Physics - Elasticity Question with Solution | TestHub

Given, $\frac{l_1}{l_2}=a,\frac{r_1}{r_2}=b,\frac{Y_1}{Y_2}=c$ Free body diagram of the two blocks brass and steel are
Let Young's modulus of steel is $Y_1$ and of brass is $Y_2$. $\therefore Y_1=\frac{F_1\cdot l_1}{A_1\cdot \Delta l_1}...(i)$ and $Y_2=\frac{F_2\cdot l_2}{A_2\cdot \Delta l_2}...(ii)$ Dividing Eq. (i) by(ii), $$\begin{array}{l}\frac{Y_1}{Y_2}=\frac{\frac{F_1\cdot l_1}{A_1\cdot \Delta l_1}}{\frac{F_2\cdot l_2}{A_2\cdot \Delta l_2}}\\ \text{ or }\frac{Y_1}{Y_2}=\frac{F_1\cdot A_2\cdot l_1\cdot \Delta l_2}{F_2\cdot A_1\cdot l_2\cdot \Delta l_1}...(ii)\end{array}$$ Force on steel wire from free body diagram $T=F_1=(2g)$ newton Force on brass wire from free body diagram $F_2=T^{\prime }=T+2g=(4g)$ newton Now, putting the value of $F_1,F_2$, in Eq. (iii), we get $$\frac{Y_1}{Y_2}=\left(\frac{2g}{4g}\right)\cdot \left(\frac{\pi r_2^2}{\pi r_1^2}\right)\cdot \left[\frac{l_1}{l_2}\right]\cdot \left(\frac{\Delta l_2}{\Delta l_1}\right)$$ or $c=\frac{1}{2}\left(\frac{1}{b^2}\right)\cdot a\left(\frac{\Delta l_2}{\Delta l_1}\right)$ or $\frac{\Delta l_1}{\Delta l_2}=\left(\frac{a}{2b^{2}c}\right)$