Two wires of diameter 0.25 cm, one made of steel and other made of...

Given, diameter of wires (2r)=0.25 cm r=0.125 cm=1.25 × 10^{-3} m For steel wire, Load (F_{1})=(4 + 6) kgf=10 × 9.8 N=98 N Length of steel wire (l_{1})=1.5 m Young's modulus (Y_{1})=2.0 × 10^{11} Pa $\text{Young's modulus (Y)}=\dfrac{{F}_{1} \times {l }_{1}}{{A}_{1} \times \text{\Delta} {l }_{1}} \text{Change in length} \left(\text{\Delta} {l }_{1}\right)=\dfrac{{F}_{1} \times {l }_{1}}{{A}_{1} \times {Y}_{1}} =\dfrac{{F}_{1} \times {l }_{1}}{\text{\pi} {r}_{1}^{2} \times {Y}_{1}} =\dfrac{\text{98} \times \text{1.5}}{\text{3.14} \times {\left(\text{1.25} \times {10}^{- 3}\right)}^{2} \times \text{2.0} \times {10}^{11}}$ =1.5 × 10^{-4} m For brass wire, Load (F_{2})=6 kgf=6 × 9.8 N=58.8 N Length of brass wire (l_{2})=1.0 m Young's modulus (Y_{2})=0.91 × 10^{11} Pa $\text{Change in length} \left({l }_{2}\right)=\dfrac{{F}_{2} \times {l }_{2}}{\text{\pi} {r}_{2}^{2} \times {Y}_{2}} =\dfrac{\text{58.8} \times \text{1.0}}{\text{3.14} \times {\left(\text{1.25} \times {10}^{- 3}\right)}^{2} \times \text{0.91} \times {10}^{11}}$ =1.3 × 10^{-4} m