Question

The length of a rubber cord is $l_{1}\,m$ when the tension is $4 \,N $ and $l_{2} \,m$ when the tension is $6\, N$. The length when the tension is $9 \,N$, is

• A. $\left(2.5 l_{2}-1.5l_{1}\right)m$
• B. $\left(6l_{2}-1.5l_{1}\right)m$
• C. $\left(3l_{1}-2l_{2}\right)m$
• D. $\left(3.5l_{2}-2.5l_{1}\right)m$

Answer 1

Answer: (A)

Let the original unstretched length be $l$.
$\therefore \quad$ $Y$ $=\frac{T}{A}$ $\frac{l}{\Delta l}$ Now $Y$ $=\frac{4}{A}$ $\frac{l}{\left(l_{1}-l\right)}$ $=\frac{6}{A}$ $\frac{l}{\left(l_{2}-l\right)}$ $=\frac{9}{A} \frac{l}{\left(l_{3}-l\right)}$
$\therefore \quad$ $4\left(l_{3}-l\right)=9\left(l_{1}-l\right)$ or $4l_{3}+5l=9l_{1}$ $\quad\ldots\left(i\right)$
Again, $6\left(l_{3}-l\right)$ $=9\left(l_{2}-l\right)$or $2l_{3}+l=3l_{2}$ $\quad\ldots\left(ii\right)$
To obtain $l_{3}$, solve $\left(i\right) and \left(ii\right)$.
$\therefore \quad$ $l_{3}$ $=\left(2.5l_{2}-1.5l_{1}\right)m$