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A thick rope of density rho and length L...

A thick rope of density `rho` and length `L` is hung from a rigid support. The increase in length of the rope due to its own weight is (`Y` is the Young's modulus)

A

`0.1/(4Y)rhoL^(2)g`

B

`1/(2Y)rhoL^(2)g`

C

`(rhoL^(2)g)/Y`

D

`(rhoLg)/Y`

Text Solution

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The correct Answer is:
To solve the problem of finding the increase in length of a thick rope due to its own weight, we will follow these steps: ### Step 1: Understanding the Problem We have a thick rope of density \( \rho \) and length \( L \) that is hung from a rigid support. We need to determine the increase in length of the rope due to its own weight, using Young's modulus \( Y \). ### Step 2: Identify the Force Acting on the Rope The weight of the rope acts downward due to gravity. The total weight \( W \) of the rope can be expressed as: \[ W = mg \] where \( m \) is the mass of the rope and \( g \) is the acceleration due to gravity. ### Step 3: Calculate the Mass of the Rope The mass \( m \) of the rope can be calculated using its density \( \rho \) and volume \( V \). The volume of the rope can be expressed as the product of its cross-sectional area \( A \) and its length \( L \): \[ m = \rho \cdot V = \rho \cdot A \cdot L \] ### Step 4: Substitute Mass into Weight Equation Substituting the expression for mass into the weight equation gives: \[ W = \rho \cdot A \cdot L \cdot g \] ### Step 5: Calculate the Stress on the Rope Stress \( \sigma \) is defined as force per unit area: \[ \sigma = \frac{W}{A} = \frac{\rho \cdot A \cdot L \cdot g}{A} = \rho \cdot L \cdot g \] ### Step 6: Calculate the Strain in the Rope Strain \( \epsilon \) is defined as the change in length \( \Delta L \) divided by the original length \( L \): \[ \epsilon = \frac{\Delta L}{L} \] ### Step 7: Relate Stress and Strain Using Young's Modulus According to the definition of Young's modulus \( Y \): \[ Y = \frac{\sigma}{\epsilon} \] Substituting the expressions for stress and strain: \[ Y = \frac{\rho \cdot L \cdot g}{\Delta L / L} \] ### Step 8: Rearranging to Find Change in Length Rearranging the equation to solve for \( \Delta L \): \[ \Delta L = \frac{\rho \cdot L^2 \cdot g}{Y} \] ### Step 9: Considering the Effective Length Since the weight acts at the center of gravity, we consider only half the length of the rope for the extension: \[ \Delta L = \frac{1}{2} \cdot \frac{\rho \cdot L^2 \cdot g}{Y} = \frac{\rho \cdot L^2 \cdot g}{2Y} \] ### Final Expression Thus, the increase in length of the rope due to its own weight is: \[ \Delta L = \frac{\rho \cdot L^2 \cdot g}{2Y} \]

To solve the problem of finding the increase in length of a thick rope due to its own weight, we will follow these steps: ### Step 1: Understanding the Problem We have a thick rope of density \( \rho \) and length \( L \) that is hung from a rigid support. We need to determine the increase in length of the rope due to its own weight, using Young's modulus \( Y \). ### Step 2: Identify the Force Acting on the Rope The weight of the rope acts downward due to gravity. The total weight \( W \) of the rope can be expressed as: \[ ...
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