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The following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied ? [NEET 2013]
- a)Length = 100 cm, diameter = 1 mm
- b)Length = 200 cm, diameter = 2 mm
- c)Length = 300 cm, diameter = 3 mm
- d)Length = 50 cm, diameter = 0.5 mm
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
The following four wires are made of the same material. Which of these...
So, extension, 
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The ratio of 
is maximum for case (d).
Hence, option (d) is correct.
is maximum for case (d).Hence, option (d) is correct.
Most Upvoted Answer
The following four wires are made of the same material. Which of these...
Explanation:
To determine which wire will have the largest extension when the same tension is applied, we need to consider the concept of stress and strain.
Stress:
Stress is defined as the force acting on a material per unit area. It is represented by the formula:
Stress = Force/Area
In this case, the tension applied to each wire is the same.
Strain:
Strain is defined as the change in length of a material per unit length. It is represented by the formula:
Strain = Change in length/Original length
Hooke's Law:
According to Hooke's Law, the extension of a wire is directly proportional to the force applied, as long as the elastic limit is not exceeded. The formula is:
Extension = Force × Length/Original length × Area × Young's modulus
From the above formula, we can see that the extension is directly proportional to the length and inversely proportional to the area.
Analysis:
Let's analyze each wire and calculate their extensions when the same tension is applied.
Wire A:
Length = 100 cm = 1 m
Diameter = 1 mm = 0.001 m
Radius (r) = Diameter/2 = 0.001/2 = 0.0005 m
Area (A) = πr² = 3.14 × (0.0005)² = 7.85 × 10^(-7) m²
Wire B:
Length = 200 cm = 2 m
Diameter = 2 mm = 0.002 m
Radius (r) = Diameter/2 = 0.002/2 = 0.001 m
Area (A) = πr² = 3.14 × (0.001)² = 3.14 × 10^(-6) m²
Wire C:
Length = 300 cm = 3 m
Diameter = 3 mm = 0.003 m
Radius (r) = Diameter/2 = 0.003/2 = 0.0015 m
Area (A) = πr² = 3.14 × (0.0015)² = 7.07 × 10^(-6) m²
Wire D:
Length = 50 cm = 0.5 m
Diameter = 0.5 mm = 0.0005 m
Radius (r) = Diameter/2 = 0.0005/2 = 0.00025 m
Area (A) = πr² = 3.14 × (0.00025)² = 1.96 × 10^(-7) m²
Calculating Extensions:
As the tension applied is the same, the force can be considered constant.
Using the formula for extension, we can calculate the extensions for each wire:
Extension = Force × Length/Original length × Area × Young's modulus
Since the force and Young's modulus are constant, we can compare the extensions by simply comparing the values of Length/Original length × Area.
Comparing Extensions:
Wire A:
Length/Original length × Area = (1/1) × (7.85 ×
To determine which wire will have the largest extension when the same tension is applied, we need to consider the concept of stress and strain.
Stress:
Stress is defined as the force acting on a material per unit area. It is represented by the formula:
Stress = Force/Area
In this case, the tension applied to each wire is the same.
Strain:
Strain is defined as the change in length of a material per unit length. It is represented by the formula:
Strain = Change in length/Original length
Hooke's Law:
According to Hooke's Law, the extension of a wire is directly proportional to the force applied, as long as the elastic limit is not exceeded. The formula is:
Extension = Force × Length/Original length × Area × Young's modulus
From the above formula, we can see that the extension is directly proportional to the length and inversely proportional to the area.
Analysis:
Let's analyze each wire and calculate their extensions when the same tension is applied.
Wire A:
Length = 100 cm = 1 m
Diameter = 1 mm = 0.001 m
Radius (r) = Diameter/2 = 0.001/2 = 0.0005 m
Area (A) = πr² = 3.14 × (0.0005)² = 7.85 × 10^(-7) m²
Wire B:
Length = 200 cm = 2 m
Diameter = 2 mm = 0.002 m
Radius (r) = Diameter/2 = 0.002/2 = 0.001 m
Area (A) = πr² = 3.14 × (0.001)² = 3.14 × 10^(-6) m²
Wire C:
Length = 300 cm = 3 m
Diameter = 3 mm = 0.003 m
Radius (r) = Diameter/2 = 0.003/2 = 0.0015 m
Area (A) = πr² = 3.14 × (0.0015)² = 7.07 × 10^(-6) m²
Wire D:
Length = 50 cm = 0.5 m
Diameter = 0.5 mm = 0.0005 m
Radius (r) = Diameter/2 = 0.0005/2 = 0.00025 m
Area (A) = πr² = 3.14 × (0.00025)² = 1.96 × 10^(-7) m²
Calculating Extensions:
As the tension applied is the same, the force can be considered constant.
Using the formula for extension, we can calculate the extensions for each wire:
Extension = Force × Length/Original length × Area × Young's modulus
Since the force and Young's modulus are constant, we can compare the extensions by simply comparing the values of Length/Original length × Area.
Comparing Extensions:
Wire A:
Length/Original length × Area = (1/1) × (7.85 ×
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The following four wires are made of the same material. Which of these...
Option C
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The following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied ? [NEET 2013]a)Length = 100 cm, diameter = 1 mmb)Length = 200 cm, diameter = 2 mmc)Length = 300 cm, diameter = 3 mmd)Length = 50 cm, diameter = 0.5 mmCorrect answer is option 'D'. Can you explain this answer?
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