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4CP

Expert-verifiedFound in: Page 45

Book edition
4th edition

Author(s)
Ruth W. Chabay, Bruce A. Sherwood

Pages
1135 pages

ISBN
9781118875865

**You hang a 10 kg mass from a copper wire, and the wire stretches by . **

**(a) ****If you suspend the same mass from two copper wires, identical to the original wire, what happens? **

**(b) ****If you suspend the same mass from a copper wire with half the cross-sectional area but the same length as the original wire, what happens? **

**(c) ****If you suspend the same mass from a copper wire with the same cross-sectional area but twice the length of the original wire, what happens?**

(a) The stretch in the wire when two wires are used instead of one is **4 mm**.

(b) The stretch in the wire when the cross-sectional area is halved is **16 mm**.

(b) The stretch in the wire when the initial length is doubled is **16 mm**.

A 10 kg mass is hung from a copper wire, and the wire stretches by 8 mm.

**The change in length of a wire of initial length L**** and cross-sectional area** A** when force** F** is applied to it can be expressed as,**

** ${\mathbf{\u2206}}{\mathbf{L}}{\mathbf{=}}\frac{\mathbf{FL}}{\mathbf{AY}}$ (I)**

**Here L**** is Young's modulus of the material of the wire.**

a)

When two wires are used, the cross-sectional area doubles. From equation (I), the new change in length can be calculated as,

$\begin{array}{rcl}\u2206\mathrm{L}& =& \frac{\mathrm{FL}}{2\mathrm{AY}}\\ & =& \frac{1}{2}\left(\frac{\mathrm{FL}}{\mathrm{AY}}\right)\\ & =& \frac{1}{2}\times 8\mathrm{mm}\\ & =& 4\mathrm{mm}\end{array}$

Thus, the new change in length is 4 mm.

b)

From equation (I), the new change in length when the cross-sectional area is halved can be calculated as,

$\begin{array}{rcl}\u2206\mathrm{L}& =& \frac{\mathrm{FL}}{\frac{\mathrm{A}}{2}\mathrm{Y}}\\ & =& 2\left(\frac{\mathrm{FL}}{\mathrm{AY}}\right)\\ & =& 2\times 8\mathrm{mm}\\ & =& 16\mathrm{mm}\\ & & \end{array}$

Thus, the new change in length is 16 mm.

c)

From equation (I), the new change in length when the initial length is doubled can be calculated as,

$\begin{array}{rcl}\u2206\mathrm{L}& =& \frac{\mathrm{F}2\mathrm{L}}{\mathrm{AY}}\\ & =& 2\left(\frac{\mathrm{FL}}{\mathrm{AY}}\right)\\ & =& 2\times 8\mathrm{mm}\\ & =& 16\mathrm{mm}\end{array}$

Thus, the new change in length is 16 mm.

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