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Two wires of the same material and length, but diameters in the ratio 1 : 2 are stretched by the same force. The elastic potential energy stored per unit volume for the two wires when stretched, will be in the ratio of
- a)2 : 1
- b)4 : 1
- c)8 : 1
- d)16 : 1
Correct answer is option 'D'. Can you explain this answer?
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Two wires of the same material and length, but diameters in the ratio ...
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Two wires of the same material and length, but diameters in the ratio ...
Given:
- Two wires of the same material and length
- Diameters in the ratio 1:2
- Stretched by the same force
To find:
The ratio of elastic potential energy stored per unit volume for the two wires when stretched
Solution:
Step 1: Calculate the cross-sectional area of each wire
- Let the diameter of the first wire be 'd'
- Therefore, the radius of the first wire is 'r = d/2'
- The cross-sectional area of the first wire is 'A1 = πr^2 = π(d/2)^2 = πd^2/4'
- According to the given ratio, the diameter of the second wire is '2d'
- Therefore, the radius of the second wire is 'r = 2d/2 = d'
- The cross-sectional area of the second wire is 'A2 = πr^2 = πd^2'
Step 2: Calculate the volume of each wire
- Since the wires have the same length, the volume of each wire is directly proportional to the cross-sectional area
- Let the length of the wires be 'L'
- The volume of the first wire is 'V1 = A1 * L = (πd^2/4) * L = (π/4) * d^2 * L'
- The volume of the second wire is 'V2 = A2 * L = (πd^2) * L = π * d^2 * L'
Step 3: Calculate the ratio of elastic potential energy stored per unit volume
- Elastic potential energy stored per unit volume is given by the formula '1/2 * stress * strain'
- Stress is the force applied divided by the cross-sectional area, and strain is the change in length divided by the original length
- Since the force applied and the length are the same for both wires, the ratio of elastic potential energy stored per unit volume is equal to the ratio of stress to strain
- The stress for both wires is the same because they are stretched by the same force
- The strain for both wires is the same because they have the same change in length
- Therefore, the ratio of elastic potential energy stored per unit volume is equal to the ratio of the cross-sectional areas of the wires
- The ratio of A2 to A1 is (πd^2) / (πd^2/4) = 4/1 = 4:1
Hence, the correct answer is option 'D', 16:1.
- Two wires of the same material and length
- Diameters in the ratio 1:2
- Stretched by the same force
To find:
The ratio of elastic potential energy stored per unit volume for the two wires when stretched
Solution:
Step 1: Calculate the cross-sectional area of each wire
- Let the diameter of the first wire be 'd'
- Therefore, the radius of the first wire is 'r = d/2'
- The cross-sectional area of the first wire is 'A1 = πr^2 = π(d/2)^2 = πd^2/4'
- According to the given ratio, the diameter of the second wire is '2d'
- Therefore, the radius of the second wire is 'r = 2d/2 = d'
- The cross-sectional area of the second wire is 'A2 = πr^2 = πd^2'
Step 2: Calculate the volume of each wire
- Since the wires have the same length, the volume of each wire is directly proportional to the cross-sectional area
- Let the length of the wires be 'L'
- The volume of the first wire is 'V1 = A1 * L = (πd^2/4) * L = (π/4) * d^2 * L'
- The volume of the second wire is 'V2 = A2 * L = (πd^2) * L = π * d^2 * L'
Step 3: Calculate the ratio of elastic potential energy stored per unit volume
- Elastic potential energy stored per unit volume is given by the formula '1/2 * stress * strain'
- Stress is the force applied divided by the cross-sectional area, and strain is the change in length divided by the original length
- Since the force applied and the length are the same for both wires, the ratio of elastic potential energy stored per unit volume is equal to the ratio of stress to strain
- The stress for both wires is the same because they are stretched by the same force
- The strain for both wires is the same because they have the same change in length
- Therefore, the ratio of elastic potential energy stored per unit volume is equal to the ratio of the cross-sectional areas of the wires
- The ratio of A2 to A1 is (πd^2) / (πd^2/4) = 4/1 = 4:1
Hence, the correct answer is option 'D', 16:1.
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Two wires of the same material and length, but diameters in the ratio 1 : 2 are stretched by the same force. The elastic potential energy stored per unit volume for the two wires when stretched, will be in the ratio ofa)2 : 1b)4 : 1c)8 : 1d)16 : 1Correct answer is option 'D'. Can you explain this answer?
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Two wires of the same material and length, but diameters in the ratio 1 : 2 are stretched by the same force. The elastic potential energy stored per unit volume for the two wires when stretched, will be in the ratio ofa)2 : 1b)4 : 1c)8 : 1d)16 : 1Correct answer is option 'D'. 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 Two wires of the same material and length, but diameters in the ratio 1 : 2 are stretched by the same force. The elastic potential energy stored per unit volume for the two wires when stretched, will be in the ratio ofa)2 : 1b)4 : 1c)8 : 1d)16 : 1Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two wires of the same material and length, but diameters in the ratio 1 : 2 are stretched by the same force. The elastic potential energy stored per unit volume for the two wires when stretched, will be in the ratio ofa)2 : 1b)4 : 1c)8 : 1d)16 : 1Correct answer is option 'D'. Can you explain this answer?.
Two wires of the same material and length, but diameters in the ratio 1 : 2 are stretched by the same force. The elastic potential energy stored per unit volume for the two wires when stretched, will be in the ratio ofa)2 : 1b)4 : 1c)8 : 1d)16 : 1Correct answer is option 'D'. 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 Two wires of the same material and length, but diameters in the ratio 1 : 2 are stretched by the same force. The elastic potential energy stored per unit volume for the two wires when stretched, will be in the ratio ofa)2 : 1b)4 : 1c)8 : 1d)16 : 1Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two wires of the same material and length, but diameters in the ratio 1 : 2 are stretched by the same force. The elastic potential energy stored per unit volume for the two wires when stretched, will be in the ratio ofa)2 : 1b)4 : 1c)8 : 1d)16 : 1Correct answer is option 'D'. Can you explain this answer?.
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Two wires of the same material and length, but diameters in the ratio 1 : 2 are stretched by the same force. The elastic potential energy stored per unit volume for the two wires when stretched, will be in the ratio ofa)2 : 1b)4 : 1c)8 : 1d)16 : 1Correct answer is option 'D'. Can you explain this answer?, a detailed solution for Two wires of the same material and length, but diameters in the ratio 1 : 2 are stretched by the same force. The elastic potential energy stored per unit volume for the two wires when stretched, will be in the ratio ofa)2 : 1b)4 : 1c)8 : 1d)16 : 1Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of Two wires of the same material and length, but diameters in the ratio 1 : 2 are stretched by the same force. The elastic potential energy stored per unit volume for the two wires when stretched, will be in the ratio ofa)2 : 1b)4 : 1c)8 : 1d)16 : 1Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
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