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Two wire $A$ and $B$ are stretched by same force. If, for $A$ and $B, Y_A: Y_B=1: 2, r_A: r_B=3: 1$ and $L_A: L_B=4: 1$, then ratio of their extension $\left(\frac{\Delta L_A}{\Delta L_B}\right)$ will be .............
$10: 13$
$8: 9$
$11: 7$
$6: 5$
Solution
(b)
$\Delta x=\frac{F L}{A Y}$
For wire $A$
$\Delta L_A=\frac{F \cdot L_A}{\pi r_A^2 \cdot Y_A} \ldots (1)$
For wire $B$
$\Delta L_B=\frac{F \cdot L_B}{\pi r_B^2 \cdot Y_B} \ldots (2)$
Divide $(1)$ by $(2)$
$\frac{\Delta L_A}{\Delta L_B}=\frac{F \cdot L_A}{\pi r_A^2 \cdot Y_A} \times \frac{\pi r_B^2 \cdot Y_B}{F \times L_B}=\frac{L_A}{L_B} \times\left(\frac{r_B}{r_A}\right)^2 \times \frac{Y_B}{Y_A}$
Substituting the value of ratio's
$\frac{\Delta L_A}{\Delta L_B}=\frac{4}{1} \times\left(\frac{1}{3}\right)^2 \times \frac{2}{1}=\frac{8}{9}$
