Discussion Forum

One end of a horizontal thick copper wire of length 2 L and radius 2R is welded to an end of another horizontal thin copper wire of length L  and radius R. When the arrangement is treated by applying forces at two ends, the ratio of the elongation in the thin wire to that in the thick wire is 

Answer:

Explanation:

$\triangle l= \frac{FL}{AY}=\frac{FL}{(\pi r^{2})Y}$

  $\therefore$      $\triangle l\propto \frac{L}{r^{2}}$

 $therefore$    $\frac{\triangle l_{1}}{\triangle l_{2}}=\frac{L/R^{2}}{2L/(2R)^{2}}=2$