Question

To wires $A$ and $B$ have the same length and area of cross section. But Young's modulus of $A$ is two times the Young's modulus of $B$. Then the ratio of force constant of $A$ to that of $B$ is

• A. $1$
• B. $2$
• C. $\frac{1}{2}$
• D. $122$

Answer 1

The correct option is B $2$

$\begin{array}{c}\text{Given,}{Y}_A=2Y_B\\ \text{Same length and area of cross - section of wires}A\text{\&}B\\ \text{According to the spring equivalent concept ,}\end{array}$

$\begin{array}{c}F=kx(x=\text{extension in spring})\\ F=\left(\frac{YA}{L}\right)\left(\Delta L\right)\\ \text{From above equations, we conclude}\\ k=\frac{YA}{L}=\text{force constant}\\ \therefore \frac{Y_A}{Y_B}=\frac{k_a}{k_b}=2(\text{area and length}\\ \text{are same.})\end{array}$