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There are two wire of same material and same length while the diameter of second wire is two times the diameter of first wire, then the ratio of extension produced in the wires by applying same load will be
- a)1 : 1
- b)2 : 1
- c)1 : 2
- d)4 : 1
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
There are two wire of same material and same length while the diameter...
∵ Both wires are same materials so both will have same Young’s modulus, and let it be Y.
A = area of cross-section of wire Now,
Since load and length are same for both
ΔL1 : ΔL2= 4 :1
A = area of cross-section of wire Now,
Since load and length are same for both
ΔL1 : ΔL2= 4 :1
Most Upvoted Answer
There are two wire of same material and same length while the diameter...
Explanation:
When a load is applied to a wire, it extends. This extension is directly proportional to the length of the wire, the force applied, and inversely proportional to the cross-sectional area of the wire. The formula for the extension produced in a wire is:
Extension = (Force x Length) / (Area x Young's modulus)
Where Young's modulus is a constant for a material and represents its stiffness.
Given that the two wires are of the same material and same length, we can assume that their Young's modulus is the same. Let the diameter of the first wire be d and the diameter of the second wire be 2d. Therefore, the cross-sectional area of the first wire is πd^2/4 and the cross-sectional area of the second wire is π(2d)^2/4 = 4πd^2.
Now, let the same force be applied to both wires. The extension produced in the first wire is:
Extension1 = (F x L) / (πd^2/4 x E)
The extension produced in the second wire is:
Extension2 = (F x L) / (4πd^2/4 x E)
Simplifying both equations, we get:
Extension1 = (4F x L) / (πd^2 x E)
Extension2 = (F x L) / (πd^2 x E)
Dividing Extension2 by Extension1, we get:
Extension2 / Extension1 = (F x L) / (πd^2 x E) / (4F x L) / (πd^2 x E)
Extension2 / Extension1 = 1/4
Therefore, the ratio of extension produced in the wires is 4:1. Hence, option D is the correct answer.
When a load is applied to a wire, it extends. This extension is directly proportional to the length of the wire, the force applied, and inversely proportional to the cross-sectional area of the wire. The formula for the extension produced in a wire is:
Extension = (Force x Length) / (Area x Young's modulus)
Where Young's modulus is a constant for a material and represents its stiffness.
Given that the two wires are of the same material and same length, we can assume that their Young's modulus is the same. Let the diameter of the first wire be d and the diameter of the second wire be 2d. Therefore, the cross-sectional area of the first wire is πd^2/4 and the cross-sectional area of the second wire is π(2d)^2/4 = 4πd^2.
Now, let the same force be applied to both wires. The extension produced in the first wire is:
Extension1 = (F x L) / (πd^2/4 x E)
The extension produced in the second wire is:
Extension2 = (F x L) / (4πd^2/4 x E)
Simplifying both equations, we get:
Extension1 = (4F x L) / (πd^2 x E)
Extension2 = (F x L) / (πd^2 x E)
Dividing Extension2 by Extension1, we get:
Extension2 / Extension1 = (F x L) / (πd^2 x E) / (4F x L) / (πd^2 x E)
Extension2 / Extension1 = 1/4
Therefore, the ratio of extension produced in the wires is 4:1. Hence, option D is the correct answer.
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There are two wire of same material and same length while the diameter of second wire is two times the diameter of first wire, then the ratio of extension produced in the wires by applying same load will bea)1 : 1b)2 : 1c)1 : 2d)4 : 1Correct answer is option 'D'. Can you explain this answer?
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