question_answer

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 \[\Delta x\] on applying force F, how much force is needed to stretch wire 2 by the same amount?

A)  4 F                                        

B)  6 F

C)  9 F                                        

D)  F

Correct Answer: C

Solution :

As shown in the figure, the wires will have the same Young's modulus (same material) and the length of the wire of area of cross-section 3A will be \[\ell /3\] (same volume as wire 1). For wire 1, \[Y=\frac{F/A}{\Delta x/\ell }\]                                            ?(i) For wire 2, \[Y=\frac{F'/3A}{\Delta x/(\ell /3)}\]                 ?(ii) From (i) and (ii), \[\frac{F}{A}\times \frac{\ell }{\Delta x}=\frac{F'}{3A}\times \frac{\ell }{3\Delta x}\Rightarrow F'=9F\]