then height from which the weight is dropped (in mm) will be (take G = 85000 MPa)a)37b)38Correct answer is between '37,38'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A close-coiled helical spring, made out of a 10 mm diameter steel wire, has 20 complete coils each of mean diameter of 100 mm. A weight of 10 kg is dropped onto the spring, compresses it 60 mm, then height from which the weight is dropped (in mm) will be (take G = 85000 MPa)a)37b)38Correct answer is between '37,38'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A close-coiled helical spring, made out of a 10 mm diameter steel wire, has 20 complete coils each of mean diameter of 100 mm. A weight of 10 kg is dropped onto the spring, compresses it 60 mm, then height from which the weight is dropped (in mm) will be (take G = 85000 MPa)a)37b)38Correct answer is between '37,38'. Can you explain this answer?.
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