Question

Two wires are of the same material have masses in the ratio 3:4. What is the ratio of their extensions under the same load , if their lengths are in the ratio 9:10?

Answer 1

Dear Student
using Hooke's law
$$\begin{gathered} Y=\frac{FL}{A∆L} \\ ∆L=\frac{FL}{AY}=\frac{FL}{{\displaystyle \frac{m}{L\rho }}Y}=\frac{FL}{{\displaystyle \frac{m}{L\rho }Y}}=\frac{F{L}^2\rho }{{\displaystyle mY}}=\frac{F\rho }{{\displaystyle Y}}\times \frac{L^2}{m}=c\frac{L^2}{m}(wherec=\frac{F\rho }{{\displaystyle Y}}=cons\tan t) \\ \frac{∆L_1}{∆L_2}=\frac{\frac{{L_1}^2}{m_1}}{\frac{{L_2}^2}{m_2}}={\left(\frac{L_1}{L_2}\right)}^2\times \frac{m_2}{m_1}={\left(\frac{9}{10}\right)}^2\times \frac{4}{3}=\frac{81}{100}\times \frac{4}{3}=\frac{27}{25} \end{gathered}$$
Regards