Question

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?

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

Answer 1

The correct option is C $2.00$

Change in length, $\Delta L=\frac{F_{rest}l}{yA}$

Since the $2$ rods are in series.

$(F_1{)}_{rest}=(F_2{)}_{rest}$

$\Delta l\propto \frac{l}{A}$

or, $\Delta l=\frac{l}{R^2}(\because A=\pi R^2)$

$\therefore \frac{\Delta l_1}{\Delta l_2}=\frac{l_1}{l_2}\times \frac{R_2^2}{R_1^2}$

$⟹\frac{\Delta l_1}{\Delta l_2}=\frac{L}{2L}\times \frac{4R^2}{R^2}=\frac{2}{1}=2.00$