Q. Two wires are made of the same material and have the same volume. However, wire 1 has cross-sectional area A and wire 2 has cross-sectional area 3A. If the length of wire 1 increases by $\mathrm{\Delta}x$ on applying force F, how much force is needed to stretch wire 2 by the same amount?
A) F
B) 4F
C) 6F
D) 9F
Explanation
(d) $A_{1}l_{1} = A_{2}l_{2}$
$\Longrightarrow \ l_{2} = \frac{A_{2}l_{1}}{A_{1}} = \frac{A \times l_{1}}{3A} = \frac{l}{3}$
$\Longrightarrow \ \ \frac{l_{1}}{l_{2}} = 3$
$\ \mathrm{\Delta}x_{1} = \frac{F_{1}}{A\gamma}\ l_{1}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...(i)$
$\ \ \ \mathrm{\Delta}x_{2} = \frac{F_{2}}{3A\gamma}l_{2}$ …..(ii)
Here $\mathrm{\Delta}x_{1} = \mathrm{\Delta}x_{2}$
$\frac{F_{2}}{3A\gamma}l_{2} = \ \frac{F_{1}}{A\gamma}l_{1}$
$F_{2} = 3F_{1} \times \frac{l_{1}}{l_{2}}$
$= 3F_{1} \times 3 = 9F$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.