Question

The length of an elastic string is $Xm$ when the tension is $8N$, and $Ym$ when the tension is $10N$. The length in metres when the tension is $18N$ in terms of X and Y.

Answer 1

Consider unstretched length of string be $L$.

Hooke's law states that , $F=K△x$ $[△x:elongation]$

apply in two conditions,

$8=K(X-L)$ $10=K(Y-L)$ - $\left(1\right)$

divide noth and get $L$ in terms of $X$ & $Y$

$\Rightarrow \frac{X-L}{Y-L}=\frac{4}{5}\Rightarrow L=5x-4y$ - $\left(2\right)$

Substitute $\left(2\right)$ in any $L$ to get $K$

$\Rightarrow K=\frac{8}{(X-L)}=\frac{8}{X-(5x-4y))}=\frac{2}{Y-X}$ - $\left(3\right)$

and under tension $T=18N$ It elongates to $L$, given as :

$L8=K(L^{\prime }-L)\Rightarrow L^{\prime }=\frac{\perp 8(Y-X)}{2}+(5X-4Y)$

$\Rightarrow L^{\prime }=5Y-4X$