Question

Two wires $A$ and $B$ of same material have radii in ratio $2:1$ and length $4:1$. The ratio of the normal forces required to produce the same change in the length of these two wires is:

• A. 1 : 4
• B. 2 : 1
• C. 1 : 2
• D. 1 : 1

Answer 1

The correct option is D 1 : 1

We know, elongation in wire $\left(\Delta L\right)=\frac{FL}{AY}$
or $F=\frac{AY\Delta L}{L}(\because$ F = normal force)
Given:
Material of both wires is same $\therefore Y_A=Y_B$
Elongation in both wires $A$ and $B$ are equal $\therefore \Delta L_A=\Delta L_B$
So, $F\propto \frac{A}{L}$

$\Rightarrow \frac{F_A}{F_B}=\frac{A_A}{L_A}\times \frac{L_B}{A_B}$ But$\frac{A_A}{A_B}=\frac{\pi r_A^2}{\pi r_B^2}=\frac{r_A^2}{r_B}$
$\Rightarrow \frac{F_A}{F_B}=\frac{r_A^2}{r_B^2}\times \frac{L_B}{L_A}$ $=(2{)}^2\times \frac{1}{4}=1$ $[\because \frac{r_A}{r_B}=\frac{2}{1}and\frac{L_A}{L_B}=\frac{4}{1}\text{(given)}]$