Question

One end of a horizongal 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 then

• A. $0.25$
• B. $0.50$
• C. $2.00$
• D. $4.00$.

Answer 1

Answer: (C)

$2.00$

Both the rods are made up of same kind of material i.e., their Young's modulus is same.

$K=\frac{F}{\Delta L}=\frac{YA}{L}$

$k_1=\frac{\pi 4R^{2}Y}{2L},k_2=\frac{\pi 4R^{2}Y}{L}$
$\therefore F=k_{1}x=k_{2}y\Rightarrow \frac{y}{x}=\frac{k_1}{k_2}=2$