Question

Two cylindrical wires $A$ and $B$ are of the same material. Their lengths are in the ratio $1:2$ and the diameters are in the ratio $2:1$. If they are pulled by the same force, then increase in their respective lengths will be in the ratio

• A. $2:1$
• B. $1:4$
• C. $1:8$
• D. $8:1$

Answer 1

The correct option is C $1:8$

We know that Young's modulus
$Y=\frac{F}{A}\times \frac{L}{\Delta L}$
$\because A=\pi r^2$
$\Rightarrow Y=\frac{F}{\pi r^2}\times \frac{L}{\Delta L}$
Since $Y,F$ are same for both the wires,
$Y_1=Y_2$
$\Rightarrow \frac{L_1}{\Delta L_1{r}_1^2}=\frac{L_2}{\Delta L_2{r}_2^2}$
$\Rightarrow \frac{\Delta L_1}{\Delta L_2}=\frac{r_2^2\times L_1}{r_1^2\times L_2}=\frac{{\left(\frac{D_2}{2}\right)}^2\times L_1}{{\left(\frac{D_1}{2}\right)}^2\times L_2}$
$\Rightarrow \frac{\Delta L_1}{\Delta L_2}=\frac{D_2^2\times L_1}{D_1^2\times L_2}=\frac{D_2^2}{(2D_2{)}^2}\times \frac{\left(\frac{L_2}{2}\right)}{L_2}=\frac{1}{8}$
$\therefore \frac{\Delta L_1}{\Delta L_2}=\frac{1}{8}$
$\Rightarrow \Delta L_1:\Delta L_2=1:8$