Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

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

Mechanical Properties of Solids

Solution:

image
Both the rods are made up of same kind of material, i.e. their young’s modulus is same.
$K_{1}$ $=\frac{\pi4R^{2}Y}{2L}$, $k_{2}$ $=\frac{\pi R^{2}Y}{L}$
Force constants of two rods are,
$\therefore $ $\quad$ $F=k_{1}x$$=k_{2}Y$
$\Rightarrow $$\quad$ $\frac{y}{x}=\frac{k_{1}}{k_{2}}=2$