Physics - Elasticity Question with Solution | TestHub

Force$\mathrm{F}=\frac{\mathrm{Y}\mathrm{A}∆\mathrm{l}}{l}$

Substituting area$\mathrm{A}=\mathrm{\pi }{\left(\frac{\mathrm{d}}{2}\right)}^2=\frac{\mathrm{\pi }{\mathrm{d}}^2}{4}$

So, we have$\mathrm{F}=\frac{\mathrm{Y}\mathrm{\pi }{\mathrm{d}}^2∆l}{4\mathrm{l}}$

$\mathrm{F}\propto \frac{{\mathrm{d}}^2∆\mathrm{l}}{l}$$\left[\because \frac{\mathrm{Y}\mathrm{\pi }}{4}=\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}\right]\mathrm{ }$

Hence, using equation (i), we have

${\mathrm{F}}_{\mathrm{A}}\propto \frac{{\mathrm{d}}_{\mathrm{A}}^2∆l_{\mathrm{A}}}{l_{\mathrm{A}}}$...(i)

${\mathrm{F}}_{\mathrm{B}}\propto \frac{{\mathrm{d}}_{\mathrm{B}}^2∆l_{\mathrm{B}}}{l_{\mathrm{B}}}$...(ii)

Form equations (i) and (ii), we get

$\therefore \frac{{\mathrm{F}}_{\mathrm{A}}}{{\mathrm{F}}_{\mathrm{B}}}=\frac{{\mathrm{d}}_{\mathrm{A}}^2∆l_{\mathrm{A}}}{l_{\mathrm{A}}}\times \frac{l_{\mathrm{B}}}{{\mathrm{d}}_{\mathrm{B}}^2∆l_{\mathrm{B}}}$

As$∆l_{\mathrm{A}}=\mathrm{ }∆l_{\mathrm{B}}=∆l$

Therefore$\frac{{\mathrm{F}}_{\mathrm{A}}}{{\mathrm{F}}_{\mathrm{B}}}=\mathrm{ }\frac{{\mathrm{d}}_{\mathrm{A}}^2}{{\mathrm{d}}_{\mathrm{B}}^2}\times \frac{l_B}{l_{\mathrm{A}}}$

Putting$\frac{{\mathrm{d}}_{\mathrm{A}}}{{\mathrm{d}}_{\mathrm{B}}}=\frac{2}{1}$

And$\frac{l_{\mathrm{B}}}{l_{\mathrm{A}}}=\frac{2}{1}$

We get${\mathrm{F}}_{\mathrm{A}}\mathrm{ }:{\mathrm{F}}_{\mathrm{B}}=8\mathrm{ }:1$