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 ${\ell }_3=5{\ell }_2-4{\ell }_1$