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A steel wire of length 4.7 m and cross-sectional area 3.0 × 10−5 m2 stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10−5 m2 under a given load. What is the ratio of the Young’s modulus of steel to that of copper?
- a)1.2
- b)1.6
- c)1.8
- d)2.0
Correct answer is option 'C'. Can you explain this answer?
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A steel wire of length 4.7 m and cross-sectional area 3.0 × 10...
Given:
Length of steel wire, L1 = 4.7 m
Cross-sectional area of steel wire, A1 = 3.0 × 10^5 m^2
Length of copper wire, L2 = 3.5 m
Cross-sectional area of copper wire, A2 = 4.0 × 10^5 m^2
To find:
Ratio of Young's modulus of steel to that of copper
Formula:
Young's modulus (Y) = Stress / Strain
Stress = Force / Area
Strain = Change in length / Original length
Calculation:
Let the force applied on both wires be F.
For steel wire:
Stress1 = F / A1
Strain1 = Change in length1 / Original length1 = x / L1
For copper wire:
Stress2 = F / A2
Strain2 = Change in length2 / Original length2 = x / L2
Since both wires stretch by the same amount under the given load, we can equate the strains.
Strain1 = Strain2
x / L1 = x / L2
Cross multiplying, we get:
x * L2 = x * L1
L2 = L1
Therefore, the ratio of the lengths of the two wires is 3.5 / 4.7 = 1.4.
Now, substituting the values in the stress equation:
Stress1 = F / A1
Stress2 = F / A2
Since the forces applied are the same for both wires, we can equate the stresses.
Stress1 = Stress2
F / A1 = F / A2
Cross multiplying, we get:
F * A2 = F * A1
A2 = A1
Therefore, the ratio of the cross-sectional areas of the two wires is 4.0 × 10^5 / 3.0 × 10^5 = 1.33.
Now, substituting the values in the Young's modulus equation:
Young's modulus of steel (Y1) = Stress1 / Strain1 = (F / A1) / (x / L1)
Young's modulus of copper (Y2) = Stress2 / Strain2 = (F / A2) / (x / L2)
We can cancel out the applied force and the change in length:
Y1 = 1 / (A1 / L1)
Y2 = 1 / (A2 / L2)
Substituting the values:
Y1 = 1 / (3.0 × 10^5 / 4.7)
Y2 = 1 / (4.0 × 10^5 / 3.5)
Calculating the ratios:
Y1 / Y2 = [(4.7 × 10^5) / 3.0] / [(3.5 × 10^5) / 4.0]
= (4.7 × 10^5 × 4.0) / (3.0 × 3.5 × 10^5)
= 1.8
Therefore, the ratio of the Young's modulus of steel to that of copper is 1.8.
Length of steel wire, L1 = 4.7 m
Cross-sectional area of steel wire, A1 = 3.0 × 10^5 m^2
Length of copper wire, L2 = 3.5 m
Cross-sectional area of copper wire, A2 = 4.0 × 10^5 m^2
To find:
Ratio of Young's modulus of steel to that of copper
Formula:
Young's modulus (Y) = Stress / Strain
Stress = Force / Area
Strain = Change in length / Original length
Calculation:
Let the force applied on both wires be F.
For steel wire:
Stress1 = F / A1
Strain1 = Change in length1 / Original length1 = x / L1
For copper wire:
Stress2 = F / A2
Strain2 = Change in length2 / Original length2 = x / L2
Since both wires stretch by the same amount under the given load, we can equate the strains.
Strain1 = Strain2
x / L1 = x / L2
Cross multiplying, we get:
x * L2 = x * L1
L2 = L1
Therefore, the ratio of the lengths of the two wires is 3.5 / 4.7 = 1.4.
Now, substituting the values in the stress equation:
Stress1 = F / A1
Stress2 = F / A2
Since the forces applied are the same for both wires, we can equate the stresses.
Stress1 = Stress2
F / A1 = F / A2
Cross multiplying, we get:
F * A2 = F * A1
A2 = A1
Therefore, the ratio of the cross-sectional areas of the two wires is 4.0 × 10^5 / 3.0 × 10^5 = 1.33.
Now, substituting the values in the Young's modulus equation:
Young's modulus of steel (Y1) = Stress1 / Strain1 = (F / A1) / (x / L1)
Young's modulus of copper (Y2) = Stress2 / Strain2 = (F / A2) / (x / L2)
We can cancel out the applied force and the change in length:
Y1 = 1 / (A1 / L1)
Y2 = 1 / (A2 / L2)
Substituting the values:
Y1 = 1 / (3.0 × 10^5 / 4.7)
Y2 = 1 / (4.0 × 10^5 / 3.5)
Calculating the ratios:
Y1 / Y2 = [(4.7 × 10^5) / 3.0] / [(3.5 × 10^5) / 4.0]
= (4.7 × 10^5 × 4.0) / (3.0 × 3.5 × 10^5)
= 1.8
Therefore, the ratio of the Young's modulus of steel to that of copper is 1.8.
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A steel wire of length 4.7 m and cross-sectional area 3.0 × 10−5 m2 stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10−5 m2 under a given load. What is the ratio of the Young’s modulus of steel to that of copper?a)1.2b)1.6c)1.8d)2.0Correct answer is option 'C'. Can you explain this answer?
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A steel wire of length 4.7 m and cross-sectional area 3.0 × 10−5 m2 stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10−5 m2 under a given load. What is the ratio of the Young’s modulus of steel to that of copper?a)1.2b)1.6c)1.8d)2.0Correct answer is option 'C'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A steel wire of length 4.7 m and cross-sectional area 3.0 × 10−5 m2 stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10−5 m2 under a given load. What is the ratio of the Young’s modulus of steel to that of copper?a)1.2b)1.6c)1.8d)2.0Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A steel wire of length 4.7 m and cross-sectional area 3.0 × 10−5 m2 stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10−5 m2 under a given load. What is the ratio of the Young’s modulus of steel to that of copper?a)1.2b)1.6c)1.8d)2.0Correct answer is option 'C'. Can you explain this answer?.
A steel wire of length 4.7 m and cross-sectional area 3.0 × 10−5 m2 stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10−5 m2 under a given load. What is the ratio of the Young’s modulus of steel to that of copper?a)1.2b)1.6c)1.8d)2.0Correct answer is option 'C'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A steel wire of length 4.7 m and cross-sectional area 3.0 × 10−5 m2 stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10−5 m2 under a given load. What is the ratio of the Young’s modulus of steel to that of copper?a)1.2b)1.6c)1.8d)2.0Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A steel wire of length 4.7 m and cross-sectional area 3.0 × 10−5 m2 stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10−5 m2 under a given load. What is the ratio of the Young’s modulus of steel to that of copper?a)1.2b)1.6c)1.8d)2.0Correct answer is option 'C'. Can you explain this answer?.
Solutions for A steel wire of length 4.7 m and cross-sectional area 3.0 × 10−5 m2 stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10−5 m2 under a given load. What is the ratio of the Young’s modulus of steel to that of copper?a)1.2b)1.6c)1.8d)2.0Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 11.
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