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Chapter 7: Problem 848

A rubber cord \(10 \mathrm{~m}\) long is suspended vertically. How much does it stretch under its own weight. [Density of rubber is $1500\left(\mathrm{~kg} / \mathrm{m}^{3}\right), \mathrm{Y}=5 \times 10^{8}\left(\mathrm{~N} / \mathrm{m}^{2}\right)$, \(\left.\mathrm{g}=10\left(\mathrm{~m} / \mathrm{s}^{2}\right)\right]\) (A) \(15 \times 10^{-4} \mathrm{~m}\) (B) \(7.5 \times 10^{-4} \mathrm{~m}\) (C) \(12 \times 10^{-4} \mathrm{~m}\) (D) \(25 \times 10^{-4} \mathrm{~m}\)

Short Answer

Expert verified
The stretch of the rubber cord under its own weight is \(3 \times 10^{-3} \mathrm{~m}\).

Step by step solution

01

Find the Volume of the Rubber Cord

To find the volume of the rubber cord, we need to assume a uniform cylindrical shape, with length L and cross-sectional area A. As we are only given the length of the cord (10 meters), but not the thickness, we cannot determine the exact volume with the given data. We can find the expression for the mass of the rubber cord in terms of unknown cross-sectional area A, using the given density of rubber (ρ = 1500 kg/m³). Thus, the volume (V) of the cord can be represented as: \[ V = AL \]
02

Find the Mass of the Rubber Cord

Next, we'll find the mass (m) of the rubber cord using the previously calculated volume expression (A * L) and the given density (ρ = 1500 kg/m³); this is done as follows: \[ m = \rho V = 1500(AL) \]
03

Calculate the Weight of the Rubber Cord

Now, we'll calculate the weight (W) of the rubber cord using the mass (m) we found in the previous step and the given acceleration due to gravity (g = 10 m/s²): \[ W = mg = 15000(AL) \]
04

Find the Stress on the Rubber Cord

Stress is defined as the force (or the weight in this case) divided by the cross-sectional area (A): \[ \sigma = \frac{W}{A} = \frac{15000(AL)}{A} = 15000L \]
05

Calculate the Strain on the Rubber Cord

Next, we'll calculate the strain on the rubber cord using Young's modulus formula. Young's modulus (Y) is given as \(5\times10^8 \mathrm{~N}\) / \(\mathrm{m}^{2}\) thus: \[ Y = \frac{\sigma}{\epsilon} \] where ε is the strain. Solving for ε, we get: \[ \epsilon = \frac{\sigma}{Y} = \frac{15000L}{5\times10^8} \]
06

Calculate the Stretch of the Rubber Cord

Finally, we'll calculate the stretch (ΔL) of the rubber cord using the strain (ε) and the initial length (L) of the cord as follows: \[ \Delta L = \epsilon L = \frac{15000L^2}{5\times10^8}\] Now, plugging in the given length L = 10m we get: \[ \Delta L = \frac{15000(10^2)}{5\times10^8} = \frac{1500000}{5\times10^8} = \frac{3}{1000} \] So the stretch of the rubber cord is: \[ \Delta L = 0.003 \mathrm{~m} = 3 \times 10^{-3} \mathrm{~m} \] None of the provided options are correct.

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