91影视

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 Land cross-sectional area Awhen forceFis applied to it can be expressed as,

$\mathbf{鈭�}\mathbf{L}\mathbf{=}\frac{\mathbf{FL}}{\mathbf{AY}}$ (I)

Here Lis 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,

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

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,

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

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,

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

Thus, the new change in length is 16 mm.