VITEEE 2009 Paper

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 $Y_2$. ∴ $Y_1=\frac{F_1\cdot l_1}{A_1\cdot \Delta l_1}$ ...(i) and $Y_2=\frac{F_2\cdot l_2}{A_2\cdot \Delta l_2}$ ...(ii) Dividing Eq.(i) by (ii), $\frac{Y_1}{Y_2}=\frac{\frac{F_1\cdot l_1}{A_1\cdot \Delta l_1}}{\frac{F_2\cdot l_2}{A_2\cdot \Delta l_2}}$ or $\;\;\frac{Y_1}{Y_2}=\frac{F_1\cdot A_2\cdot l_1\cdot \Delta l_2}{F_2\cdot A_1\cdot l_2\cdot \Delta l_1}$...(iii) Force on steel wire from free body diagram $T=F_1=(2g)$ newton Force on brass wire from free body diagram $F_2=T'=T+2g=(4g)$ 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 $\;\;\frac{\Delta l_1}{\Delta l_2} = \left(\frac{a}{2b^2 c}\right)$