Question

The length of elastic string, obeying Hooke’s law is $\ell_{1}$ metres when the tension $4N$ and $\ell_{2}$ metres when the tension is $5N$. The length in metres when the tension is $9N$ is -

• A. $5\ell_{1} - 4\ell_{2}$
• B. $5\ell_{2} - 4\ell_{1}$
• C. $9\ell_{1} - 8\ell_{2}$
• D. $9\ell_{2} - 8\ell_{1}$

Answer 1

Answer: (B)

Let $\ell_{0}$ be the unstretched length and $\ell_{3}$ be the length under a tension of $9N$.
Then $Y = \frac{4\ell_{0}}{A\left(\ell _{1}-\ell _{0}\right)} = \frac{5\ell _{0}}{A\left(\ell _{2}-\ell _{0}\right)}$
$ = \frac{9\ell _{0}}{A\left(\ell _{3}-\ell _{0}\right)}$
These give
$\frac{4}{\ell _{1}-\ell _{0}} = \frac{5}{\ell _{2}-\ell _{0}} $
$\Rightarrow \ell_{0} = 5\ell_{1} -4\ell_{2}$
Further, $\frac{4}{\ell _{1}-\ell _{0}} = \frac{9}{\ell _{2}-\ell _{0}} $
Substituting the value of $\ell_{0}$ and solving,
we get $\ell _{3} = 5\ell _{2} -4\ell _{1}$