Question

There are two wire of same material and same length while the diameter of second wire is two times the diameter of first wire, then the ratio of extension produced in the wires by applying same load will be

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

Answer 1

Answer: (D)

$\because$ Both wires are same materials so both will have same Young's modulus, and let it be $Y$.
$Y=\frac{\text { stress }}{\text { strain }}=\frac{F}{A \cdot\left(\Delta \frac{L}{L}\right)}$
$F=$ applied force
$A=$ area of cross-section of wire
Now,
$Y_{1}=Y_{2} \Rightarrow \frac{F L}{\left(A_{1}\right)\left(\Delta L_{1}\right)}=\frac{F L}{\left(A_{2}\right)\left(\Delta L_{2}\right)}$
Since load and length are same for both
$\Rightarrow r_{1}^{2} \Delta L_{1}=r_{2}^{2} \Delta L_{2},\left(\frac{\Delta L_{1}}{\Delta L_{2}}\right)=\left(\frac{r_{2}}{r_{1}}\right)^{2}=4$
$\Delta L_{1}: \Delta L_{2}=4: 1$