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A thick rope of density rho and length L is. Hung from a rigid support . The increase in length of the rope and due to it's own weight is (Y - Young's modulus) Ans: rho×L sq.×g /2Y?
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A thick rope of density rho and length L is. Hung from a rigid support...
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A thick rope of density rho and length L is. Hung from a rigid support...
Calculation of Increase in Length of the Rope due to its own Weight
Given:
Density of the rope (ρ) = ?
Length of the rope (L) = ?
Young's modulus (Y) = ?
Acceleration due to gravity (g) = 9.8 m/s²
To Find:
Increase in length of the rope due to its own weight = ?
Solution:
The weight of the rope can be calculated as follows:
Weight of the rope = Mass of the rope x Acceleration due to gravity
= Density of the rope x Volume of the rope x Acceleration due to gravity
= ρALg, where A is the cross-sectional area of the rope
The tension in the rope at any point can be calculated as follows:
Tension at any point = (Weight of the rope below that point + Weight of any suspended load)
Since the rope is in equilibrium, the tension at any point is constant. Therefore, the tension at the upper end of the rope is equal to the weight of the entire rope.
Tension at the upper end of the rope = ρALg
Using Hooke's law, we know that the increase in length of the rope is proportional to the tension in the rope and inversely proportional to the Young's modulus of the rope.
Therefore, increase in length of the rope = (Tension at the upper end of the rope / (Young's modulus of the rope)) x (Length of the rope)
Substituting the values, we get:
Increase in length of the rope = (ρALg / Y) x L
= (ρL²g / 2Y) x A (since A = Lρ, cross-sectional area of the rope)
= (ρL²g / 2Y)
Therefore, the increase in length of the rope due to its own weight is ρL²g / 2Y.
Answer:
Increase in length of the rope due to its own weight = ρL²g / 2Y.
Given:
Density of the rope (ρ) = ?
Length of the rope (L) = ?
Young's modulus (Y) = ?
Acceleration due to gravity (g) = 9.8 m/s²
To Find:
Increase in length of the rope due to its own weight = ?
Solution:
The weight of the rope can be calculated as follows:
Weight of the rope = Mass of the rope x Acceleration due to gravity
= Density of the rope x Volume of the rope x Acceleration due to gravity
= ρALg, where A is the cross-sectional area of the rope
The tension in the rope at any point can be calculated as follows:
Tension at any point = (Weight of the rope below that point + Weight of any suspended load)
Since the rope is in equilibrium, the tension at any point is constant. Therefore, the tension at the upper end of the rope is equal to the weight of the entire rope.
Tension at the upper end of the rope = ρALg
Using Hooke's law, we know that the increase in length of the rope is proportional to the tension in the rope and inversely proportional to the Young's modulus of the rope.
Therefore, increase in length of the rope = (Tension at the upper end of the rope / (Young's modulus of the rope)) x (Length of the rope)
Substituting the values, we get:
Increase in length of the rope = (ρALg / Y) x L
= (ρL²g / 2Y) x A (since A = Lρ, cross-sectional area of the rope)
= (ρL²g / 2Y)
Therefore, the increase in length of the rope due to its own weight is ρL²g / 2Y.
Answer:
Increase in length of the rope due to its own weight = ρL²g / 2Y.
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A thick rope of density rho and length L is. Hung from a rigid support . The increase in length of the rope and due to it's own weight is (Y - Young's modulus) Ans: rho×L sq.×g /2Y?
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A thick rope of density rho and length L is. Hung from a rigid support . The increase in length of the rope and due to it's own weight is (Y - Young's modulus) Ans: rho×L sq.×g /2Y? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A thick rope of density rho and length L is. Hung from a rigid support . The increase in length of the rope and due to it's own weight is (Y - Young's modulus) Ans: rho×L sq.×g /2Y? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A thick rope of density rho and length L is. Hung from a rigid support . The increase in length of the rope and due to it's own weight is (Y - Young's modulus) Ans: rho×L sq.×g /2Y?.
A thick rope of density rho and length L is. Hung from a rigid support . The increase in length of the rope and due to it's own weight is (Y - Young's modulus) Ans: rho×L sq.×g /2Y? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A thick rope of density rho and length L is. Hung from a rigid support . The increase in length of the rope and due to it's own weight is (Y - Young's modulus) Ans: rho×L sq.×g /2Y? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A thick rope of density rho and length L is. Hung from a rigid support . The increase in length of the rope and due to it's own weight is (Y - Young's modulus) Ans: rho×L sq.×g /2Y?.
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